6:04 - If you don't have an accurate low resistance ohm meter you can also get by with a current meter and either another volt meter or the ammeter (panel meter) that you are using. Also keep in mind when you are creating the shunt from multiple resistors that you will also need to confirm that you don't go over the maximum power dissipation of each of the resistors. I would stay under 1/2 the power rating of each for continuous use but for quick tests you could go to the full power rating.
This and your last video were very well timed for me. I had a meter on a Johnson Viking 2 I am restoring that was not working and I was able troubleshoot it with your info. I enjoy your videos, and look forward to future boat anchors content.
Thanks for the very helpful and illuminating video. I recently picked up a nice old Triplett panel meter for a project. It looked good if a little gunked up, but measured open, and upon disassembly I found that the coiled-up length of ~34AWG magnet wire that they used for the shunt was badly corroded and in several pieces. I needed an excuse to buy some magnet wire anyway and will have a functioning meter soon. Thanks again!
Occasionally, when I needed a very low ohm shunt resistor (not having a precision low-ohm meter), I have just calculated how many parallel strands, how long, and of what gauge, of nichrome wire were needed. If you are making a new scale, getting the shunt value close and then measuring the actual deflection for a given Amperage will let you design the final scale. For instance, if you want a 50 Amp meter, making a shunt that gets you close (maybe 53 Amps full scale) will let you make a 50 Amp scale for slightly less than full deflection.
That was perfect. Thank you. I purchased a digital ammeter to add to a pwr supply I'm building. Going on my experience with automotive ammeters I did not realize a shunt was required. I tied the meter into a simple 5v LED bread board circuit, and released some smoke from my YwRobot P.S. 5V regulator. Now it produces 6.9V. I Went on line to find the specs, before blowing anything else up. I found the exact same unit, and the manufacturer. The device is available in green and blue as well as the red which I have. All of them show shipped with the Shunt included. I'm a little bit perturbed. What happened to the shunt? Clearly the device was repackaged. zip lock bag. 0 to 10 Amps, 5 to 12V is all I'll be using, device I believe goes to 30amp. I know a shunt is made specifically for this thing but I can't find the value, only math classes. I think I can easily figure it out now, even with nothing but multimeter. I see Mr. Caralson was last one to drop by. I spend a lot of time over at his place as well.
Short video, but very good explanation. Due to a final exam, I didnt see it until now. Please make more videos. I would like small applications from a novice point of view. Again thanks. I look forward to more :)
There hasn't been much said about checking the linearity of these meters, in this or the previous video. It may have been helpful to set the power supply at various intermediate currents, after verifying full scale, to make sure the meter movement is linear and accurate over its full range.
Alan, Nice video. I needed a shunt and didn't have any nichrome wire, so I used a copper wire and filed down a small area in the center section until I had the full scale reading I needed.
When using a Fluke DVM meter in ammeter mode, is there internal input and output buffers circuits or DC decoupling capacitors on the input ports and output ports while in ammeter mode? There is protection circuits to block DC and buffers even while in ammeter mode.
Guitar amplifiers biasing using a transformer shunt method to measure the output tubes plate current flowing through the output transformer. The DVM meters impedance is LESS than the output transformers primary windings so the current will Flow through the DVM meter instead to measure the biasing current. You connect the DVM meter in current mode across the output tubes PLATE to the output transformer primary High End terminal to measure the biasing current. Can you do this type of biasing for FETs amplifiers and transistor amplifiers in power amplifiers to use the shunt method? Its rare to see a DVM meter in current mode placed in parallel across a WIRE. Maybe you can think or find other applications that can use a DVM meter in current mode using it to measure bias current in a SHUNT Method?
What lab test equipment can I use to test how much current my AC outlet can handle BEFORE it Pops the mains circuit breaker. If you didn't have or know the buildings wiring or main fuse panel and just wanted to measure how much current each AC outlet can handle what lab test equipment would you use to measure how much current it can handle before popping circuit breakers?
+fabts4 In the previous video, I mentioned the typical ranges for the meter resistance. In this case, we have 0.185V/200uA, or about 925 ohms. In most references I've seen, they use the meter resistance when calculating a shunt resistor, which can be a little tough mathematically for the novice - I thought this approach was more straight-forward.
+w2aew Maybe I missed the previous video. You're right to think about the novice, that's also why I thought it was worth mentioning the impact of removing the meter was negligible (BECAUSE the 10k resistance is so high). Maybe the safest for the novice is to leave the amp meter in series while using another multimeter to check the voltage? Or use the voltmeter mode only to check the voltage drop of the 10k and the gauge?
+fabts4 You'll notice in the video that I *re-adjusted* the power supply after removing the ammeter. I guess I should've stated that I did this to account for the removal of the burden voltage of the ammeter. It would NOT have been negligible. Safest/easiest would have been to make both measurements with two meters - but I wanted to use my most accurate DMM for both measurements (and I wanted to show how to make each measurement for folks that only have one DMM). Using the DMM to measure the drop across the 10K is also a good way to go - provided you accurately measure the 10K value too. I mention this last point in my previous video also.
Great illustration. Dont you think that a carefully chosen wheatstone bridge 4 arms values has an equivalent benefit or probably adds to this method of separating the applied current leads from the voltage measuring leads (aka 4-wire method)?
Another thing than works well for making a current shunt is some pcb. You can calculate the value pretty closely, then trim it after etching. Of course copper has a much higher temperature coefficient than a real shunt resistor, so it won't be as accurate when it heats up.
Before the 50s they didn't use the resistor Color Code system. Each manufacture on the schematic would indicate a color code system for the resistors and capacitors. What are these other color code systems called and were used before the 50's?
How would you calculate full scale deflection current given the resistances, the meter current, and the leads between 2 points? No voltage given. This question is from the arco electrician exam book in the back of the book. (Exams)
@@w2aew So does the analog panel meter have a built-in resistor, or is it just the inherent resistance of the meter itself ( the windings and such)? An "ideal" ammeter would have zero resistance, so any (parallel) shunt would have no effect on its full scale current handling ability, right? So, I guess I'm asking if you only rely on the inherent resistance of the meter itself, or can you add to it a bit with an additional series resistor... Like if I wanted to use a 100mA meter for a 1A @ 24V project I could set up a 250m Ohm (shunt) / 2.5 Ohm (series) divider... Or is this series component not required since the meter itself is already acting as a 2.5R resistor? I apologize for my ignorance and sloppy math.
@@andrewferg8737 In the case of analog meters like these, the meter's coil has resistance and determines the voltage drop when measuring current. Any additional series resistance would simply add to the ammeter's burden voltage.
@@w2aew Thanks so much for your patient reply. I did a bit a experimenting last evening and was able to get a 3mA meter working with 100m Ohm shunt for my 24V @ 1A project. I've got bunch of these .1R resistors in my junk box, and have always wondered why anyone would manufacture a such a tiny value resistor. Thanks again!
Hi, I need some insight on how to make the minor adjustment to get a 50 microamp panel meter extent to work in place of a panel meter specified for 60 microamps. This is for an old Lafayette TE-46 capacitor analyzer. The original meter must have been damaged and someone replaced the original with a 1 Milliamp meter. Using your process The original meter is reading full scale at 1.14 milliamp and a new 50 microamp meter is running full scale at 48 microamps... Thanks for the video as I now know both work and what gets them to full scale.. Note my 100K resistor used for this tests at 9.946k on a 4 wire HP 3478a... That said I'm not sure what to do next. do I pursue working on extending the range of the 50 MA meter or shunting the 1 Milliamp meter? I'd rather use the 50 Microamp meter. Along with this question I've seen some discussion regarding adding diodes across these meters to protect them from going overscale and I'm not sure what or how to add this into the mix... One gentleman added two schottky diodes and another used two 1N4001 diodes reversed against each other. Honestly I don't really understand how these help yet but I'm sure there is a voltage drop issue I need to be aware of...Any insight on how to proceed and certainly some assistance on the shunt values to extend the 50microamp meter would be appreciated... (Would it be possible to use a small resistor in series with the meter instead of a shunt across the meter??) Oh I hope you are still out here... Of all the videos I've seen (and there have been many) yours are the easiest to understand for a novice...
The process shown in this video is *exactly* what you should be able to do to in order to use the 50uA FSD meter in the application that needs a 60uA movement. The shunt resistor would be carrying 10uA when the meter is at full scale. If you have measured the voltage across the 50uA meter when it is reading full scale, simply take that voltage and divide it by 10uA, and you'll have the shunt resistor value that you need. For example, when the 50uA meter is at full scale, suppose you measured 100mV. Then, the shunt resistor you would need would be 100mv divided by 10uA, or 10kohms.
@@w2aew Thank you, exactly as suggested the 10K resistor across the meter provided a full scale 60 uA across the meter. I also experimented with adding two schottky diodes across the meter along with the 10k resistor but these added an additional voltage drop causing the full scale to be more like 80 uA. (78.97 ua and 101.772 mv across the meter) However, I added two 1N4001 diodes across the meter along with the 10K and there was no impact. With the 10K resistor and the two 1N4001 diodes I have 59.52 uA and 101.158 mv across the meter. Without any shunt resistor the measurements are 48.57 uA and .098889 mv. Using the calculations I learned from you it appears that the meter itself must be 2036 ohms... I'm confused why there is no voltage drop impact by the 1n4001 diodes and there was an impact with the 1N5817 schottky diodes. Either way, considering I'm at full scale at 100 mv I imagine the meter will be destroyed long before either of these diode solutions start to short and pull current away from the meter.... Anyway, thanks for you help and awesome videos. I've watched many and find them well thought out and very insightful!! Thx, Ben. NJ
@@bendec2969 Well, I supposed I had a lucky guess by suggesting that the voltage across the meter was 100mV at full scale! Glad it worked out. The reason the schottky had a problem is likely because you had it forward biased and it was turning on slightly as the meter reached full scale.
@@w2aew Which diodes will provide the best protection... Just seeking your advise and not holding you responsible for any decision I make! Again thanks for responding... I really appreciate the education. I'm having a lot of fun fixing my old toys with this new found hobby. Several heathkits sig gens, vtvms, Fisher Amp improvements - 400, 500c, (500B is next). Very happy with the result on a restoration of a very old Precision Model 68 VTVM and a PACO model 70 both with a DC voltage accuracy that is blowing my mind... I have a couple of their tube testers (610 and a 1012) and I have to say they were really a quality focused company... I've made minor adjustments to both and they are working fine although the 610 is at its limits and really needs a bunch of precision resistors replaced but I'm nervous about digging into this as the wiring is intimidating... My favorite piece is a fully professionally restored and beautiful Hickok model 605A (which I won't touch). Precise, Precise, Precise.... I've noticed that several tube purchases from reputable ebay sellers who used precise metrics for their tube sales I almost mirror....And several new tubes test at the exact recommended value listed in the documentation... Anyway, thanks for the help. I think I'm going to watch one of your videos and relax for the evening... Ben
Thank you for a great explanation. I recently purchased a piece of equipment that has a 500A ammeter. I opened it up and there is a shunt resistor of course for measuring the current, however there is a small potentiometer in series with the leads going to the panel ammeter. So it is exactly like your drawing in #3 but with the addition of a potentiometer in series with the meter. Is the purpose of this to reduce the amount of current actually flowing through the ammeter?
since they are in series, it really doesn't matter. It is most common to connect the ammeter to the positive connector so that the load can remain connected to / referenced to the lower rail (usually ground).
The lab supply is feeding a 10k resistor as well as the meter, in series. With 200 uA flowing through the 10k resistor, there is a 2 volt drop across the resistor, and then we expect the 0.185 volts across the meter. We saw a little bit less than 2.185, first because the power supply was being adjusted in 10mV increments to 2.180mV, and then there were several other factors that made it no so perfect. It's off by about 2.7%; plenty close enough for a tutorial!
Hi Sir, I do have a little question. Do you know if a car fuel gauge in a 90s car is generally an ammeter? If so I can apply the knowledge of your video to test my fuel gauge as well as the sending unit. Thanks for your time!
When using Light bulbs as current limiters to test for short circuits its really confusing because if you use one light bulb rated at 60 watts divided by 120VAC = 0.5 amps of current. If you use 2 light bulb wired "in parallel" will be a total of 120 watts divided by 120VAC = 1 amp of current. Logically I would think wiring up resistors in parallel would decrease and lower the current limiting. If you used a 120 ohm/60 watt resistor as a current limiter that is 120ohm /120vac = 1amp of current, if you wire two 120 ohms/60watts that is a total of 60 ohms/120 watts = 0.5amps of current. Its confusing because the light bulbs are inverse of a resistor? or what am I missing and not understanding?
Your math for the resistors is wrong. If you have a 120ohm resistor with 120VAC across it, that resistor will dissipate 120W and the current will be 1A. If you wire two of these resistors in parallel, then *each* resistor will draw 1A and dissipate 120W - so the combination draws 2A total, and dissipates 240W total.
@@w2aew 120 ohm resistors in parallel with another 120 ohm resistor = total resistance is 60 ohms. So 120vac divided by 60 ohms = 2 amps which is Doubled. A 60watt lamp bulb is 60watts divided by 120vac = 0.5amps. Using two 60 watt lamp bulbs in parallel is 120 Watts divided by 120vac = 1 amp. What I'm confused about is a 60ohm resistor is NOT Equivalent to a 60 watt bulb, any reasons why? because 120vac/60 ohms = 2 amps compared to 60watts/120vac = 0.5amps. See how they are inverse the resistor and lamp bulb?
@@waynegram8907 You're mixing up the *resistance* and the *power dissipation*. A 60W bulb operated at 120VAC results in 0.5A current. Thus, the bulb's resistance is 120VAC/0.5A = 240 ohms. Put two of them in parallel and you get 120ohms, which will draw 1A at 120VAC, each one dissipating 60W for a total of 120W dissipation.
@@w2aew The filament inside the 60 watt lamp bulb will always have a resistance at 240 ohms at 120vac at Full Brightness like if the circuit under test has a short circuit, but when the circuits current draw is LESS than 0.5A the resistance of the filament changes to a different value. The Filament is not equivalent to a resistor. A 60 watt Filament is a 240 ohm resistor at 120vac but its not the same as a 60 watt/240 ohm resistor, any reasons why? A Filament is like a variable temperature potentiometer, as the circuit under test draws the maximum current of the Filament which is 120vac at 60 watts = 0.5 amps short circuit max current draw which is 240 ohms resistance of the filament. If the circuit under test is only drawing 100mA the 60watt lamp bulb resistance has changed to 1200 ohms and only using 12 watts of the 60watt lamp bulb . 120vac/100mA = 1200 ohms
@@waynegram8907 If a bulb is operated at a lower current, thus the filament is not heated up to normal operating temperature, then its resistance is LOWER than the fully heated value. For your 60W bulb, the filament is 240ohms at full 120VAC. But, at a lower current, it is lower than 240 ohms (you'd have to measure it). The idea with a dim-bulb tester is to put the bulb in SERIES with the device you're testing, so that it presents very little voltage drop during normal operating of your device, but if your device shorts out, then more current flows in the bulb, heating the filament, increasing it's resistance which then limits the current to save your device from damage.
Alan, do you have to remove/disable any original internal shunt before doing a mod like this? I have a 0-1A AC ammeter that I want to make dual range (0-10A) with a switchable external shunt. 73 KCØHST
Does any lab current ammeter measure MESH current? because most current ammeter just only measure Branch current. What are the differences are between MESH current and Branch current because you get different current values for MESH current compared to Branch Current. When would an EE engineer or technician want to know and measure the MESH current because most technician are just measuring Branch current?
@@w2aew I'm confused why the MESH current value is different than the branch current. Why is the current have different values and what's the point of knowing the MESH current?
@@waynegram8907 As I stated, it is simply a MATHEMATICAL ANALYTICAL METHOD to solve for the total circuit current values. Mesh currents are computed for each "loop". When loops intersect, meaning that they share the same path or part, the mesh currents from each loop add up (or subtract, depending on the direction). You don't measure MESH currents, they're just a computational tool. Branch currents are the sum of all of the mesh currents that flow in that branch.
@@w2aew yes I understand its a computational tool but its totally different current values compared to using Kirchhoffs current law. What I'm confused about is why isn't kirchhoffs current values the same current values as MESH current. I would think logically that current should have the same value, because if the network or circuit is 1 amp using Kirchhoff current law but the MESH current is not going to be 1 amp it will be a totally different current value which doesn't make sense to me.
@@w2aew Ohms law uses V/R = I so the MESH current is not going to be V/R=I it will be a totally different current value, any reasons why the MESH current it totally different compared to ohm law?
Easier to use a low value resistor (1 Ohm for example) in series with the circuit, then turn the meter into a voltmeter to simply read the voltage drop across the low value resistor. MUCH easier than all those small value resistors. Also, since the meter is multi scale, it is easier to use it as a voltmeter too. Not just dedicate it to reading current.
Your videos are so Succulent ...I never understood why some single cell brain creatures are giving thumbs down to your videos. There are very, very, very, few people like you on YT who give useful info for the mankind. Keep up good the work.
+rrangana11 I think some people are so disgusted with their own boring lives that they can't help but try to make others miserable. Even when I complain to someone about their videos, I don't click dislike.
ok, sth have you converted an ammeter-to- a voltmeter? this is not an easy concept for the average viewer. Also, sth like shunt resistors may need explaining further. no?!!
I guess the point here is to do this is thoroughly as you could, but in this case you could safely ignore the Ifs term and just calculate Rs=Vfs/Ifsd. The simplest way to see this is probably that you're doing the calculations to 3 significant digits, (assuming this is a satisfactory amount of precision) whereas 200 µA would only change the 4th significant digit when added to 600 mA. As is to be expected, if the meter is meant to be useful as a voltmeter.
This one already had a scale that I wanted, but that scale was only valid with the proper shunt. Full scale deflection was at 200uA, so the shunt was needed to get back to the 600mA scale.
Once again an excellent presentation and it ties in so well with the previous meter clip, your Bird meter clips & also with something I've been trying to figure out. Has the shunt in my wattmeter changed value or why would I be getting measurements shown in my clip.th-cam.com/video/S1Ew8wQwSI4/w-d-xo.html. My lab equipment is a little more basic...I need to find out voltage required for FSD next I believe for the equation.
+Christopher Larwood The meter used in the Bird 43 (and I presume in your meter as well) is a 30uA full scale meter movement. It does not have an external shunt. Thus, if it doesn't read 30uA at full scale, the meter movement itself if bad - there is nothing to can "add" to it externally to correct for this short of designing a current amplifier - and there's no guarantee that this will work correctly with the proper linearity.
All of the panel mount ammeters seem to be from China. 2.5% accuracy. So your calculating the current flowing through the meter the 200 micro amps would be useless. Also the standard for the meters is 75mV. So all of the different models of that ammeter can be made identical. Only the paper with the printed scale would have to be different.
Really helpful and clear explanation! I hope you will continue making videos, you're a great teacher
I think your educational videos are most informative videos in electronics on youtube. Im impressed. finaly I found sb whos speaking human language :)
This is the first place I come. Always great videos cheers
6:04 - If you don't have an accurate low resistance ohm meter you can also get by with a current meter and either another volt meter or the ammeter (panel meter) that you are using. Also keep in mind when you are creating the shunt from multiple resistors that you will also need to confirm that you don't go over the maximum power dissipation of each of the resistors. I would stay under 1/2 the power rating of each for continuous use but for quick tests you could go to the full power rating.
Grateful to have this explained so clearly. Appreciated and thank you
Very nice retro subject.Very well explained, brings back memories 30 years + ago.I really appreciate the slow pace style of your videos.Thanks!
This and your last video were very well timed for me. I had a meter on a Johnson Viking 2 I am restoring that was not working and I was able troubleshoot it with your info. I enjoy your videos, and look forward to future boat anchors content.
Thanks for the very helpful and illuminating video. I recently picked up a nice old Triplett panel meter for a project. It looked good if a little gunked up, but measured open, and upon disassembly I found that the coiled-up length of ~34AWG magnet wire that they used for the shunt was badly corroded and in several pieces. I needed an excuse to buy some magnet wire anyway and will have a functioning meter soon. Thanks again!
Occasionally, when I needed a very low ohm shunt resistor (not having a precision low-ohm meter), I have just calculated how many parallel strands, how long, and of what gauge, of nichrome wire were needed.
If you are making a new scale, getting the shunt value close and then measuring the actual deflection for a given Amperage will let you design the final scale. For instance, if you want a 50 Amp meter, making a shunt that gets you close (maybe 53 Amps full scale) will let you make a 50 Amp scale for slightly less than full deflection.
+Peter W. Meek Sounds like I need to get some nichrome wire!
Handy stuff to have around.
Looking to use a panel meter on my power supply and this explained everything I needed. Long live Ohms Law !! Thank you very much.
That was perfect. Thank you. I purchased a digital ammeter to add to a pwr supply I'm building. Going on my experience with automotive ammeters I did not realize a shunt was required. I tied the meter into a simple 5v LED bread board circuit, and released some smoke from my YwRobot P.S. 5V regulator. Now it produces 6.9V.
I Went on line to find the specs, before blowing anything else up. I found the exact same unit, and the manufacturer. The device is available in green and blue as well as the red which I have. All of them show shipped with the Shunt included. I'm a little bit perturbed. What happened to the shunt? Clearly the device was repackaged. zip lock bag.
0 to 10 Amps, 5 to 12V is all I'll be using, device I believe goes to 30amp. I know a shunt is made specifically for this thing but I can't find the value, only math classes.
I think I can easily figure it out now, even with nothing but multimeter.
I see Mr. Caralson was last one to drop by. I spend a lot of time over at his place as well.
Short video, but very good explanation. Due to a final exam, I didnt see it until now. Please make more videos. I would like small applications from a novice point of view. Again thanks. I look forward to more :)
There hasn't been much said about checking the linearity of these meters, in this or the previous video. It may have been helpful to set the power supply at various intermediate currents, after verifying full scale, to make sure the meter movement is linear and accurate over its full range.
Scott Allen . I tried old VU meters, and they are not linear. I had to hand draw the scale.
Alan,
Nice video. I needed a shunt and didn't have any nichrome wire, so I used a copper wire and filed down a small area in the center section until I had the full scale reading I needed.
Congratulations, you've won the award of being my new fuse maker!
Excellent video, really easy to understand your explanations.
Another winner Alan! More fodder for the next solder-melting session.
Really good video with detailed explanation and demonstration!
When using a Fluke DVM meter in ammeter mode, is there internal input and output buffers circuits or DC decoupling capacitors on the input ports and output ports while in ammeter mode? There is protection circuits to block DC and buffers even while in ammeter mode.
As always great video explanation and illustration!!! Thank you 73s
Guitar amplifiers biasing using a transformer shunt method to measure the output tubes plate current flowing through the output transformer. The DVM meters impedance is LESS than the output transformers primary windings so the current will Flow through the DVM meter instead to measure the biasing current. You connect the DVM meter in current mode across the output tubes PLATE to the output transformer primary High End terminal to measure the biasing current. Can you do this type of biasing for FETs amplifiers and transistor amplifiers in power amplifiers to use the shunt method? Its rare to see a DVM meter in current mode placed in parallel across a WIRE. Maybe you can think or find other applications that can use a DVM meter in current mode using it to measure bias current in a SHUNT Method?
Always an education. Thanks again.
Stainless steel works good for making shunts.
I have used some 1/8" thick stock.
What lab test equipment can I use to test how much current my AC outlet can handle BEFORE it Pops the mains circuit breaker. If you didn't have or know the buildings wiring or main fuse panel and just wanted to measure how much current each AC outlet can handle what lab test equipment would you use to measure how much current it can handle before popping circuit breakers?
There is nothing that I know of that can *predict* the tripping level for a circuit breaker.
Very nice video.
Maybe you could have mentioned the multimeter series resistance on that caliber and said it was negligible compared to the 10k.
+fabts4 In the previous video, I mentioned the typical ranges for the meter resistance. In this case, we have 0.185V/200uA, or about 925 ohms. In most references I've seen, they use the meter resistance when calculating a shunt resistor, which can be a little tough mathematically for the novice - I thought this approach was more straight-forward.
+w2aew Maybe I missed the previous video. You're right to think about the novice, that's also why I thought it was worth mentioning the impact of removing the meter was negligible (BECAUSE the 10k resistance is so high).
Maybe the safest for the novice is to leave the amp meter in series while using another multimeter to check the voltage?
Or use the voltmeter mode only to check the voltage drop of the 10k and the gauge?
+fabts4 You'll notice in the video that I *re-adjusted* the power supply after removing the ammeter. I guess I should've stated that I did this to account for the removal of the burden voltage of the ammeter. It would NOT have been negligible. Safest/easiest would have been to make both measurements with two meters - but I wanted to use my most accurate DMM for both measurements (and I wanted to show how to make each measurement for folks that only have one DMM). Using the DMM to measure the drop across the 10K is also a good way to go - provided you accurately measure the 10K value too. I mention this last point in my previous video also.
+w2aew I would be willing to bet that my Brymen 869s will do quite well with the measurements.
Great illustration. Dont you think that a carefully chosen wheatstone bridge 4 arms values has an equivalent benefit or probably adds to this method of separating the applied current leads from the voltage measuring leads (aka 4-wire method)?
Another thing than works well for making a current shunt is some pcb. You can calculate the value pretty closely, then trim it after etching. Of course copper has a much higher temperature coefficient than a real shunt resistor, so it won't be as accurate when it heats up.
Another useful video, Alan. Thank you.
Great video. Super useful.
great 4-wire usage, who needs Kelvin clips ?
Before the 50s they didn't use the resistor Color Code system. Each manufacture on the schematic would indicate a color code system for the resistors and capacitors. What are these other color code systems called and were used before the 50's?
How would you calculate full scale deflection current given the resistances, the meter current, and the leads between 2 points? No voltage given. This question is from the arco electrician exam book in the back of the book. (Exams)
Sounds like a little application of ohms law and linear extrapolation...
Curious if I would also need a series resistor going to the meter, or is that already contained inside the meter?
For an ammeter, a series resistor would just add unneeded (and unwanted) voltage drop.
@@w2aew So does the analog panel meter have a built-in resistor, or is it just the inherent resistance of the meter itself ( the windings and such)?
An "ideal" ammeter would have zero resistance, so any (parallel) shunt would have no effect on its full scale current handling ability, right?
So, I guess I'm asking if you only rely on the inherent resistance of the meter itself, or can you add to it a bit with an additional series resistor...
Like if I wanted to use a 100mA meter for a 1A @ 24V project I could set up a 250m Ohm (shunt) / 2.5 Ohm (series) divider...
Or is this series component not required since the meter itself is already acting as a 2.5R resistor?
I apologize for my ignorance and sloppy math.
@@andrewferg8737 In the case of analog meters like these, the meter's coil has resistance and determines the voltage drop when measuring current. Any additional series resistance would simply add to the ammeter's burden voltage.
@@w2aew Thanks so much for your patient reply. I did a bit a experimenting last evening and was able to get a 3mA meter working with 100m Ohm shunt for my 24V @ 1A project.
I've got bunch of these .1R resistors in my junk box, and have always wondered why anyone would manufacture a such a tiny value resistor. Thanks again!
Adding two resistors (series-parallel or parallel- series) with a little more simple algebra give more flexibility to the uses of any given meter.
Hi, I need some insight on how to make the minor adjustment to get a 50 microamp panel meter extent to work in place of a panel meter specified for 60 microamps. This is for an old Lafayette TE-46 capacitor analyzer. The original meter must have been damaged and someone replaced the original with a 1 Milliamp meter. Using your process The original meter is reading full scale at 1.14 milliamp and a new 50 microamp meter is running full scale at 48 microamps... Thanks for the video as I now know both work and what gets them to full scale.. Note my 100K resistor used for this tests at 9.946k on a 4 wire HP 3478a... That said I'm not sure what to do next. do I pursue working on extending the range of the 50 MA meter or shunting the 1 Milliamp meter? I'd rather use the 50 Microamp meter. Along with this question I've seen some discussion regarding adding diodes across these meters to protect them from going overscale and I'm not sure what or how to add this into the mix... One gentleman added two schottky diodes and another used two 1N4001 diodes reversed against each other. Honestly I don't really understand how these help yet but I'm sure there is a voltage drop issue I need to be aware of...Any insight on how to proceed and certainly some assistance on the shunt values to extend the 50microamp meter would be appreciated... (Would it be possible to use a small resistor in series with the meter instead of a shunt across the meter??) Oh I hope you are still out here... Of all the videos I've seen (and there have been many) yours are the easiest to understand for a novice...
The process shown in this video is *exactly* what you should be able to do to in order to use the 50uA FSD meter in the application that needs a 60uA movement. The shunt resistor would be carrying 10uA when the meter is at full scale. If you have measured the voltage across the 50uA meter when it is reading full scale, simply take that voltage and divide it by 10uA, and you'll have the shunt resistor value that you need. For example, when the 50uA meter is at full scale, suppose you measured 100mV. Then, the shunt resistor you would need would be 100mv divided by 10uA, or 10kohms.
@@w2aew Thank you, exactly as suggested the 10K resistor across the meter provided a full scale 60 uA across the meter. I also experimented with adding two schottky diodes across the meter along with the 10k resistor but these added an additional voltage drop causing the full scale to be more like 80 uA. (78.97 ua and 101.772 mv across the meter) However, I added two 1N4001 diodes across the meter along with the 10K and there was no impact. With the 10K resistor and the two 1N4001 diodes I have 59.52 uA and 101.158 mv across the meter. Without any shunt resistor the measurements are 48.57 uA and .098889 mv. Using the calculations I learned from you it appears that the meter itself must be 2036 ohms... I'm confused why there is no voltage drop impact by the 1n4001 diodes and there was an impact with the 1N5817 schottky diodes. Either way, considering I'm at full scale at 100 mv I imagine the meter will be destroyed long before either of these diode solutions start to short and pull current away from the meter.... Anyway, thanks for you help and awesome videos. I've watched many and find them well thought out and very insightful!! Thx, Ben. NJ
@@bendec2969 Well, I supposed I had a lucky guess by suggesting that the voltage across the meter was 100mV at full scale! Glad it worked out. The reason the schottky had a problem is likely because you had it forward biased and it was turning on slightly as the meter reached full scale.
@@w2aew Which diodes will provide the best protection... Just seeking your advise and not holding you responsible for any decision I make!
Again thanks for responding... I really appreciate the education. I'm having a lot of fun fixing my old toys with this new found hobby. Several heathkits sig gens, vtvms, Fisher Amp improvements - 400, 500c, (500B is next). Very happy with the result on a restoration of a very old Precision Model 68 VTVM and a PACO model 70 both with a DC voltage accuracy that is blowing my mind... I have a couple of their tube testers (610 and a 1012) and I have to say they were really a quality focused company... I've made minor adjustments to both and they are working fine although the 610 is at its limits and really needs a bunch of precision resistors replaced but I'm nervous about digging into this as the wiring is intimidating... My favorite piece is a fully professionally restored and beautiful Hickok model 605A (which I won't touch). Precise, Precise, Precise.... I've noticed that several tube purchases from reputable ebay sellers who used precise metrics for their tube sales I almost mirror....And several new tubes test at the exact recommended value listed in the documentation... Anyway, thanks for the help. I think I'm going to watch one of your videos and relax for the evening... Ben
Thank you for a great explanation. I recently purchased a piece of equipment that has a 500A ammeter. I opened it up and there is a shunt resistor of course for measuring the current, however there is a small potentiometer in series with the leads going to the panel ammeter. So it is exactly like your drawing in #3 but with the addition of a potentiometer in series with the meter. Is the purpose of this to reduce the amount of current actually flowing through the ammeter?
This is typically done as a fine-tune calibration adjustment.
In a DC circuit should the Ammeter or Load be connected to the Positive connector?
since they are in series, it really doesn't matter. It is most common to connect the ammeter to the positive connector so that the load can remain connected to / referenced to the lower rail (usually ground).
@@w2aew Yes...I saw that too....but, wanted to know if there was any specific reason?
@@akmt123 Yes, as stated above, it is so that the circuit under test can remain connected to it's reference potential - usually ground.
I have one question 2:12 why 0.185 when lab supply shows 2.180v can you please clarify. thanks
The lab supply is feeding a 10k resistor as well as the meter, in series. With 200 uA flowing through the 10k resistor, there is a 2 volt drop across the resistor, and then we expect the 0.185 volts across the meter. We saw a little bit less than 2.185, first because the power supply was being adjusted in 10mV increments to 2.180mV, and then there were several other factors that made it no so perfect. It's off by about 2.7%; plenty close enough for a tutorial!
Alan what size nichrome wire do you suggest to buy
Thanks Alan. I need a better power supply.
Hi Sir, I do have a little question. Do you know if a car fuel gauge in a 90s car is generally an ammeter? If so I can apply the knowledge of your video to test my fuel gauge as well as the sending unit. Thanks for your time!
Hi, did you ever do a "four wire" vid please ?
Search is your friend.
th-cam.com/video/kqkluABmJhQ/w-d-xo.html
Thanks for sharing. Did you choose that particular analog meter because it was made in New Jersey? Great channel!
+PelDaddy Hmmm - could be....
This is great stuff. Thank you.
Really cool, thank you
Excellent, thanks Alan!
When using Light bulbs as current limiters to test for short circuits its really confusing because if you use one light bulb rated at 60 watts divided by 120VAC = 0.5 amps of current. If you use 2 light bulb wired "in parallel" will be a total of 120 watts divided by 120VAC = 1 amp of current. Logically I would think wiring up resistors in parallel would decrease and lower the current limiting. If you used a 120 ohm/60 watt resistor as a current limiter that is 120ohm /120vac = 1amp of current, if you wire two 120 ohms/60watts that is a total of 60 ohms/120 watts = 0.5amps of current. Its confusing because the light bulbs are inverse of a resistor? or what am I missing and not understanding?
Your math for the resistors is wrong. If you have a 120ohm resistor with 120VAC across it, that resistor will dissipate 120W and the current will be 1A. If you wire two of these resistors in parallel, then *each* resistor will draw 1A and dissipate 120W - so the combination draws 2A total, and dissipates 240W total.
@@w2aew 120 ohm resistors in parallel with another 120 ohm resistor = total resistance is 60 ohms. So 120vac divided by 60 ohms = 2 amps which is Doubled. A 60watt lamp bulb is 60watts divided by 120vac = 0.5amps. Using two 60 watt lamp bulbs in parallel is 120 Watts divided by 120vac = 1 amp. What I'm confused about is a 60ohm resistor is NOT Equivalent to a 60 watt bulb, any reasons why? because 120vac/60 ohms = 2 amps compared to 60watts/120vac = 0.5amps. See how they are inverse the resistor and lamp bulb?
@@waynegram8907 You're mixing up the *resistance* and the *power dissipation*. A 60W bulb operated at 120VAC results in 0.5A current. Thus, the bulb's resistance is 120VAC/0.5A = 240 ohms. Put two of them in parallel and you get 120ohms, which will draw 1A at 120VAC, each one dissipating 60W for a total of 120W dissipation.
@@w2aew The filament inside the 60 watt lamp bulb will always have a resistance at 240 ohms at 120vac at Full Brightness like if the circuit under test has a short circuit, but when the circuits current draw is LESS than 0.5A the resistance of the filament changes to a different value. The Filament is not equivalent to a resistor. A 60 watt Filament is a 240 ohm resistor at 120vac but its not the same as a 60 watt/240 ohm resistor, any reasons why? A Filament is like a variable temperature potentiometer, as the circuit under test draws the maximum current of the Filament which is 120vac at 60 watts = 0.5 amps short circuit max current draw which is 240 ohms resistance of the filament. If the circuit under test is only drawing 100mA the 60watt lamp bulb resistance has changed to 1200 ohms and only using 12 watts of the 60watt lamp bulb . 120vac/100mA = 1200 ohms
@@waynegram8907 If a bulb is operated at a lower current, thus the filament is not heated up to normal operating temperature, then its resistance is LOWER than the fully heated value. For your 60W bulb, the filament is 240ohms at full 120VAC. But, at a lower current, it is lower than 240 ohms (you'd have to measure it). The idea with a dim-bulb tester is to put the bulb in SERIES with the device you're testing, so that it presents very little voltage drop during normal operating of your device, but if your device shorts out, then more current flows in the bulb, heating the filament, increasing it's resistance which then limits the current to save your device from damage.
Simple but beautiful!
Dear Sir, can you explain to me how to change ammeter scale from 50 A
to 5 A ?
It may or may not be possible, depending on the meter design. If the meter has a built-in shunt, you'd have to change the shunt value.
Alan, do you have to remove/disable any original internal shunt before doing a mod like this? I have a 0-1A AC ammeter that I want to make dual range (0-10A) with a switchable external shunt. 73 KCØHST
As long as you're adding a higher range, you can simply add an external shunt as needed.
Does any lab current ammeter measure MESH current? because most current ammeter just only measure Branch current. What are the differences are between MESH current and Branch current because you get different current values for MESH current compared to Branch Current. When would an EE engineer or technician want to know and measure the MESH current because most technician are just measuring Branch current?
MESH current refers to an ANALYSIS method, not a measurement method. www.allaboutcircuits.com/textbook/direct-current/chpt-10/mesh-current-method/
@@w2aew I'm confused why the MESH current value is different than the branch current. Why is the current have different values and what's the point of knowing the MESH current?
@@waynegram8907 As I stated, it is simply a MATHEMATICAL ANALYTICAL METHOD to solve for the total circuit current values. Mesh currents are computed for each "loop". When loops intersect, meaning that they share the same path or part, the mesh currents from each loop add up (or subtract, depending on the direction). You don't measure MESH currents, they're just a computational tool. Branch currents are the sum of all of the mesh currents that flow in that branch.
@@w2aew yes I understand its a computational tool but its totally different current values compared to using Kirchhoffs current law. What I'm confused about is why isn't kirchhoffs current values the same current values as MESH current. I would think logically that current should have the same value, because if the network or circuit is 1 amp using Kirchhoff current law but the MESH current is not going to be 1 amp it will be a totally different current value which doesn't make sense to me.
@@w2aew Ohms law uses V/R = I so the MESH current is not going to be V/R=I it will be a totally different current value, any reasons why the MESH current it totally different compared to ohm law?
brilliant guru
But is the response still fairly linear?
+Matthew Humphrey Yes - the ratio of currents through each side of the parallel circuit is fixed/linear.
Thanks Al!
73!
I like this video. Thanks
Thanks Alan..
Great and much fun as always, but why are they called " analog" meters? Have always wanted to know. Dave W4GSM
Because it can indicate any value within its range, as opposed to discrete values / digits like a digital meter (DVM DMM)
I'm trying to implement using a digital panel meter without success.
Digital panel meters often have auto-ranging, so there isn't one constant sense resistance for the current.
Easier to use a low value resistor (1 Ohm for example) in series with the circuit, then turn the meter into a voltmeter to simply read the voltage drop across the low value resistor.
MUCH easier than all those small value resistors. Also, since the meter is multi scale, it is easier to use it as a voltmeter too. Not just dedicate it to reading current.
Of course - it all depends on how much burden voltage you're willing to accept when measuring current.
Your videos are so Succulent ...I never understood why some single cell brain creatures are giving thumbs down to your videos. There are very, very, very, few people like you on YT who give useful info for the mankind.
Keep up good the work.
+rrangana11 I think some people are so disgusted with their own boring lives that they can't help but try to make others miserable. Even when I complain to someone about their videos, I don't click dislike.
good stuff but isn't a bit too much squeezed in time of 6 minutes ?
I like to keep my videos short, with high content density, due to the average youtube view duration of about 5 minutes.
ok, sth have you converted an ammeter-to- a voltmeter? this is not an easy concept for the average viewer. Also, sth like shunt resistors may need explaining further. no?!!
I can expand on these topics in future videos. The previous video (#235) gives more background on the meters...
@@w2aew You will be glad to hear that this is actually a 20 min video after I watched it three times ;) Great video!
All calculators should use RPN :)
+Michael Lloyd ;-)
@@w2aew Hey what about an Abacus ? LOL
I guess the point here is to do this is thoroughly as you could, but in this case you could safely ignore the Ifs term and just calculate Rs=Vfs/Ifsd. The simplest way to see this is probably that you're doing the calculations to 3 significant digits, (assuming this is a satisfactory amount of precision) whereas 200 µA would only change the 4th significant digit when added to 600 mA. As is to be expected, if the meter is meant to be useful as a voltmeter.
You're not really changing the scale here, are you? The meter's scale is 0.6A, which I think is why you had such a low shunt resistor value, no?
This one already had a scale that I wanted, but that scale was only valid with the proper shunt. Full scale deflection was at 200uA, so the shunt was needed to get back to the 600mA scale.
Once again an excellent presentation and it ties in so well with the previous meter clip, your Bird meter clips & also with something I've been trying to figure out. Has the shunt in my wattmeter changed value or why would I be getting measurements shown in my clip.th-cam.com/video/S1Ew8wQwSI4/w-d-xo.html. My lab equipment is a little more basic...I need to find out voltage required for FSD next I believe for the equation.
+Christopher Larwood The meter used in the Bird 43 (and I presume in your meter as well) is a 30uA full scale meter movement. It does not have an external shunt. Thus, if it doesn't read 30uA at full scale, the meter movement itself if bad - there is nothing to can "add" to it externally to correct for this short of designing a current amplifier - and there's no guarantee that this will work correctly with the proper linearity.
All of the panel mount ammeters seem to be from China. 2.5% accuracy. So your calculating the current flowing through the meter the 200 micro amps would be useless. Also the standard for the meters is 75mV. So all of the different models of that ammeter can be made identical. Only the paper with the printed scale would have to be different.
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