14:20 to 14:50 showing your use of copper clad board to make circuit platform good teachable moment. Thanks for posting. On my next build, I will incorporate that technique.
Cool design. FYI: you may want to individually fuse each of the pass transistors to prevent a cascading failure. So if one of the transistors or emitter resistors fails, the rest of the transistors/resistors don't get damaged when they try to make up for the extra current.
I haven't even gotten through the video and this thing is excellent I've been wanting someone to show me how to make a better DC to DC power supply so thank you so much this video is excellent keep going I'm hoping I can drop a couple questions to you and get an answer all thumbs up.
Nice video!!!! I've made a linear PSU using one l7805 regulator to supply the opamp with the Reference voltage instead the combination of resistor and zenner. I used this because I think I will be more stable. I'm a diy guy and all I know Is from self learning, so maybe I'm doing a very bad thing going this way... For now it's working well powering 4 tda7498 amps on my DIY multichannel poweramp. Keep the good work!
Thanks. Using a proper regulator IC is really the best way to go, as you're doing. This project was to reinforce the previous video on zener diode theory. Eventually I'll design a board using a modern regulator to drive the pass transistors - similar to what you're doing. The good thing about a chip like the 7805 is that it already has a boatload of protection built in already. It's been around for a long time, but does what its supposed to. -Derek
@@AmRadPodcast glad to know that I'm going to the right path :). The opamp I'm using is the lm741 and it is driving the same power transistor you are using, but I ain't using the other transistors, I don't have the need of that amount of current, or I should always use some kind of output array like you did? Keep the good work. Looking forward to new videos about this psu
@@ruifaias8258 I think as long as your power transistor isn't dissipating too much power, you should be good and don't need an absurd array of transistors as I'm doing. The TDA7498 being a class D amp is very efficient and probably doesn't require all that much current. -Derek
You are using a transistor for the output of the op amp. That is better than using a diode. I forget why, but I know that you are supposed to use a transistor. I learned that from an electronics engineer with the TH-cam channel called All About Electronics. You are very smart. Are you an engineer? I am only an electronics technician, but I love studying EE and building my own electronics test instruments from schematics and redesigning them a little.
zeners are noisy and suffer from thermal drift .. functional for brute force type supplies but not so much for precision.. also .. especially when pulling such hi current, you will definitely want to explore power factor correction .. along with the mov's on the input side ... without such considerations you will generate huge amounts of heat and a ton of noise in your house power circuits.. HTH .. cheers :)
Hi Tim. Totally agree. The zener in this application is really meant as a companion to the previous video on how zeners work. In the (very) near future the prototype board will be replaced with a more stable design around a "modern day" regulator IC. MOV's will have some kind of bypass circuit once they've done their job with initial inrush current (though even bypass circuits can be unreliable). I get a bit uncomfortable knowing how hot they get during high current demands. -Derek
One thing I've wanted to see in linear supplies is a readout for how much power the pass elements are dissipating. I know it can be computed, but I'd like to see a display. That'd let me know how badly I was abusing a supply.
Hey, Jim! I guess you could use an ADC, monitor voltage across a current shunt, and voltage across the C-E junction.. then do the calculation in a microcontroller. PWM output could drive an analog meter. I don't think I'll bother with this supply.. the heatsink barely gets warm at 20A.
nice design. but if you would use 431 instead you would be better off.. No need to worry about Opamp at all.. Simple 431 and LM317 and also those transistors..and you can get much better and what is important much more stable output voltage.. Also i would suggest to use simple diode with lover voltage and ofc resistor parralel to circuit to reduce influence on input voltage change of your power supply to reduce voltage deviation ..so that would make your 431 much more stable ..even it can do well without it. Simple 431 or alternative 1431 voltage regulator and LM317 to drive few power transistors that would improve your design much more and no need to use opamp at all
An interesting add on would be a buck converter in parallel with the linear regulator. Then add a control circuit that determines the minimum load needed for the buck converter to operate. Now you get low noise at low currents and less heat (better efficiency and more power) at high currents, with the switchover point controlled by the user.
HI, nice video. i have a question, it seems that your circuit diagram shows not what you building on the bread board. Please correct me if i´m false, but in your diagram the kollektor of the darlington is conected to the bases of the other transistors and on your breadboard it is connected to V+.
Hi Julian. You are correct, on the schematic I mistakenly left out the resistor from the darlington's collector to the positive rail (this is also true on the breadboard circuit). It is physically mounted to the power transistors on the external heatsink. It is also difficult to see the collector resistor in the video, which is somewhat hidden underneath the power/ballast resistors on the heatsink. -Derek
Where does the current on the base of the main transistors come from? The TIP120 only drains the current on the bases. Something must source current to them.
My apologies sir. I omitted resistors from the schematic, and the discussion, which source current to the darlington. These are mounted on the heatsink side.
So, you do electronics from RF frequencies into frequencies of the visible spectrum of light. Most people, i think, just think of electronics dealing with RF frequencies. They forget how electronics span microwave frequencies (magnetron, klystron, semiconductor gun diode), infrared (infrared LEDs as in TV remote controls) and light (optoelectric couplers and LEDs). Frank
Is there a reason you didn't use SiC FET's in lieu of the BJT's? It'd eliminate the need for the Darlington control transistor by removing the need for the large base current (x5) for one... Just a thought from a fellow eEngineer. :)
I'm partial to older technology (just look at the ancient test equipment on my bench!), and I enjoy poking around at designs from back in the day. This one is a great example. That being said, there's no reason you couldn't use a FET - I just have stacks of various BJTs lying around. Might be a good idea to revisit this one with SiC FETs, as I've never used them. -Derek
@@AmRadPodcast Me too (partial to old as I'm 62 :). I am a converted BJT guy though and have gone to FET's almost exclusively. The SiC FET's are new to me too, but saw your BJT's, instantly thought FET's. I have just recently read up on the SiC technology and what you are doing is the perfect niche. Having had to use a BJT (or Darlington) to drive power BJT transistors was something I really disliked... That's all. I wasn't putting it down at all... just curious. If you like, I can find the material I was looking at on the SiC transistors and get you a link. BTW, I'd be right at home in your shop, and I can see you'd be right at home in mine too! :)
Perhaps this is a dumb question but I'm curious, why did you use a darlington transistor versus a MOSFET or other for switching the pass transistors? I don't really have much experience with darlington transistors.
The short answer is, I have more experience with bipolar junction transistors and have stacks of them. There's no reason you couldn't, or shouldn't use a FET. The darlington is really just two transistors on the same die, and multiplies the transistor's current gain, lessening the load on the op-amp. -Derek
Exactly what I was building.. And same vdrop as you.. Too bad I didn't found an answer in your video. What's that method of pcb you used? These were copper pads glued to the pcb?
I bought a larger, single toroidal transformer at 500VA to replace the two in parallel. This will get me closer to 30A - not quite, but close enough at around 27A. For quick prototyping I cut little double sided PCB blanks and superglue them to the main board. Great for making RF stuff on a ground plane, but both surfaces should be cleaned, and preferably roughed up with scotchbrite or they'll pop off.
i am confused your circuit diagram doesn't show can you drive npn pass transistors with npn tip120 without supplying current to the pass transistors base. npn can only sink current from the base of the pass transistors. shouldn't there be a pnp transistor supplying current to the pass transistors...
A TL431 would be a much better choice and would only cost like $0.30 more... Also no current liming (even simple "short-circuit protection") is extremely risky and although you may understand the risks and accept them, people who try to replicate this may learn an expensive lesson... Thanks for the video!
When paralleling BJTs, none of them are perfectly matched as far as current gain goes. One transistor will carry more current than the rest and heat up; this leads to an increase in current gain (beta), which causes it to heat up more.. so on and so forth. This is called thermal runaway. Short story is they force the transistors to share the current distribution equally. -Derek
It's not the zener that failed. You're not providing enough base drive for the pass transistors. Worst case calculation is 300mA each. The schematic shown actually doesn't turn on the pass transistors it only turns them off. Also you didn't account for the voltage drop in the cabling, it can be significant at those currents..
hm, where does any positive potential driving the bases off the pass transistors come from? i guess the TIP is supposed to be operated as a common emitter ... in that case there is probably a resistor missing from the collector of the TIP120 to the positive rail? But then the feedback would be positive because the TIP120 would cause a phase shift of 180 degrees and the + / - Terminals of the opamp would effectily "flip" But thinking about it again ... i think the TIP 120's Base-Collector junction is operated as a diode in this case, which would make the circuit work ... but the TIP120 would be basically useless and the opamp would have to able to provide the neccessary base currents alone.
I mistakenly left the collector resistor out of the schematic when I "finalized" the drawing in KiCAD.. the resistors are physically mounted to the pass transistors on the heatsink, hidden under the ballast resistors. If/when I do a follow up, I'll more clearly explain this point. I think it caused a lot of confusion here in the comments! -Derek
Good question. Once I've added in the current/thermal/overvoltage protection, it'll power my ham radio equipment. Most (if not all) transceivers are designed to run at 13.8V - typical automotive voltage seen when the car/truck is running and the alternator is doing its job. Some modern radios act funny if you give them only 12 volts.
Electronics is a fascinating subject but with wrong schematics and wrong concepts it's hard to teach others. Probably this circuit made no sense to you but you're not wrong. The circuit is wrong and does not work. What was presented was not the circuit used on the demonstration.
The circuit presented has a positive feedback and nothing to provide current to the transistor bases. It's unstable. Try to simulate the circuit: it does not work! Do you guys know ANYTHING on circuit analysis?! If the output voltage increases, the feedback voltage increases, which makes the output of the opamp decrease, which makes the TIP drain less current from the bases, which will make (if there was a resistor from Vcc to the bases) the voltage on the bases increase, which will increase more the output voltage. If the main transistors were PNP transistors it may work despite having no current limit protection on the TIP.
@@vioariton8510 I think it will be easier to describe it. Freeze the image at 8:55 of the video. If Vout increases, negative feedback voltage will increase also, right? Because it's just a resistor voltage divider from Vout to gnd. Let's consider that the voltage on pin 3 of the opamp (the "positive input") is constant for this analysis (it's a reasonable assumption for now). On the opamp side, if the "positive input*" go up, the output will go up. If the "negative input*" go up, the output will go down, ok? So if we assume that the "positive input" is fixed, the output will vary in a inverse way of the feedback (if it go up the out goes down and vice versa). Therefore the output of the opamp changes in the inverse way as the Vout as the feedback voltage is proportional to Vout, right? As the output of the opamp is connected on the base of the transistor Q1 (TIP120 - we can assume for now that this darlington transistors act as a single NPN transistor with a very high gain), if the output goes high Q1 conducts more and if it goes down it conduct less current, ok? So linking all together until now, if Vout increases, feedback increases, opamp output decreases and current on the collector of Q1 decreases, right? Keep that in hold. The problem on the circuit is that, as it's presented, there is no path to the current from the positive input (Vi) to reach the base of the main transistors. You need a resistor from Vi to the connected bases of the transistors to provide a path to the current to reach the base of them. It is missing on the schematic. According to a comment from "The Current Source", they omitted the resistors from the schematic. In this case, a resistor will provide current from Vi to the bases of the transistors and they will conduct Vi to the output. If Q1 (the TIP120) does not drain current from the bases, the current is max and Vout is max, right? So, to reduce Vout, Q1 must drain part of the current to give this missing resistor a voltage drop and the base be at a lower voltage, leading to a lower voltage on the emitters and therefore on the Vout, right? So, increasing current on Q1 will lead to a decrease on Vout, ok? And a decrease on Q1 collector current will lead to a increase on Vout. But, remember that an increase in Vout leads to a decrease on Q1 collector current? And if it decreases, it will lead to an increase on Vout, as we saw. So, an increase in Vout will eventually lead to more increase in Vout! And that increase will lead to more increase and so on until it reaches the maximum voltage. See? It diverges from the setpoint. The same reasoning can be done to the inverse direction. To correct that you must invert that logic at some point. One possible solution is not to use the resistor to provide current to the transistors bases but instead use the TIP120 itself to take current from Vi and deliver it to the bases of the trasistors! Connect the collector of Q1 to Vi, keep the base at the opamp output and the emitter to the bases of the transistors. Let's see if it works? An increase in Vout will lead to a decrease at the current conducted by the TIP120, right? We saw that. If the base current of the transistors (the same current that passes through the TIP120) decreases, the Vout decreases as well. So an increase in Vout will eventually lead to a decrease in Vout. THAT is static stability of a system! It shows that a dynamic system is statically stable at some point, and that is what we want: to keep Vout stable at a certain value. Got it? This config can have some instabilities like oscillations but it is a more complex discussion. You can learn on youtube about simulations using LTspice (it's free). I used it to simulate this circuit and with this modification (put Q1 as I described) solved the issue and the circuit is stable. I hope it helps you and others. If there is any doubt left there is no problem to ask. *The "positive input" of the opamp is the non-inverting input and the "negative input" is the inverting input. Those are the correct way to call them.
How are you even supposed to drive the output transistors with that topology? Not only that, but with the TIP 120 with its emitter directly connected to ground, the output of the LM324 is clamped to ground + 1.4V. Okay, to be fair, in the description you said that the collector of the TIP121 is connected to the power rail and the output (you probably mean emitter) is connected to the base of the output transistors... That's more like it... but is definitely not what you have in the schematic.
Linear supplies are inherently inefficient. I'd be surprised if it was 50%, but something worth measuring. I'll be using it to power a transceiver and doing morse code.. which during transmit means large transients, and during receive it's good to have an electrically quiet power source. Short answer, efficiency is nowhere close to a switch mode supply. -Derek
Wait a minute, your schematic doesn't match the circuit. In the schematic tip120 is connected as common emitter amp, ain't gonna work as shown in the schematic. In the real deal, tip120 is connected as a common collector amp, or emitter follower; works fine.
OMG, what a breath of fresh air to see electronics that are not just a different way how to box a rpi! A bit more of these kind of videos, please!
Thanks. No guarantees that an RPi won't make its way into some of the projects ;) I think it's more fun to use discrete components to the job! -Derek
14:20 to 14:50 showing your use of copper clad board to make circuit platform good teachable moment. Thanks for posting. On my next build, I will incorporate that technique.
Cool design. FYI: you may want to individually fuse each of the pass transistors to prevent a cascading failure. So if one of the transistors or emitter resistors fails, the rest of the transistors/resistors don't get damaged when they try to make up for the extra current.
I haven't even gotten through the video and this thing is excellent I've been wanting someone to show me how to make a better DC to DC power supply so thank you so much this video is excellent keep going I'm hoping I can drop a couple questions to you and get an answer all thumbs up.
Nice video!!!!
I've made a linear PSU using one l7805 regulator to supply the opamp with the Reference voltage instead the combination of resistor and zenner. I used this because I think I will be more stable. I'm a diy guy and all I know Is from self learning, so maybe I'm doing a very bad thing going this way...
For now it's working well powering 4 tda7498 amps on my DIY multichannel poweramp.
Keep the good work!
Thanks. Using a proper regulator IC is really the best way to go, as you're doing. This project was to reinforce the previous video on zener diode theory. Eventually I'll design a board using a modern regulator to drive the pass transistors - similar to what you're doing. The good thing about a chip like the 7805 is that it already has a boatload of protection built in already. It's been around for a long time, but does what its supposed to. -Derek
@@AmRadPodcast glad to know that I'm going to the right path :).
The opamp I'm using is the lm741 and it is driving the same power transistor you are using, but I ain't using the other transistors, I don't have the need of that amount of current, or I should always use some kind of output array like you did?
Keep the good work.
Looking forward to new videos about this psu
@@ruifaias8258 I think as long as your power transistor isn't dissipating too much power, you should be good and don't need an absurd array of transistors as I'm doing. The TDA7498 being a class D amp is very efficient and probably doesn't require all that much current. -Derek
You are using a transistor for the output of the op amp. That is better than using a diode. I forget why, but I know that you are supposed to use a transistor. I learned that from an electronics engineer with the TH-cam channel called All About Electronics. You are very smart. Are you an engineer? I am only an electronics technician, but I love studying EE and building my own electronics test instruments from schematics and redesigning them a little.
zeners are noisy and suffer from thermal drift .. functional for brute force type supplies but not so much for precision.. also .. especially when pulling such hi current, you will definitely want to explore power factor correction .. along with the mov's on the input side ... without such considerations you will generate huge amounts of heat and a ton of noise in your house power circuits.. HTH .. cheers :)
Hi Tim. Totally agree. The zener in this application is really meant as a companion to the previous video on how zeners work. In the (very) near future the prototype board will be replaced with a more stable design around a "modern day" regulator IC. MOV's will have some kind of bypass circuit once they've done their job with initial inrush current (though even bypass circuits can be unreliable). I get a bit uncomfortable knowing how hot they get during high current demands. -Derek
I actually managed to follow some of that. Even more when I replayed bits. Thank you
14:20 Redefining surface mount components, lol.
One thing I've wanted to see in linear supplies is a readout for how much power the pass elements are dissipating. I know it can be computed, but I'd like to see a display. That'd let me know how badly I was abusing a supply.
Hey, Jim! I guess you could use an ADC, monitor voltage across a current shunt, and voltage across the C-E junction.. then do the calculation in a microcontroller. PWM output could drive an analog meter. I don't think I'll bother with this supply.. the heatsink barely gets warm at 20A.
Awesome💥
nice design. but if you would use 431 instead you would be better off.. No need to worry about Opamp at all.. Simple 431 and LM317 and also those transistors..and you can get much better and what is important much more stable output voltage.. Also i would suggest to use simple diode with lover voltage and ofc resistor parralel to circuit to reduce influence on input voltage change of your power supply to reduce voltage deviation ..so that would make your 431 much more stable ..even it can do well without it. Simple 431 or alternative 1431 voltage regulator and LM317 to drive few power transistors that would improve your design much more and no need to use opamp at all
Excellent explanation.
I wonder if capacitance multiplier could have been used here.
Good video i regret a scope to watch the output waveform with heavy load
An interesting add on would be a buck converter in parallel with the linear regulator. Then add a control circuit that determines the minimum load needed for the buck converter to operate. Now you get low noise at low currents and less heat (better efficiency and more power) at high currents, with the switchover point controlled by the user.
Interesting idea. The transceivers that this will be powering require a couple of amps idle, and at max around 25A during transmit.
Thanks for the explanation
HI, nice video. i have a question, it seems that your circuit diagram shows not what you building on the bread board. Please correct me if i´m false, but in your diagram the kollektor of the darlington is conected to the bases of the other transistors and on your breadboard it is connected to V+.
Hi Julian. You are correct, on the schematic I mistakenly left out the resistor from the darlington's collector to the positive rail (this is also true on the breadboard circuit). It is physically mounted to the power transistors on the external heatsink. It is also difficult to see the collector resistor in the video, which is somewhat hidden underneath the power/ballast resistors on the heatsink. -Derek
i wonder why your feedback voltage divider equation is not Vout = Vz(1 + R2/R1 + R2)?
Nice!
Where does the current on the base of the main transistors come from? The TIP120 only drains the current on the bases. Something must source current to them.
My apologies sir. I omitted resistors from the schematic, and the discussion, which source current to the darlington. These are mounted on the heatsink side.
@@AmRadPodcast But the circuit as presented even with the resistors is unstable. What else is wrong or missing?
So, you do electronics from RF frequencies into frequencies of the visible spectrum of light. Most people, i think, just think of electronics dealing with RF frequencies. They forget how electronics span microwave frequencies (magnetron, klystron, semiconductor gun diode), infrared (infrared LEDs as in TV remote controls) and light (optoelectric couplers and LEDs).
Frank
Is there a reason you didn't use SiC FET's in lieu of the BJT's? It'd eliminate the need for the Darlington control transistor by removing the need for the large base current (x5) for one... Just a thought from a fellow eEngineer. :)
I'm partial to older technology (just look at the ancient test equipment on my bench!), and I enjoy poking around at designs from back in the day. This one is a great example. That being said, there's no reason you couldn't use a FET - I just have stacks of various BJTs lying around. Might be a good idea to revisit this one with SiC FETs, as I've never used them. -Derek
@@AmRadPodcast Me too (partial to old as I'm 62 :). I am a converted BJT guy though and have gone to FET's almost exclusively. The SiC FET's are new to me too, but saw your BJT's, instantly thought FET's. I have just recently read up on the SiC technology and what you are doing is the perfect niche. Having had to use a BJT (or Darlington) to drive power BJT transistors was something I really disliked... That's all. I wasn't putting it down at all... just curious. If you like, I can find the material I was looking at on the SiC transistors and get you a link. BTW, I'd be right at home in your shop, and I can see you'd be right at home in mine too! :)
Perhaps this is a dumb question but I'm curious, why did you use a darlington transistor versus a MOSFET or other for switching the pass transistors? I don't really have much experience with darlington transistors.
The short answer is, I have more experience with bipolar junction transistors and have stacks of them. There's no reason you couldn't, or shouldn't use a FET. The darlington is really just two transistors on the same die, and multiplies the transistor's current gain, lessening the load on the op-amp. -Derek
@@AmRadPodcast OK. Thanks for taking the time to explain that! I'm going to read up on darlington transistors.
Exactly what I was building.. And same vdrop as you..
Too bad I didn't found an answer in your video.
What's that method of pcb you used? These were copper pads glued to the pcb?
I bought a larger, single toroidal transformer at 500VA to replace the two in parallel. This will get me closer to 30A - not quite, but close enough at around 27A. For quick prototyping I cut little double sided PCB blanks and superglue them to the main board. Great for making RF stuff on a ground plane, but both surfaces should be cleaned, and preferably roughed up with scotchbrite or they'll pop off.
Really like that watch, what is it?
i am confused your circuit diagram doesn't show can you drive npn pass transistors with npn tip120 without supplying current to the pass transistors base. npn can only sink current from the base of the pass transistors. shouldn't there be a pnp transistor supplying current to the pass transistors...
A TL431 would be a much better choice and would only cost like $0.30 more... Also no current liming (even simple "short-circuit protection") is extremely risky and although you may understand the risks and accept them, people who try to replicate this may learn an expensive lesson... Thanks for the video!
Any video on linear induction motor and it's control system?
Can you talk more about the purpose of those ballast resistors?
When paralleling BJTs, none of them are perfectly matched as far as current gain goes. One transistor will carry more current than the rest and heat up; this leads to an increase in current gain (beta), which causes it to heat up more.. so on and so forth. This is called thermal runaway. Short story is they force the transistors to share the current distribution equally. -Derek
It's not the zener that failed. You're not providing enough base drive for the pass transistors. Worst case calculation is 300mA each. The schematic shown actually doesn't turn on the pass transistors it only turns them off. Also you didn't account for the voltage drop in the cabling, it can be significant at those currents..
hm, where does any positive potential driving the bases off the pass transistors come from?
i guess the TIP is supposed to be operated as a common emitter ... in that case there is probably a resistor missing from the collector of the TIP120 to the positive rail?
But then the feedback would be positive because the TIP120 would cause a phase shift of 180 degrees and the + / - Terminals of the opamp would effectily "flip"
But thinking about it again ... i think the TIP 120's Base-Collector junction is operated as a diode in this case, which would make the circuit work ... but the TIP120 would be basically useless and the opamp would have to able to provide the neccessary base currents alone.
I mistakenly left the collector resistor out of the schematic when I "finalized" the drawing in KiCAD.. the resistors are physically mounted to the pass transistors on the heatsink, hidden under the ballast resistors. If/when I do a follow up, I'll more clearly explain this point. I think it caused a lot of confusion here in the comments! -Derek
@@AmRadPodcast cool, thank's
Does this design work for high voltage power supplies?
Why 13,8V?
Why this odd voltage?
Good question. Once I've added in the current/thermal/overvoltage protection, it'll power my ham radio equipment. Most (if not all) transceivers are designed to run at 13.8V - typical automotive voltage seen when the car/truck is running and the alternator is doing its job. Some modern radios act funny if you give them only 12 volts.
I love electronics but still need more learning 😔
Electronics is a fascinating subject but with wrong schematics and wrong concepts it's hard to teach others. Probably this circuit made no sense to you but you're not wrong. The circuit is wrong and does not work. What was presented was not the circuit used on the demonstration.
Please join the element14 community if you haven't already! Links are up in the description - we'd love to have you!
The circuit presented has a positive feedback and nothing to provide current to the transistor bases. It's unstable. Try to simulate the circuit: it does not work! Do you guys know ANYTHING on circuit analysis?!
If the output voltage increases, the feedback voltage increases, which makes the output of the opamp decrease, which makes the TIP drain less current from the bases, which will make (if there was a resistor from Vcc to the bases) the voltage on the bases increase, which will increase more the output voltage.
If the main transistors were PNP transistors it may work despite having no current limit protection on the TIP.
@@vioariton8510 I think it will be easier to describe it. Freeze the image at 8:55 of the video.
If Vout increases, negative feedback voltage will increase also, right? Because it's just a resistor voltage divider from Vout to gnd. Let's consider that the voltage on pin 3 of the opamp (the "positive input") is constant for this analysis (it's a reasonable assumption for now).
On the opamp side, if the "positive input*" go up, the output will go up. If the "negative input*" go up, the output will go down, ok? So if we assume that the "positive input" is fixed, the output will vary in a inverse way of the feedback (if it go up the out goes down and vice versa). Therefore the output of the opamp changes in the inverse way as the Vout as the feedback voltage is proportional to Vout, right? As the output of the opamp is connected on the base of the transistor Q1 (TIP120 - we can assume for now that this darlington transistors act as a single NPN transistor with a very high gain), if the output goes high Q1 conducts more and if it goes down it conduct less current, ok?
So linking all together until now, if Vout increases, feedback increases, opamp output decreases and current on the collector of Q1 decreases, right? Keep that in hold.
The problem on the circuit is that, as it's presented, there is no path to the current from the positive input (Vi) to reach the base of the main transistors. You need a resistor from Vi to the connected bases of the transistors to provide a path to the current to reach the base of them. It is missing on the schematic. According to a comment from "The Current Source", they omitted the resistors from the schematic. In this case, a resistor will provide current from Vi to the bases of the transistors and they will conduct Vi to the output. If Q1 (the TIP120) does not drain current from the bases, the current is max and Vout is max, right? So, to reduce Vout, Q1 must drain part of the current to give this missing resistor a voltage drop and the base be at a lower voltage, leading to a lower voltage on the emitters and therefore on the Vout, right? So, increasing current on Q1 will lead to a decrease on Vout, ok? And a decrease on Q1 collector current will lead to a increase on Vout.
But, remember that an increase in Vout leads to a decrease on Q1 collector current? And if it decreases, it will lead to an increase on Vout, as we saw. So, an increase in Vout will eventually lead to more increase in Vout! And that increase will lead to more increase and so on until it reaches the maximum voltage. See? It diverges from the setpoint. The same reasoning can be done to the inverse direction.
To correct that you must invert that logic at some point. One possible solution is not to use the resistor to provide current to the transistors bases but instead use the TIP120 itself to take current from Vi and deliver it to the bases of the trasistors! Connect the collector of Q1 to Vi, keep the base at the opamp output and the emitter to the bases of the transistors. Let's see if it works?
An increase in Vout will lead to a decrease at the current conducted by the TIP120, right? We saw that. If the base current of the transistors (the same current that passes through the TIP120) decreases, the Vout decreases as well. So an increase in Vout will eventually lead to a decrease in Vout. THAT is static stability of a system! It shows that a dynamic system is statically stable at some point, and that is what we want: to keep Vout stable at a certain value. Got it?
This config can have some instabilities like oscillations but it is a more complex discussion. You can learn on youtube about simulations using LTspice (it's free). I used it to simulate this circuit and with this modification (put Q1 as I described) solved the issue and the circuit is stable.
I hope it helps you and others. If there is any doubt left there is no problem to ask.
*The "positive input" of the opamp is the non-inverting input and the "negative input" is the inverting input. Those are the correct way to call them.
that was my first though... like hmmm, wait a second, this will be a quite unstable and there is nothing to compensate it
How are you even supposed to drive the output transistors with that topology? Not only that, but with the TIP 120 with its emitter directly connected to ground, the output of the LM324 is clamped to ground + 1.4V. Okay, to be fair, in the description you said that the collector of the TIP121 is connected to the power rail and the output (you probably mean emitter) is connected to the base of the output transistors...
That's more like it... but is definitely not what you have in the schematic.
Yep. There's a mistake in the schematic. The darlington collector resistor is mounted on the heatsink, and I left it out of the schematic.
What is it efficiency?
Linear supplies are inherently inefficient. I'd be surprised if it was 50%, but something worth measuring. I'll be using it to power a transceiver and doing morse code.. which during transmit means large transients, and during receive it's good to have an electrically quiet power source. Short answer, efficiency is nowhere close to a switch mode supply. -Derek
Wait a minute, your schematic doesn't match the circuit. In the schematic tip120 is connected as common emitter amp, ain't gonna work as shown in the schematic. In the real deal, tip120 is connected as a common collector amp, or emitter follower; works fine.