17:00 These are high voltage transistors (>350V breakdown voltage each), and what actually happens for a negative enough voltage, is the Q211 diode starts conducting, and that clamps the base of Q204 to -0.7V, and thus the emitter of Q203 always stays at > +0.7V. This prevents the source to be pulled below +0.7V and limits the current to what the source is normally generating. The transistor Q211, used as a diode, is chosen for its ultra-low reverse current, to not interfere with a normal operation of the source.
Hi. I am just getting the hang of this stuff. Why is it that when a negative voltage is applied in the wrong direction the Q211 diode starts conducting I'm unsure as to why this happens.
@@trif169To see it more clearly, start with a simpler circuit. Replace the eight transistors with just one, with the 200K resistor between the collector and the base. Assume some fixed DC current gain, say 100. The constant current from the current source, say 1 mA, flows into the emitter. See what the voltages on the emitter and the base will be, as a function of the voltage on the collector, as you vary it.
@@trif169The diode-connected Q211 goes from AGND through four series resistors to the output terminal. AGND is higher than the negative output terminal thus Q211 turns on, and the resistors divide the output voltage drop among them. The Darlington-connected transistor pairs then copy this voltage to their emitters. Since those are high voltage transistors, they can easily each survive having 250V dropped across it. At 1mA that’s 1/4W of dissipation.
@@absurdengineering You guys are on a good path, but "missed by a hair" is also a target miss:P:) Q211 is a MMBF4117A (Transistor J-FET N-Chan D-Mode). This classic circuit (which, by the way, also appears very often in digital ICs) is called: FET constant current two-pole [1][2]. The [RTFM] shows: " Protection from large negative voltages is provided by the sum of the collector to base breakdown voltages of Q203, Q205, Q207, and Q209. Bias for these transistors is provided by Q211 and R203 to R206 while negative overvoltages are applied." [1] I will not explain here why a self-conducting junction FET (junction FET = JFET) is suitable as a current limiter. This is best read in further literature such as “The semiconductor circuitry” by U.Tietze and Ch.Schenk. [2] From the German terminology "FET-Konstantstromzweipol", see "JFET as a constant current source | All About Circuits" or "Mosfet constant current", EE-StackExchange [RTFM] Keysight 34401A 6½ Digit Multimeter, Service Guide, Edition 9, August 2014 by Keysight Technologies on Page 98.
I had some success in lowering the noise of the ohm meter current source by replacing CR201 4V7 with a low-noise type. As the zener CR201 is in series with the output of the op amp it contributes to the noise on the current source. I would also replace C210 220pf with COG capacitor to reduce any TC effects. Hope this helps someone.cheers
@@AnalogDude_ If you knew the principles of those circuit design before ... you could actually see the invisible mouse and what she was pointing at. But please forgive me this comment, which isn't very helpful and solely meant to make you guys smile. My apologies and don't hesitate to [RTFM] which explains most, if not all of your questions!:) [RTFM] Keysight 34401A 6½ Digit Multimeter, Service Guide, Edition 9, August 2014 by Keysight Technologies
The protection circuit for negative negative voltages works like a kind of cascode circuit. Q211 works like a very low leakage diode. R203, R204, R205, R206 are a voltage divider, so the reverse voltage is divided into -1000V, -750V, -500V, -250V, -0.7V. This causes also causes the corresponding voltage levels on the of collectors of each darlington transistor pairs. The BE diodes of Q203, Q204 and the reverse biased Q211 limit the minimum voltage at Q202 to -2.1 V. For normal operation with positive voltages from the current source Q211 has high impedance. The very high amplification of the four darlington pairs cause just a small voltage drop on each of the steps of the voltage divider, like 1.5 V. So the total voltage drop of the protection circuit will be around 6 V for "high" test currents of 1 mA. For the low current ranges with 100/10/5/0.5 µA it will be even lower: U= (1,4 V) + (4 * 196 k * I).
Nope, but interesting thoughts. But: The [RTFM] shows: "Protection from large negative voltages is provided by the sum of the collector to base breakdown voltages of Q203, Q205, Q207, and Q209. Bias for these transistors is provided by Q211 and R203 to R206 while negative overvoltages are applied." Q211 is a MMBF4117A (Transistor J-FET N-Chan D-Mode). This classic circuit is called: FET constant current two-pole. [RTFM] Keysight 34401A 6½ Digit Multimeter, Service Guide, Edition 9, August 2014 by Keysight Technologies on Page 98.
1. The protection diode at 15:45 have the wrong direction. For practical considerations the leakage to max current ratio is possible not good enough. 2. Without external interfering source the PNP darlington string operate in saturation. Only the Q210 have Vce(Q210)=1D and Q209 have Vce(Q209)=2D, where D denotes a Vbe. The total voltage burden is 3D plus 3x Vcesat, including the CR202. The R203-R206 are dimension for max measure current divided by beta by square and Vcesat. This structure also work with triple darlington or none. 3. The Q211 is a low leakage diode which activate negative protection at -2D with an impedance of 4x196k where the emitter of Q203 remain at +D. I guess the active circuit left to the darlington string is sensitive and worth to protect. Btw the D notation is frequently used in analog IC designs where "stacking" and "potential matching" is an impotant circuit topology construction item.
Thanks, I thought this was would be very interesting circuit when I suggested it a while back. The protection circuit had me scratching my head, finally had to SPICE the thing to see how it worked. The key is the string of resistors, the left end of the string is connected to ground through a low leak forwarded biased diode (the FET;s g-s) when a negative voltage is applied to the probe. Thus each resistor drops 1/4 the voltage. Think of the transistor pairs as darlingtons and with the resistors makes this a current limiter ( or current source if you like) when the probe is accidentally connected to a negative voltage. With -400V on the probe each resistor would drop 100V and the transistors would limit the current to limited about 1mA if I remember correctly. Thus each pair to transistor only has to manage the Vce breakdown and power of 1/4 the -1000V speck. And the 1mA current limit prevents damage to the main percision current source circuit. As IMSA explained so well. The output diode protects from positive voltages. In normal mode (probe > 0V) all the transistors are conducting and all current entering the left side exits the right side since there are not other connection paths. As mentioned, there would be some small Vbe voltage drop which doesn't matter. Thanks again for the video.
Q211 is a MMBF4117A (Transistor J-FET N-Chan D-Mode). This classic circuit is called: FET constant current two-pole [1][2]. You are very near the design principle behind the circuit with your "biased diode", but "missed by a hair" is also a target miss, hehehehe. The [RTFM] shows: " Protection from large negative voltages is provided by the sum of the collector to base breakdown voltages of Q203, Q205, Q207, and Q209. Bias for these transistors is provided by Q211 and R203 to R206 while negative overvoltages are applied." [1] I will not explain here why a self-conducting junction FET (junction FET = JFET) is suitable as a current limiter. This is best read in further literature such as “The semiconductor circuitry” by U.Tietze and Ch.Schenk. [2] From the German terminology "FET-Konstantstromzweipol", see "JFET as a constant current source | All About Circuits" or "Mosfet constant current", EE-StackExchange [RTFM] Keysight 34401A 6½ Digit Multimeter, Service Guide, Edition 9, August 2014 by Keysight Technologies on Page 98.
Thanks for the video, I just subscribed to your channel. I'm always curious on how test equipment electronics work. Back in the 1990's I worked at HP Test and Measurement Division as a firmware engineer. I was working on Automated Test Equipment that could test Flash Memory in the wafer sort in the semiconductor manufacturing process... I wished I had used my HP discount to buy my test equipment then. Anyway, I purchased a used HP34401 bench multimeter, specifically wanted it branded as HP, and what a joy it is to use. My display is dim, so I've already ordered a new one through E-Bay, a new VFD is on its way to me in 10 days. I saw your video on the subject and hope to get that meter back to its old self with the new display. There are hardly any scratches on this equipment, so I think it had an easy life.
Hi Ray, perhaps we overlapped a bit. I worked as a hardware engineer on "the hill" until about 1991 and knew the whole design team for the 34401. I'm happy to say that I did purchase one at employee discount and it is still working fine on my bench at home. My present employer is kind enough to calibrate it for me from time to time. The ADC design from the 34401 was reused in the wildly successful 34970 DAQ box. We have three or four dozen of the later in service where I work. They are used in many production cells and also in R&D.
@@davidrick959 That is so cool... I think it is so amazing to work with talented people that work on design... it takes a team, That is for sure! I was at HP/Agilent 1996 - 2000, Worked at the Santa Clara Site, Building 53. I worked in the CSTD (California Semiconductor Test Division) doing the firmware on their Test Site Controller boards for the Versatest line of FLASH memory testers primarily used in production cleanroom wafter sort testers. I have to say that those were the last days of HP that I witnessed as the company spun itself into various companies and is no longer what it was... but at least I got to see it. I visited HP Labs in the day, it was only a couple of times, but I'll always have the fond memories. I found a local company here in the Bay Area that sells used lab equipment, and the old HP gear still fetches high prices... but they had some nice HP branded equipment and bought a couple of pieces, the Arb Frequency Generator 33120A and the 34401A multimeter. They are excellent, as you well know, I put them in my home lab, and rack mounted them (somewhat on a shelf). The Display on the multimeter is a bit dimmer than the other unit, while it is still visible, I'm going to replace it anyway... I plan to keep the other display... but it is a piece of history, and I enjoy using it at my bench. Presently looking for work, Are you still with Keysight or whatever they call themselves today?
Stacked Darlingtons that distribute the voltage drop. Darlingtons are necessary so that the resistors can be high value for low power dissipation. It would be interesting to see the protection circuitry for the voltage and current ranges.
Not quite darlingtons for 3 of 4 stages, but you're on the right track. The resistors divide the voltage drop, while the transistors carry the bulk of current.
I saw your video last night, so this morning I went to analyze the circuit (at 9:07) and I think this is correct: Q201 and U102 form a voltage mirror on pin 1 of Q202. It seems confused, since U201 pin 6 is connected to Q202 pin 1 through the switches, it copies voltage from Q201 pin 1 to Q202 gate via U201, ending up Q202 working as a current mirror of Q201, it works pretty well. The current mirror at the left inside U101, promotes almost 3mA (when the switch on source of the second fet is on) over the 5k resistor, what in fact zero the voltage at pin 1 of Q202, since the Vdrop over 5k is higher than 18V-CR203 zenerV, it can calibrate the ADC for zero ohms when going into that scale. This circuit could promote 8 ohms range, 4 x 2 (R201/R202), but uses only 7 ranges.
As for me, the protection section looks more like a quadruple VBe multiplier which must drop much voltage in case of overcurrent in the forward direction.
On the protection circuit: Q211 will act as a diode for significantly large negative voltages, so putting the base of Q204 at about -0.7V As the absolute value of negative voltage at Xref increases voltage Drop on R203, R204, R205, R206 increases, so the base to base voltages of Q204,Q206,Q208,Q210 increases, making the transistors to shoulder the voltages equally. Remember the emitter of these transistors are 0.7 V above base voltage (pnp transistor). Also note that this circuit is beta multiplying circuit like Darlington structure. During normal operation only very small current from main test current is used to bias these transistors. To better see this replace this structure with Darlington Structure and connect the base of the Darlington structure connected to collector using 196K Ohm and compare it with single pnp transistor. Even though, all the currents add up in both ends, however much less voltage drop created on resistors. Also note that Q211 is connected as diode ( I wonder why they did not used regular diode, may be because of low leakage current) and is off when output voltage is positive. One should simulate this circuit with output connected to -1V, -5V, -10V etc. to see the distribution of base and collector voltages. Also it might be instructive to simulate this circuit with output connected to a resistor.
This video sorta fell off the rails the farther along you got, You forgot to mention that Q211 is used as a fast ultra-low leakage diode. That's critical to how the protection stacked Darlington's work. If possible, this video would be a good candidate for a re-do.
I already lost faith, when he brushed over the JFET without actually explaining why and how the JFET is different (but explaining the well known BJT stuff at length)
@@opros7At the lowest setting, the current source itself produces only 500 nA. To achieve good accuracy, the diode leakage must be much smaller than that. The 2N4117A transistor chosen for the Q211 is famous for its ultra-low leakage, about 0.0002 nA at room temperature. Thus the 500/0.0002 ratio preserves full 6.5 digit accuracy, even for the lowest current.
Problem is, the „standard“ opamp current source with the ref voltage against gnd is in fact a current sink. If it had to be a source (feeding current into GND), the ref voltages is against positive supply. With „sinking“ current to GND through a resistor connected to positive supply you get exactly this done: having a „5V-below positive power“ ref voltage to compare the voltage drop of the GND referenced current source‘s R against. It’s a clever use of transferring voltage into current into voltage into current, where the input ref voltage (the 7V) is GND based and the current output is sourcing into GND. (The old) HP has(/had) clever engineers. 👍🏻
34401 is not that old. Ohm meters usually employ a 'low side' current source... that puts the current source between the DUT (a resistance) and the return (call it ground)
1) Contrary to what was said below this is not a stack of darlington. In darlington configuration both collectors are tied together. Here the even numbered transistors collector is connected to base of next odd numbered transistors. 2) Q211 as connected act like a diode with heigh internal resistance and in normal operation it doesn't conduct current. But if a negagative voltage is present at CR202 then Q211 conduct current and the voltage drop between CR202 anode and Q203 emetter will be equally distributed between even numbered transistors VCE. The voltage at base of Q204 base will be negative but below Vbe breakdown voltate and Q203 emitter voltage will stay a few hundred milivolt positive. A JFET is used for Q211 instead of a diode because the internal resistance of the JFET channel limit current in Q203,Q204 bases. As long as the maximum Vce voltage of transistors stay in specifications there will be no damage to circuit. I simulated it in ltspice before commenting.
WOW a stroll down memory lane: Schottky diode Zener diode SPDT Schematics Circuit analysis Q201 HP equipment CMOS FET Over on the right a TO package I could go on and on. 😎
There's a Widlar current mirror in the lower left corner of U101-E (that's the numbering on the IC) probably a custom planar CMOS FET array, and current mirror (Notice the degeneration resistor is not there... maybe the FET matching means it isn't needed?) I think the Guy has discussed current sources in another vid? But what's the dope on the one here?
Your descriptions are very good but would be much better if instad of just sayingh "this resistor" you actually pointed out which one you are talking about. Frequent references to "here" without a pointer are not helpful....
Oh, thank you very much. Maybe you will do a series of lessons and we will examine the entire multimeter 34401 according to the scheme. It would be great to keep going
Thank you very much for taking your time to explain the whole circuit, indeed is a complex arrangement to make a "simple" current source. About the array of protection transistors, that was what got me thinking harder, although i had concluded that would be for negative voltage protection, i wasn't quite sure if it works as i thought it would. On the comments @cogoid complemented your explanation, i think that should be pinned. Once again, thank you very much! 👍😃
Hi, I guess that using transistor breakdown instead of zener is for speed. Seen many IC's with two transistors, each having their collectors shorted to base, and using the Vebo breakdown (typically 5-6v) as back to back zeners for input protection. The reason is the much higher speed of the transistor breakdown than zeners, so very fast spikes can be clamped.
the "unused" FETs on U101 (pin 42) are forming a current mirror, with current set by R207, so probably part of calibration. also, the video didn't explain why there's two current sources- why is the first one needed when you could just provide a reference voltage directly? having two in series seems like it'd introduce more errors. maybe the reason is process variations on the chips? if you used mechanical switches, the contacts generally have a known small resistance, but using FETs on a chip will have a much higher (10's of ohms or more) and worse, variable resistance that varies some chip to chip. I am curious what the zener diodes' purposes are- is it to reduce the Vdrop across resistors which would improve their temperature coefficient by reducing self heating? (re: CR201 and CR203).
If you provide a fixed reference voltage to the second current source then the temperature coefficient of the resistors over there cause the current to change. In this case the resistor which sets the 5V reference also changes to compensate for it. You can look at the second stage as a temperature independent scalable current mirror where the accuracy is determined only by the first reference current generating circuit.
@@robertvandersanden Also it allowed the voltage reference to be raised closer to the positive rail so the circuit could be used to source instead of sink current.
Even though I know what I am doing and was trying to follow along, this fairly new (for this channel) business of using the blotchy red 'electronic' markups left me cold. Often hard to read what was being written. Much more frequently, the narration says "now this thing over here" (etc) is not accompanied by any discernible red marks, so I am thinking, "OK, what is he talking about?". Sometimes the old ways are the best, and videoing some pencil marks being made real-time on a schematic work better.
I think D211, the odd FET in the bipolar chain on the output, is used as a very low leakage current diode. Leakage currents are an error, and I've seen FETs used like a diode in sample-and-hold circuits.
Thank you for the explanation. What always confuses me at this kind of complex analog circuits is the question if this wouldn’t be smarter to look for an IC to do the same job. Aren’t there any super exact constant current source ICs available ?
This interesting circuit drew my attention. I started to watch the video but it became difficult to follow without some kind of visible pointer on screen to identify the components you refer to in the verbal description.
this was great content and i thank you many time HERE was used and not referenced on screen i still learned alot but got lost many times on the word here.
16:15 I don't follow this logic and how all 8 transistors oppose to breakdown? By Kirchhoff's law breakdown voltage between Qe(203) and Qc(209) is 4 times Vceo (if all top PNPs are the same). When positive over-voltage is applied CR202 will block it. Q211 functions as a precision diode. When negative voltage is applied, Q211 will drive base of Q204 to ground and the whole transistor network will behave like an active load, limiting current and increasing its resistance to oppose increasing voltage. Voltage on the emitter of Q203 will be aroud 2Vbe (+Vgs). At extreme negative over-voltage 4 transistors on top, or in the middle will just give up and short, one by one. My two cents.
Q221 - используется как диод с очень низким уровнем утечки. Т.к. у этих полевых транзисторов очень узкий канал, а следовательно и очень маленькая площадь PN-перехода, то и утечка через него будет настолько мала, что её можно не учитывать (примерно 150-300 фА)
Their schematic symbol for a switch is absolutely for shit. You can't get any sense of whether it's a spdt or if any switch is connected or correlated in any way.
Sorry, but I keep hearing acrossT when my ears expect simply “across” (without the “T”). Other than that, thanks for the explanation for something I did not even realize built so complex!
The beta of the transistor in the current source is irrelevant so long as the op amp is working. Thus, replacing the transistor with a FET is meaningless in this case. Think about it a minute. What is important is the offset voltage of the op amp.
The load resistor is carrying both base and collector current. This could be corrected for if the beta of the transistor is constant, but real world transistors have changing beta with temperature. Opamp offset voltage should be the lesser error with a good opamp, such small offset voltage being needed to produce base (or gate) voltage.
@@hanelyp1 The beta of the transistor is IRRELEVANT so long as the op amp is working. Remember, the op amp will drive the output to turn on the transistor to whatever level necessary to keep the voltage on + and - inputs the same.
@@glasslinger placing the combined transistor drive and output current on the sense resistor. Which if you don't account for the drive current produces an error in the output.
17:00 These are high voltage transistors (>350V breakdown voltage each), and what actually happens for a negative enough voltage, is the Q211 diode starts conducting, and that clamps the base of Q204 to -0.7V, and thus the emitter of Q203 always stays at > +0.7V. This prevents the source to be pulled below +0.7V and limits the current to what the source is normally generating.
The transistor Q211, used as a diode, is chosen for its ultra-low reverse current, to not interfere with a normal operation of the source.
Hi. I am just getting the hang of this stuff. Why is it that when a negative voltage is applied in the wrong direction the Q211 diode starts conducting I'm unsure as to why this happens.
Thanks for your input, it makes sense, i hope your comment gets pinned, its relevant information.
@@trif169To see it more clearly, start with a simpler circuit. Replace the eight transistors with just one, with the 200K resistor between the collector and the base. Assume some fixed DC current gain, say 100.
The constant current from the current source, say 1 mA, flows into the emitter. See what the voltages on the emitter and the base will be, as a function of the voltage on the collector, as you vary it.
@@trif169The diode-connected Q211 goes from AGND through four series resistors to the output terminal. AGND is higher than the negative output terminal thus Q211 turns on, and the resistors divide the output voltage drop among them. The Darlington-connected transistor pairs then copy this voltage to their emitters. Since those are high voltage transistors, they can easily each survive having 250V dropped across it. At 1mA that’s 1/4W of dissipation.
@@absurdengineering You guys are on a good path, but "missed by a hair" is also a target miss:P:)
Q211 is a MMBF4117A (Transistor J-FET N-Chan D-Mode). This classic circuit (which, by the way, also appears very often in digital ICs) is called: FET constant current two-pole [1][2]. The [RTFM] shows: " Protection from large negative voltages is provided by the sum of the collector to base breakdown voltages of Q203, Q205, Q207, and Q209. Bias for these transistors is provided by Q211 and R203 to
R206 while negative overvoltages are applied."
[1] I will not explain here why a self-conducting junction FET (junction FET = JFET) is suitable as a current limiter. This is best read in further literature such as “The semiconductor circuitry” by U.Tietze and Ch.Schenk.
[2] From the German terminology "FET-Konstantstromzweipol", see "JFET as a constant current source | All About Circuits" or "Mosfet constant current", EE-StackExchange
[RTFM] Keysight 34401A 6½ Digit Multimeter, Service Guide, Edition 9, August 2014 by Keysight Technologies on Page 98.
I had some success in lowering the noise of the ohm meter current source by replacing CR201 4V7 with a low-noise type. As the zener CR201 is in series with the output of the op amp it contributes to the noise on the current source. I would also replace C210 220pf with COG capacitor to reduce any TC effects. Hope this helps someone.cheers
I think you are pointing on stuff on the screen, but it does not show up in the video...
I was starting to wonder if I needed new glasses or a better monitor.
Indeed!
He forgot to activate the mouse, but indeed it makes you wonder what he was pointing at to follow the story.
@@AnalogDude_ If you knew the principles of those circuit design before ... you could actually see the invisible mouse and what she was pointing at. But please forgive me this comment, which isn't very helpful and solely meant to make you guys smile.
My apologies and don't hesitate to [RTFM] which explains most, if not all of your questions!:)
[RTFM] Keysight 34401A 6½ Digit Multimeter, Service Guide, Edition 9, August 2014 by Keysight Technologies
The protection circuit for negative negative voltages works like a kind of cascode circuit. Q211 works like a very low leakage diode. R203, R204, R205, R206 are a voltage divider, so the reverse voltage is divided into -1000V, -750V, -500V, -250V, -0.7V. This causes also causes the corresponding voltage levels on the of collectors of each darlington transistor pairs. The BE diodes of Q203, Q204 and the reverse biased Q211 limit the minimum voltage at Q202 to -2.1 V.
For normal operation with positive voltages from the current source Q211 has high impedance. The very high amplification of the four darlington pairs cause just a small voltage drop on each of the steps of the voltage divider, like 1.5 V. So the total voltage drop of the protection circuit will be around 6 V for "high" test currents of 1 mA. For the low current ranges with 100/10/5/0.5 µA it will be even lower: U= (1,4 V) + (4 * 196 k * I).
Nope, but interesting thoughts. But: The [RTFM] shows: "Protection from large negative voltages is provided by the sum of the collector to base breakdown voltages of Q203, Q205, Q207, and Q209. Bias for these transistors is provided by Q211 and R203 to R206 while negative overvoltages are applied."
Q211 is a MMBF4117A (Transistor J-FET N-Chan D-Mode). This classic circuit is called: FET constant current two-pole.
[RTFM] Keysight 34401A 6½ Digit Multimeter, Service Guide, Edition 9, August 2014 by Keysight Technologies on Page 98.
1. The protection diode at 15:45 have the wrong direction. For practical considerations the leakage to max current ratio is possible not good enough.
2. Without external interfering source the PNP darlington string operate in saturation. Only the Q210 have Vce(Q210)=1D and Q209 have Vce(Q209)=2D, where D denotes a Vbe. The total voltage burden is 3D plus 3x Vcesat, including the CR202. The R203-R206 are dimension for max measure current divided by beta by square and Vcesat. This structure also work with triple darlington or none.
3. The Q211 is a low leakage diode which activate negative protection at -2D with an impedance of 4x196k where the emitter of Q203 remain at +D. I guess the active circuit left to the darlington string is sensitive and worth to protect.
Btw the D notation is frequently used in analog IC designs where "stacking" and "potential matching" is an impotant circuit topology construction item.
Thanks, I thought this was would be very interesting circuit when I suggested it a while back. The protection circuit had me scratching my head, finally had to SPICE the thing to see how it worked. The key is the string of resistors, the left end of the string is connected to ground through a low leak forwarded biased diode (the FET;s g-s) when a negative voltage is applied to the probe. Thus each resistor drops 1/4 the voltage.
Think of the transistor pairs as darlingtons and with the resistors makes this a current limiter ( or current source if you like) when the probe is accidentally connected to a negative voltage. With -400V on the probe each resistor would drop 100V and the transistors would limit the current to limited about 1mA if I remember correctly. Thus each pair to transistor only has to manage the Vce breakdown and power of 1/4 the -1000V speck. And the 1mA current limit prevents damage to the main percision current source circuit. As IMSA explained so well. The output diode protects from positive voltages. In normal mode (probe > 0V) all the transistors are conducting and all current entering the left side exits the right side since there are not other connection paths. As mentioned, there would be some small Vbe voltage drop which doesn't matter.
Thanks again for the video.
Q211 is a MMBF4117A (Transistor J-FET N-Chan D-Mode). This classic circuit is called: FET constant current two-pole [1][2]. You are very near the design principle behind the circuit with your "biased diode", but "missed by a hair" is also a target miss, hehehehe. The [RTFM] shows: " Protection from large negative voltages is provided by the sum of the collector to base breakdown voltages of Q203, Q205, Q207, and Q209. Bias for these transistors is provided by Q211 and R203 to R206 while negative overvoltages are applied."
[1] I will not explain here why a self-conducting junction FET (junction FET = JFET) is suitable as a current limiter. This is best read in further literature such as “The semiconductor circuitry” by U.Tietze and Ch.Schenk.
[2] From the German terminology "FET-Konstantstromzweipol", see "JFET as a constant current source | All About Circuits" or "Mosfet constant current", EE-StackExchange
[RTFM] Keysight 34401A 6½ Digit Multimeter, Service Guide, Edition 9, August 2014 by Keysight Technologies on Page 98.
Thanks for the video, I just subscribed to your channel. I'm always curious on how test equipment electronics work. Back in the 1990's I worked at HP Test and Measurement Division as a firmware engineer. I was working on Automated Test Equipment that could test Flash Memory in the wafer sort in the semiconductor manufacturing process... I wished I had used my HP discount to buy my test equipment then. Anyway, I purchased a used HP34401 bench multimeter, specifically wanted it branded as HP, and what a joy it is to use. My display is dim, so I've already ordered a new one through E-Bay, a new VFD is on its way to me in 10 days. I saw your video on the subject and hope to get that meter back to its old self with the new display. There are hardly any scratches on this equipment, so I think it had an easy life.
Hi Ray, perhaps we overlapped a bit. I worked as a hardware engineer on "the hill" until about 1991 and knew the whole design team for the 34401. I'm happy to say that I did purchase one at employee discount and it is still working fine on my bench at home. My present employer is kind enough to calibrate it for me from time to time. The ADC design from the 34401 was reused in the wildly successful 34970 DAQ box. We have three or four dozen of the later in service where I work. They are used in many production cells and also in R&D.
@@davidrick959 That is so cool... I think it is so amazing to work with talented people that work on design... it takes a team, That is for sure! I was at HP/Agilent 1996 - 2000, Worked at the Santa Clara Site, Building 53. I worked in the CSTD (California Semiconductor Test Division) doing the firmware on their Test Site Controller boards for the Versatest line of FLASH memory testers primarily used in production cleanroom wafter sort testers.
I have to say that those were the last days of HP that I witnessed as the company spun itself into various companies and is no longer what it was... but at least I got to see it. I visited HP Labs in the day, it was only a couple of times, but I'll always have the fond memories.
I found a local company here in the Bay Area that sells used lab equipment, and the old HP gear still fetches high prices... but they had some nice HP branded equipment and bought a couple of pieces, the Arb Frequency Generator 33120A and the 34401A multimeter. They are excellent, as you well know, I put them in my home lab, and rack mounted them (somewhat on a shelf). The Display on the multimeter is a bit dimmer than the other unit, while it is still visible, I'm going to replace it anyway... I plan to keep the other display... but it is a piece of history, and I enjoy using it at my bench.
Presently looking for work, Are you still with Keysight or whatever they call themselves today?
The explanation of the protection network was rather unsatisfactory. Thankfully, some smart comments helped me to understand how it really works.
Stacked Darlingtons that distribute the voltage drop. Darlingtons are necessary so that the resistors can be high value for low power dissipation.
It would be interesting to see the protection circuitry for the voltage and current ranges.
Not quite darlingtons for 3 of 4 stages, but you're on the right track. The resistors divide the voltage drop, while the transistors carry the bulk of current.
Thank you so much . As often your videos come at a great time
Thanks for the knowledge dude your prolific and mellow style is a hit with me :)
Those are your most interesting videos, keep doing them!
Thanks, for the explanation. I really in enjoy these videos about HP equipment.
Yes. Me too. HP made the best stuff 👍
I saw your video last night, so this morning I went to analyze the circuit (at 9:07) and I think this is correct: Q201 and U102 form a voltage mirror on pin 1 of Q202. It seems confused, since U201 pin 6 is connected to Q202 pin 1 through the switches, it copies voltage from Q201 pin 1 to Q202 gate via U201, ending up Q202 working as a current mirror of Q201, it works pretty well. The current mirror at the left inside U101, promotes almost 3mA (when the switch on source of the second fet is on) over the 5k resistor, what in fact zero the voltage at pin 1 of Q202, since the Vdrop over 5k is higher than 18V-CR203 zenerV, it can calibrate the ADC for zero ohms when going into that scale. This circuit could promote 8 ohms range, 4 x 2 (R201/R202), but uses only 7 ranges.
As for me, the protection section looks more like a quadruple VBe multiplier which must drop much voltage in case of overcurrent in the forward direction.
On the protection circuit:
Q211 will act as a diode for significantly large negative voltages, so putting the base of Q204 at about -0.7V
As the absolute value of negative voltage at Xref increases voltage Drop on R203, R204, R205, R206 increases, so the base to base voltages of Q204,Q206,Q208,Q210 increases, making the transistors to shoulder the voltages equally. Remember the emitter of these transistors are 0.7 V above base voltage (pnp transistor).
Also note that this circuit is beta multiplying circuit like Darlington structure. During normal operation only very small current from main test current is used to bias these transistors.
To better see this replace this structure with Darlington Structure and connect the base of the Darlington structure connected to collector using 196K Ohm and compare it with single pnp transistor. Even though, all the currents add up in both ends, however much less voltage drop created on resistors.
Also note that Q211 is connected as diode ( I wonder why they did not used regular diode, may be because of low leakage current) and is off when output voltage is positive.
One should simulate this circuit with output connected to -1V, -5V, -10V etc. to see the distribution of base and collector voltages. Also it might be instructive to simulate this circuit with output connected to a resistor.
I’m not really interested in radio, but I’m following this channel for videos like this. Is brilliant the knowledge shared here.
This video sorta fell off the rails the farther along you got, You forgot to mention that Q211 is used as a fast ultra-low leakage diode. That's critical to how the protection stacked Darlington's work. If possible, this video would be a good candidate for a re-do.
He wings it... but, we still watch... don't we.
page 98 of service manual the sum of the reverse collector break down voltages, q211 is for voltage biasing the network also having a low leakage
I already lost faith, when he brushed over the JFET without actually explaining why and how the JFET is different (but explaining the well known BJT stuff at length)
Your mouse pointer is not showing.
Thank you for everthing you present t o us. I love your show. from Bill
The real question is what Q211 does. As you get this, the rest of that circuit is relatively easy to understand.
I have the same doubt. For me, it does make sense!
@@opros7At the lowest setting, the current source itself produces only 500 nA. To achieve good accuracy, the diode leakage must be much smaller than that. The 2N4117A transistor chosen for the Q211 is famous for its ultra-low leakage, about 0.0002 nA at room temperature. Thus the 500/0.0002 ratio preserves full 6.5 digit accuracy, even for the lowest current.
Q211 is a diode-connected FET. It is a diode, low leakage one, sure, but just a diode.
Problem is, the „standard“ opamp current source with the ref voltage against gnd is in fact a current sink.
If it had to be a source (feeding current into GND), the ref voltages is against positive supply. With „sinking“ current to GND through a resistor connected to positive supply you get exactly this done: having a „5V-below positive power“ ref voltage to compare the voltage drop of the GND referenced current source‘s R against.
It’s a clever use of transferring voltage into current into voltage into current, where the input ref voltage (the 7V) is GND based and the current output is sourcing into GND.
(The old) HP has(/had) clever engineers. 👍🏻
34401 is not that old. Ohm meters usually employ a 'low side' current source... that puts the current source between the DUT (a resistance) and the return (call it ground)
1) Contrary to what was said below this is not a stack of darlington. In darlington configuration both collectors are tied together. Here the even numbered transistors collector is connected to base of next odd numbered transistors.
2) Q211 as connected act like a diode with heigh internal resistance and in normal operation it doesn't conduct current. But if a negagative voltage is present at CR202 then Q211 conduct current and the voltage drop between CR202 anode and Q203 emetter will be equally distributed between even numbered transistors VCE. The voltage at base of Q204 base will be negative but below Vbe breakdown voltate and Q203 emitter voltage will stay a few hundred milivolt positive. A JFET is used for Q211 instead of a diode because the internal resistance of the JFET channel limit current in Q203,Q204 bases.
As long as the maximum Vce voltage of transistors stay in specifications there will be no damage to circuit.
I simulated it in ltspice before commenting.
WOW a stroll down memory lane:
Schottky diode
Zener diode
SPDT
Schematics
Circuit analysis
Q201
HP equipment
CMOS
FET
Over on the right a TO package
I could go on and on. 😎
There's a Widlar current mirror in the lower left corner of U101-E (that's the numbering on the IC) probably a custom planar CMOS FET array, and current mirror (Notice the degeneration resistor is not there... maybe the FET matching means it isn't needed?) I think the Guy has discussed current sources in another vid? But what's the dope on the one here?
Your descriptions are very good but would be much better if instad of just sayingh "this resistor" you actually pointed out which one you are talking about. Frequent references to "here" without a pointer are not helpful....
Oh, thank you very much. Maybe you will do a series of lessons and we will examine the entire multimeter 34401 according to the scheme. It would be great to keep going
Maybe just me but I followed the video along fine. Really interesting stuff, thanks 🙂
Thank you very much for taking your time to explain the whole circuit, indeed is a complex arrangement to make a "simple" current source. About the array of protection transistors, that was what got me thinking harder, although i had concluded that would be for negative voltage protection, i wasn't quite sure if it works as i thought it would. On the comments @cogoid complemented your explanation, i think that should be pinned. Once again, thank you very much! 👍😃
Hi,
I guess that using transistor breakdown instead of zener is for speed. Seen many IC's with two transistors, each having their collectors shorted to base, and using the Vebo breakdown (typically 5-6v) as back to back zeners for input protection. The reason is the much higher speed of the transistor breakdown than zeners, so very fast spikes can be clamped.
the "unused" FETs on U101 (pin 42) are forming a current mirror, with current set by R207, so probably part of calibration. also, the video didn't explain why there's two current sources- why is the first one needed when you could just provide a reference voltage directly? having two in series seems like it'd introduce more errors. maybe the reason is process variations on the chips? if you used mechanical switches, the contacts generally have a known small resistance, but using FETs on a chip will have a much higher (10's of ohms or more) and worse, variable resistance that varies some chip to chip. I am curious what the zener diodes' purposes are- is it to reduce the Vdrop across resistors which would improve their temperature coefficient by reducing self heating? (re: CR201 and CR203).
If you provide a fixed reference voltage to the second current source then the temperature coefficient of the resistors over there cause the current to change. In this case the resistor which sets the 5V reference also changes to compensate for it. You can look at the second stage as a temperature independent scalable current mirror where the accuracy is determined only by the first reference current generating circuit.
@@robertvandersanden Also it allowed the voltage reference to be raised closer to the positive rail so the circuit could be used to source instead of sink current.
Even though I know what I am doing and was trying to follow along, this fairly new (for this channel) business of using the blotchy red 'electronic' markups left me cold. Often hard to read what was being written. Much more frequently, the narration says "now this thing over here" (etc) is not accompanied by any discernible red marks, so I am thinking, "OK, what is he talking about?". Sometimes the old ways are the best, and videoing some pencil marks being made real-time on a schematic work better.
The pointing device was not visible most of the time and part identifies were not readable which made following your explanation difficult to follow.
I think D211, the odd FET in the bipolar chain on the output, is used as a very low leakage current diode.
Leakage currents are an error, and I've seen FETs used like a diode in sample-and-hold circuits.
Thank you for the explanation. What always confuses me at this kind of complex analog circuits is the question if this wouldn’t be smarter to look for an IC to do the same job. Aren’t there any super exact constant current source ICs available ?
not that I know of
That is a good answer and justifies the need of the circuit. Thanks !🙏
cool, circuit.
Would be nice if cursor is recorded too, anyway interesting video.
Rather elegant .... typical of older HP / Agilent / Keysight designs ...
Cool!
This interesting circuit drew my attention. I started to watch the video but it became difficult to follow without some kind of visible pointer on screen to identify the components you refer to in the verbal description.
this was great content and i thank you
many time HERE was used and not referenced on screen
i still learned alot but got lost many times on the word here.
Something in your pointer & display system is not working. Consequently, verbal reference to circuit locations is meaningless. Nice topic, though.
it is called a cascode circuit.
16:15 I don't follow this logic and how all 8 transistors oppose to breakdown? By Kirchhoff's law breakdown voltage between Qe(203) and Qc(209) is 4 times Vceo (if all top PNPs are the same).
When positive over-voltage is applied CR202 will block it. Q211 functions as a precision diode. When negative voltage is applied, Q211 will drive base of Q204 to ground and the whole transistor network will behave like an active load, limiting current and increasing its resistance to oppose increasing voltage. Voltage on the emitter of Q203 will be aroud 2Vbe (+Vgs). At extreme negative over-voltage 4 transistors on top, or in the middle will just give up and short, one by one. My two cents.
Q221 - используется как диод с очень низким уровнем утечки. Т.к. у этих полевых транзисторов очень узкий канал, а следовательно и очень маленькая площадь PN-перехода, то и утечка через него будет настолько мала, что её можно не учитывать (примерно 150-300 фА)
Without the pointer made it extra hard for me so I bailed ! but I got a lot of fun worrying about how daft I am.....cheers.
Thank you for comprehensive explanation..
Thank you, PLEASE EVPLAIN MORE ABOIUT LOW VOLTAGE AMPLIFIER AMPLIFIERS. thanks, don
7v/40k*28.57k is exactly 5v. i dont understand where that 5.14v comes from
two more than PI? fat fingers?
HP-3456A RMS converter next, please ;-)
Their schematic symbol for a switch is absolutely for shit. You can't get any sense of whether it's a spdt or if any switch is connected or correlated in any way.
Sorry, but I keep hearing acrossT when my ears expect simply “across” (without the “T”). Other than that, thanks for the explanation for something I did not even realize built so complex!
Interesting because I hear the extra 'T' all the time and its more prevalent in USA than in UK and I don't get it !! its ACROSS not ACROSST !! LOL :)
Well the explanation of the protection circuit is somewhat flawed, but luckily other commenters have already set things straight.
Very hard to follow without a pointer. Please edit and repost.
The beta of the transistor in the current source is irrelevant so long as the op amp is working. Thus, replacing the transistor with a FET is meaningless in this case. Think about it a minute. What is important is the offset voltage of the op amp.
The load resistor is carrying both base and collector current. This could be corrected for if the beta of the transistor is constant, but real world transistors have changing beta with temperature.
Opamp offset voltage should be the lesser error with a good opamp, such small offset voltage being needed to produce base (or gate) voltage.
@@hanelyp1 The beta of the transistor is IRRELEVANT so long as the op amp is working. Remember, the op amp will drive the output to turn on the transistor to whatever level necessary to keep the voltage on + and - inputs the same.
@@glasslinger placing the combined transistor drive and output current on the sense resistor. Which if you don't account for the drive current produces an error in the output.
@@hanelyp1 Please get a book on op amp theory and read it! More important, understand what you are reading!
That is fed back to the opamp for correction.@@hanelyp1