Power transistors on linear bench power supply - Page 1
Forum section: electronics / beginners

author: TomCrow
author_rank: Newbie
timestamp: May 05, 2021, 11:04:40 am
content: Hello,Iam newbie when it comes to electronics. Just got my hands on APS 3005D chinese cheap power supply, which doesnt work. It doesnt output any voltage and its always in CC mode. So I begin learning - until that moment I didnt know anything about electricity, voltage, current, etc..Now I got to this point where I looked into many many similiar bench psu schematics with same principles.One thing that I couldnt get was how the rectified voltage is regulated with the operation amplifiers. If you look on this video:at 28:55 he says that these transistors that pass the voltage to output via the power transistors, they work in ON / OFF mode.So what I got from that is if the current control op-amp or the voltage control op-amp results in negative voltage, they will redirect all the current, that was ment to be for the bases of the transistors, to the ground (and lit its LED on its way).But then there wouldnt be any voltage at all at the output of the psu.Then I come around of this video:Where he says that as I thought the transistors cannot work only in ON / OFF mode in this case.I think I know that these controlling op-amps are just comparing feedback with some reference values, but I dont know how the output is regulated both in voltage or current.Could anyone point me in the right direction please?Thank you.

author: Zero999
author_rank: Super Contributor
timestamp: May 05, 2021, 11:47:02 am
content: Firstly, is it in warranty? Was it sold to you as working? If so, don't bother trying to repair it, just return it! Sellers shouldn't be allowed to get away with pushing faulty products like this!To answer the question. In a linear power supply, the output transistors are biased in the active region, so are neither fully on, nor off. Power is dissipated across the transistors, equal to the difference between the input and output voltage, multiplied by the current. On power supplies with a higher output voltage than around 15V, it's quite common for the transformer to be tapped, so the output transistor doesn't have to dissipate the full voltage, at lower output voltage settings.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 05, 2021, 11:57:07 am
content: The first video is not really accurate and more misleading than helping in many aspects. I have not whatched much but that 2 minutes around 19:00 have several mistakes. The nice point is that they seem to have the diagrams.

author: TomCrow
author_rank: Newbie
timestamp: May 05, 2021, 12:09:45 pm
content: Thank you, so the first video was what mislead my learning.So one last thing I dont get. I uploaded in attachments random similiar psu schematics.There are 3 power transistors and 2 control transistors that make sure current goes to the power transistors bases. These 2 transistors are powered by 12V rail before 2.4V zener diode.Then there are current and voltage controlling op-amps. Which can set the voltage on the rail after zener diode from 12V to -6V. If any of these op-amps set the voltage to less than ground (0V) the current wont go into the transistor base, but instead down to the op-amp and to the negative voltage and litting the LED on its way.So its still like >0V ON and <0V OFF to me. So how can than be CV led on with current going up to the transistors for psu to output voltage?Something is really missing in my knowledge.Thank you for any help.

author: xavier60
author_rank: Super Contributor
timestamp: May 05, 2021, 12:54:43 pm
content: The CV opamp needs to draw current through LED V9 so that the voltage at the Base of V13A can be controlled to control the current passed by the output transistors to achieve voltage regulation.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 05, 2021, 01:09:11 pm
content: The current to the point before the diodes is more than what is needed to drive the power transistors. So under normal conditions part of the current goes to 1 of the OPs and thus one LED on and only the rest drives the transistor. The OPs finds a delicate balance to that the transitor is not all the way on, but also no all the way of.

author: TomCrow
author_rank: Newbie
timestamp: May 05, 2021, 01:15:59 pm
content: Thank you all for answers. Its a bit confusing for me.Since no current goes backwards the led diodes so no current comes from the op-amp power source. Then only source of current is from the +12V through the zener diode.So the op-amp only sets the voltage for the current coming through the zener diode to the base of transistor?I didnt know current could go both to the base of transistor and to the ground through LED. Since i thought that if op-amp set the voltage a bit less than ground, it would take all the current and if it set the voltage a bit more than ground, no current would go through the LED, only to the base of transistor.

author: pqass
author_rank: Super Contributor
timestamp: May 05, 2021, 03:18:52 pm
content: That schematic is too cluttered and overly complex for a beginner to understand what's going on.A key idea is that the base of the NPN series pass transistor(s) is defaulted to fully-on.Then, the voltage and current amps shut it off enough to meet the voltage and current parameters set.Read the following article to understand the basics, then return to the schematic.http://powersupply.blogs.keysight.com/2012/07/how-does-power-supply-regulate-its.htmlSimplified schematic in CV mode:Simplified schematic in CC mode:

author: TomCrow
author_rank: Newbie
timestamp: May 05, 2021, 03:25:31 pm
content: Thank you very much. Ill look onto it as soon as I got some more time. After then if something still wont be clear to me, Ill let you guys know.

author: pqass
author_rank: Super Contributor
timestamp: May 05, 2021, 03:49:24 pm
content: Okay, I think I see how the series pass transisors are even defaulted to on.V13A,V14A,R6A,R7A form a darlington arrangement which is turned-on by default via R4A,V11A zener.The darlington arrangement directly controls current from +12V to the main V22,V21,V20 series pass transistors against R8A turning them off.The two op amps then turn off the darlington  transistors by lowering the voltage against the zener's weaker pull-up.I'm not sure why V11A zener and V12A diode are even necessary!Except of course the obvious error in the schematic; the broken line above R7A!!!And where does (P2) connected to the transistor bases go off to?Does it continue to (P2) at the lower right? To an inverting amp to trigger a relay? Whaaaat?Nevermind.One more thing...Notice the ground symbol (⏚) attached to the positive ouput terminal X2 (middle right).That may look bizarre to the beginner but what it means is that the entire power supply section above uses the positive output terminal as its zero reference point. Why? A:Because the series pass transistors are NPN and require a positive voltage on their base (with respect to their emitter) to turn on regardless where the voltage control knob is set to (possibly 0V output). In fact, you can think of this lab supply as having variable negative output; X2=0V, X4=0...minus limit.Also, ⏚ does not equal "GND" ground. The latter is earth ground and shouldn't be internally connected to ⏚; only by the user on the front panel if desired.

author: xavier60
author_rank: Super Contributor
timestamp: May 05, 2021, 10:59:33 pm
content: The zener V11A should be for Under Voltage Lockout protection to prevent the Darlington from being driven when the +12V control rail is too low for proper operation of the control circuitry. This avoids possible output surges during power up and power down.

author: TomCrow
author_rank: Newbie
timestamp: May 06, 2021, 07:52:44 am
content: Thank you guys. I almost got everything, but these is still some confusion for me and that is:- Op amps voltage varies from +12 to -6 right?- Current flows from positive to negative, and from +12 rail if it goes up to the darlington to the ground 0VNow what I imagine are these 3 working modes.1. Opamp voltage is bigger than 0V, so no current goes through the LED, but all current goes to the darlinghton and the psu outputs with no LED lit, which cant be right.2. Opamp voltage is lesser then 0V, so no current goes to the darlington which would be way to 0V ground, but all current is drawn to the opamp minus voltage, which lits up the LED, but psu has no output voltage, since no transformers are open.3. The only mode I think could be possible for current to flow both into the darlington and to the LED and opamp is if opamp voltage would be same as ground so 0V, but in that case the opamp wouldnt be able to change anything.If anyone could please explain to me why my thinking is wrong, I would be really thankful.

author: xavier60
author_rank: Super Contributor
timestamp: May 06, 2021, 11:20:08 am
content: Opamps output voltage signal, so it's best to think in terms of voltage up to the Base of V13A.The Darlington as a whole will start passing current when some threshold voltage at the Base of V13A is reached, increasing to full current over say another 500mV increase. The actual voltages don't matter because the action of the negative feedback loop causes the opamp to adjust the voltage at the Base of V13A to whatever is required that will cause the Darlington to pass the exact current to maintain the set voltage across the load.

author: ledtester
author_rank: Super Contributor
timestamp: May 06, 2021, 01:10:10 pm
content: QuoteThe op-amp can precisely control the current through the LED - it's not an all or nothing affair.The attached graph shows the typical current vs. voltage curve for a red LED. Even though the current increases dramatically between 1V and 2V, an op-amp is able to output any intermediate value to precisely set the current through the LED.

author: pqass
author_rank: Super Contributor
timestamp: May 06, 2021, 06:34:55 pm
content: Quote from: TomCrow on May 06, 2021, 07:52:44 amThe voltage and current op amps (N1 and N2) have supply voltages V+ = +12, V- = -6. However, since the op amps are 741's, at best, their output won't reach that; maybe ~1.5 below V+ and above V-; ie. +10.5V to -4.5V. Better op amps are "rail-to-rail" and will output close to their supply rails.QuoteThere is the default (or always on) flow: +12V through R4A (10K) through the base of first transistor of the darlington, through the second transistor, through the base of the series pass (main power) transistor, out its emitter, back through the ⏚ ground/center-tap of the +12V,0V,-6V,-12V opamp supply.Against the default flow, either op amp (N1 or N2) has the opportunity to lower the voltage at the base of the first darlington transistor until it is off which shuts off the next two transistors. Q:How does the op amp lower the voltage? A:By providing a lower resistance to its V- (-6V) rail. The op amp is like the lower resistor in a resistor divider network; +12V, R4A, base, op amp, -6V.QuoteIf both LEDs are off, then there is no regulation. The default flow (see my expanation above) will keep the main series pass transistor always on so the output voltage follows the voltage on the main filter capacitor less Vce transistor loss, R10A,R12A,R14A loses, and shunt (Rc) voltage drop; the output will be an unregulated ripple.QuoteTo shutoff a transistor, you just drive the base below about 650mV; the base-emitter is like a diode. So If there are 3 transistors between the first base and last emitter then we have 650mV*3=1.95V as the minimum. The op amp's output lower limit will go as low as it can such that its output plus the LED voltage drop will cause a 1.95V difference between the first base and final emitter to produce a 0V at the power supply output. Anything higher and we'll be increasing the power supply output voltage."...lights up the LED, but psu has no output voltage..." I see my note earlier about the meaning of ⏚ on the positive terminal hasn't clicked for you yet. It IS POSSIBLE to be -6V below the positive output terminal at the same time the difference between the positive and negative output terminals is 0V. ie. the negative terminal is above -6V!!!!QuoteI've simplified the power supply and simulated it.See attached image where the waveforms are: red, voltage across the main filter capacitor, green, voltage across the load.From left to right on the waveform:a. I start with NO LEDs lit (unregulated output), you see output follows input (ripple).b. then lowering the VOLTAGE slider (on right margin) until it begins regulation, voltage LED is lit, output is solid,c. then lowering the VOLTAGE slider down to zero and back up, you see a "U" dip in the output voltage, also notice ripple is less because less current (over fixed 15ohm load),d. then lower the CURRENT slider below 1A, the current LED is now lit, voltage LED is off, Q:why? A:because the voltage op amp output is greater than the current op amp output and one of the LEDs is now in reverse! It also means the op amps don't fight each other since the LEDs also provide "or" logic; whichever opamp output is lower drives the base.Play with it here: ...watch how the various volt meters change as you play with the VOLTAGE and CURRENT pots.http://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3AWAnC1b0DYQFZoO0gBxjZwDMZYCCATAjpDgzgKYC0YYAUAC4gKFGYAOxZsYGuFFRwINjWgYyGbGSTYMw5UkrDhURXHUCahJDQmFC+sHBAATFgDMAhgFcANj144J4czjSYAFCcmTQNGTCyAhqGDRIkBhJ9DAYRESciQiQpDSQSOB2jq6e3nzC+SD5hCDChPQ1MhBsCHj49ekodJC2kjB6GBiEQ2SChNi0SBBVJe5eXABOgVhNGDnVkLVCkEu+kjR0K9VHO3BcAO77J-TiksH9l8eHt9JNu1d34MnHYD8fvx+2CC0l2yy+f0YXyQWDOXAAStcYdd1jJGBs-IwsdBsHtKowenUGtUNmcAfVGhsKTcoE98STGPT3ntuiAyJB6KyNFjwLiooziQIifQHrIsVwAG7fdF-aU4f6yH40bBogy4+xyhCy9n0LWrBzOebeDXKsR+SH8WWSOZlLgai3c6qVfhC62G20anXyoRApUG0oLADGbI5lqwXsd2LgBwUI1ESCm+EgFCiBjg3Cu1KFXqFAJztRzst2AAcihsvbYNo6IOKrqbrhH-k8aM7qAXQ3raVcuT9K1Cm92AhWOYwvbt+fw7CIsAIhEE0QjJ4xkTlGKieRj+micXtVyAV1PQUV4FwAObCkP0alkKriifrAlbOpVJo12kuOod5GTXXI2p+NgsFseA7EYNgYBAyBhHEBM6GQZQrDoCBgJPEQr2wSRHWEDtBGaSCDBGJRIOIiAADUAHkABkABUAEEAHEAFEuD+SQ92QGdUQ46p-SNNgPBYe1VQg7gJwQLjv2ofhkTfAE9wQpcpFhLhSwoIDpDUtkqlkp5NOnRT9IBBpakJalCV2RERl1OwrLZYZVU3YSd2WWzxPoVzQLZdkuDIegAC8WAAOxYRZ5BxWQwGgYQqDIDDiHMYh4Bodh9AJWBiIy4kFBVbg-JAWyvlc1FJAC4LQuyp4ivQtj10qldUSqrs6gw719wCSM8SyqliTHcAVPy10NmMy8xSa4avVZMcfOw-hUQUn87P-BdWMUhS3MWnibS8fjBNGqNuDQ5q2JvT96FircUNA6AUFIDKQIgABhABVeF4UYgA5aiuA-bCzvshaFP-SRAOPEDwGgEgMMqMx1j+ARGhxG9IgQYR1HMG9plB84rkbRlQw6zNQxvAVKVrRTuPk+yS3y+r3O-c0F3PGHrmZuKF1LB9NlqTnib2-rfrZYkBdie5Gfy1FCol0wF0Js6qgF87aVLYbRWGwkdKuFWAmpUVyR60NxtDPNlC0qE-F5gEeflonb35-G-AF6sxeg47JCzE6yZdrysEa6ngVWI5-dNvnPg0qorfFUsvheBttJl2OoTDz2OxOr2Raa4XU5w5Pbh+L2OoAexJEB10nIhJwMN8DhrLggA

author: TomCrow
author_rank: Newbie
timestamp: May 10, 2021, 08:51:38 am
content: Thank you so much for the help.I think I finally understand it a bit more. And one main point to me is that the -6V for the op-amp voltage input is needed only because of the limits of the op-amp. As you mentioned if it had +12V and 0V, it couldnt control the transistors, since the real output range would be smaller. Right?Ill play with it more when I got time.The good news is that I managed to fix the bench psu, there was a broken pcb line, that I didnt notice before and now everything works.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 10, 2021, 01:41:08 pm
content: Depending on the details of the circuit (one may have to move the LEDs to a different place and maybe need some extra circuit for this) and OP used, one may just get away without the -6 V neg supply for the regulator.For the circuit with the floating supply a nice feature is that the voltage range for the output power is independen form the regulator supply. So with the same -6 v and 12 V supply for the regulator part the main power may vary from some 3 V to maybe 1000 V, with relatively little change (different trsistors and transisstors of cause) to the circuit. Especially the output range can be larger than the supply to the OPs.

author: pqass
author_rank: Super Contributor
timestamp: May 11, 2021, 02:35:49 pm
content: Quote from: TomCrow on May 10, 2021, 08:51:38 amYes.If you replace the LEDs with regular diodes then the opamp output will be higher for the same voltage on the PS output terminals (due to different Vf voltage drop). If you put a voltmeter on the output of the opamp and ⏚ (in the simulator), you'll see that when output terminals are dialed down to 0V, the opamp output is about 150mV with an LED and 1.2V with a plain diode. If you put two LEDs in series, then the opamp output is -1.4V. So, to support single LED in series, a better opamp that can get close to its negative supply rail would only need a +12V and 0V supply instead; the -6V supply is extra margin. The opamp doesn't know what ⏚ is because it only has two supply pins and can swing anywhere between those two potentials as long as they are below its absolute max. limits.QuoteGlad to hear.