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I'm trying to troubleshoot a delinquent power supply I have. Most recently one of the transistors (Q1) actually exploded on it. This has led me into investigating the schematic, but try as I might, I can't make heads nor tails of what's going on. I'm an electrical engineer (but optics) and will admit being a complete noob with electronics, but I'm trying to change that.

I've attached the circuit diagram. For starters, rather than "HOW DOES THIS THING WORK," my more specific question is how the UA/LM723 operates in the circuit; that is, what is its job, and how does it do it? I have seen another LM723 in almost the exact same setup elsewhere, but the site was in Russian. I have also spent about half a day playing around with the various configurations on the datasheet to try and get a feel for how it works. However, I don't understand:

a) why the zener is shorted to ground, and the effect of this (I have seen it either greatly limit the range of Vo, or make the whole thing unstable) b) what the role of the LM723 is in this circuit. I understand that it is responsible for varying the voltage (and regulating it), how is it accomplishing this? c) BONUS, if not too strenuous, I know that Q3/4 is likely the main pass transistor and thus responsible for full current handling, but what are Q1 and Q2 doing?

I'm quite lost after spending some time monkeying with this, so any direction is greatly appreciated.

Schematic

Parts List

UPDATE: I checked the 3055, and it has 550 ohms from emitter to base (i.e., black lead on emitter); and about the same from collector to base (black lead on collector). Everything else HI. I think that my pinout is correct (used http://www.rmcybernetics.com/images/users/diy_coildriverSd2gH5.jpg).

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  • \$\begingroup\$ Unless I am missing it, I don't see a zener diode on that schematic anywhere. Also, Q1 is a transistor - which I'm sure was just a mistake. \$\endgroup\$
    – sherrellbc
    Jun 22 '14 at 21:03
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    \$\begingroup\$ Votes to close are disappointing. This is very much a design question and of great value to any who want to move beyond the "plug in a black box component" type of "design".[That said: My response below suggests a plug-in choice :-) ]. \$\endgroup\$
    – Russell McMahon
    Jun 22 '14 at 23:06
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    \$\begingroup\$ We are heading down the sad "vote to close" road again on a very interesting question that COULD teach more design to those who wish to learn than any amount of "plug in an LM317" questions ever will. This is a design from the dawn of solid state time that has much to teach. Closing would be very sad. \$\endgroup\$
    – Russell McMahon
    Jun 22 '14 at 23:08
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    \$\begingroup\$ Don't worry, this question won't stay closed. It is a good question and clearly on-topic. \$\endgroup\$
    – markrages
    Jun 22 '14 at 23:19
  • \$\begingroup\$ First, thank you kindly to those who edited my original post for errors and clarity. Secondly, apologies to those who feel this question wasn't researched enough nor novel. I will try to search more thoroughly next time before asking. To those of you who have offered assistance thank you very kindly for your time. I am just reading through all the responses now. As for the zener diode question I'm sorry that wasn't very clear. On the functional block diagram of the 723, pin 9 is the output directly below a zener diode. This circuit is the first time I've seen it used in any schematic. \$\endgroup\$
    – MJXS
    Jun 23 '14 at 12:36
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See at end for specific details of your circuit's operation.

If you need to repair this circuit with original components then it will be possible to do so. But you can probably achieve an easier result by using the existing Q3/Q4 (IF they are still undamaged) plus an LM317 (or LM317HV higher voltage version). The LM317 is used to provide a variable output voltage supply AND to drive Q3/Q4 to provide more current than the LM317 can by itself.

Easier still, you can get LM317's that provide up to about 1.5A so one of these may do what you need by itself. If so, it would need to use the heatsink that Q3/Q4 use at present. You can parallel LM317's - ideally add a small output resistor between Vout and feedback resistor network to achieve current sharing.


ADDED:

1 x LM317 will possibly do what you want.
Two in parallel may well work if you add say about 1/4 Ohm in output of each to help share the current. Put the sense resistor divider on the Vout side of the resistors.

You can get a high current version of the LM317 = LM350

LM350 data sheet here
In stock Digikey $2.09/1. 35V 3A About 2V headroom (= Vin-Vout min) at 2A. See fig 8. in data sheet. 1.2V out minimum.Be sure Vin does not exceed 35V!. With 35V in you can get about 1.2 - 33V out. See data sheet for dissipation ratings which set max Vdrop x Iout you can use.


Original design:

LM723 was the bees knees voltage regulator controller and maybe also the only one readily available 'way back when'.

You mention 2N3055 in comments but not on diagram or in question.
2N3055 was the workhorse power transistor of the day.

uA723 = LM723.
LM723 data sheet here

LM723 pinout below. I've added DIP pin numbers as those match what you show.

Pin 6 provides a buffered reference voltage - the forerunner of a modern bandgap reference.

The error amplifier (4 5 inputs) is used to compare a reference voltage with the voltage to be controlled.

The output transistor (10,11) drives the external pass element.

The zener (10,9) provides a stable voltage that the output can pull down to.

Q1 is driven on by the '723's output transistor and in turn it drives Q3.

Q2 is a current limiter. When the load voltage across R10 exceeds about 0.6V it turns on Q2 which clamps Q1 Vbe to reduce voltage. Current limit is about I= V/R = 0.6/R10. For 2A current limit R10 is about 0.3 Ohm - probably somewhat less to avoid starting to clamp too early.

Your circuit:

Vref appears on pin 6.
This is divided by R3/R4 and fed to the error amp non-inv input pin 5. This will be compared to a sample of the Vout voltage. Pin 4 = error amp inverting input is fed with an interesting arrangement. Vin is divided down and stood on top of as "pedestal' made from Vref scaled down by Vr and R5.


The LM723 internal error amp has 6.5V applied to the + (non inverting) input.
It acts to produce 6.5V at the - input. Vpot can be shown to vary from 7.15V max (top of pot connected to Vref = 7.15V) to notionally 2.85V (bottom of pot with 3k3 to ground.) I say notionally as the 10k feed from OA- + resistor from Vout (100k?) makes voltage on wiper not quite stiff as pot varies. Close enough. Vpot increasing DECREASES Vout.
Call ratio of R6:R7 = K R6 value is not shown on diagram but is probably 100k so K = 10.

enter image description here

With opamp acting to keep R6/R7 junction at 6.5 V (= OA+)
Vout must be k x (6.5-Vpot).
Rearrange this to give
Vout = 6.5 x (k+1) - Vpot x k
At pot = 7.15 V Vout = 0
At pot = 2.85V = min Vout = 43V.
ie pot has more than enough range.
Above formula gives this graph

enter image description here


The circuit is MUCH easier to understand if you draw it with the 723 internals shown and arrange it around the 723 in logical fashion.

enter image description here


Changing voltage range:

Above I named the ratio of R6:R7 as "k". Is set k=10 making R6 = 100k as that produced exactly zero Volts out at maximum pot value (ignoring pot loading) which is likely the designers intent.

If you change k in steps of 2.5 from 2.5 to 20:1 you get the family of curves shown below. ie the supply range can be altered by changing R6, subject to their being enough Vin and nothing blowing up from overvoltage.
The negative Vout will of course not happen with a ground referenced positive input supply, but would be possible if Vin was referenced against some negative value.
The common zero point of 6.5V occurs when Vout = Vpot = 6.5V = divided reference input to opamp.

R6 = k x R7 = k x 10k

enter image description here

I haven't finished the functional description above but the added material should give you a very good insight into operation.


ADDED:

So how do the transistors function? I have a basic grasp on their operation, getting more familiar with small signal characteristics, etc, but besides knowing the 3055 is the pass transistor and power transistor, that's about it. What went into the design of that portion, especially with the feedback transistor? Why use that instead of the current sense/limit?

Design: They started with a 2N3055 because it was available & cheap and & high powered. The Model T of the power transistor world.

BUT it was NPN so to turn on it needed current from V+.
BUT the 723 sinks current to ground as it turns on (although you can invert the error amp sense and use a pullup resistor to get over this)
and-but the 723 only provides 150 mA max.
For a 2N3055 at highish current you usually assume a beta (current gain) of 10, so for 2A it needs 200 mA (2A/10) so the 723 is about right but pulling to ground. If you use a resistor to turn the 3055 on and the 723 to turn it off, the 723 will need to be sinking most of the current when idle and so must be used near its current rating often.
SO you add Q1 a PNP with emitter to V+ so when 723 turns on and pin 11 goes low it turns on Q1 and this provides base drive AND the 723 needs to provided less current as if Q1 has even only a gain of 10 then the 723 needs to provide 20 mA to get 200 m base drive for the 3055.
Q1 can provide as much 3055 base drive as you want in fact, as there is no series resistor between Q1 and the 3055 base - this is, if not a design "error", then at least a design weakness. If things go wrong and Vout is not as high as it should be due to a failure somewhere then Q1 tries to tear the arms off the 3055 (ie apply too much base voltage) - if the 3055 is tougher than Q1 then Q1 may lose instead. As Q1 is a TO92 package with dissipation well under 1 Watt this is probably why it exploded - summat went aglae, the 723 tried to turn on to cause Vout to rise by turning on Q1 to turn on Q3, the cct did not comply so it kept turning Q1 on and it gave up dur to the 3055's base current.
R10 going OC would probably cause this as no V+ would then get to the 3055 so Q1 keeps trying. At 2A R10 dissipates about V x I = 0.6 x 2A = 1.2 W. It should be at least 2W rated and I'd use a 5W as they are about as cheap and much safer. If it is a 1 Watt it might last a long time. Check to see if it is O/C.

The current limit is crude in that it has no "foldback" action - it limits current AT 2A if you try to draw too much. If you eg short Vout then the 3055 will dissipate about 2A x ~30V = ~60 W. A 3055 will probably survive that if the heatsink is good enough.

The current limit works by dropping voltage across R10. When this reaches ~~~= 0.6V it will turn on Q2 which shorts the base drive to Q1 so the 3055 shuts off enough to maintain balance.

The 723 has an internal current limiter that works exactly the same way with b & e on pins 2 & 3 and the collector internally connected to shut down the internal pass transistor. HOWEVER, this relies on the internal pass transistor being the main switching transistor and being able to route Iout via a sense resistor on pins 2 & 3 which is not done here due to the external 3055 pass transistor.

Retrospective: The design followed relatively logically from what was available at that time. The 2N3055 was the logical choice of pass transistor, and as it was NPN you needed Q1 to invert the drive polarity and give current gain. If you wanted current limit the arrangement used was simple obvious and standard fare. If you wanted to go to zero Vout you needed some cunning due to the non zero reference voltage so the upside down pot polarity system was used. I'm not overly familiar with such things but I suspect it was probably standard practice in that day.

IF you want to abandon this cct but keep it working If you can tolerate 1.25V min Vout Use 2 x LM317 (or 1) or an LM350 as above. But, if you'd done that before working out how it worked you'd have learned far less :-).

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  • \$\begingroup\$ Thanks very much. I am familiar with the 317 but don't think I've ever actually used one so now I'm going to look into them as well (as an interim PS would be nice...). Don't know why your reply got downvoted, definitely of value. \$\endgroup\$
    – MJXS
    Jun 23 '14 at 12:51
  • \$\begingroup\$ Agreed - the inverting pin is fed by a very "interesting" arrangement - in fact that's the exact part that confuses me besides the Vz being pulled to ground. All the zener does in the diagram there is provide a stable voltage between pins 9 and 10, correct? What's the purpose of this? (the Q3/4 means 'only one is used' and in my case it's Q4, which is a 2N3055) When are Q1, Q2 and Q4 off/on? Are they unity voltage gain? If so, how do they get the LM723 to drive 0-30V instead of say 2V-30V? Sorry for the barrage of questions, answer at your earliest convenience. \$\endgroup\$
    – MJXS
    Jun 23 '14 at 17:57
  • \$\begingroup\$ -1 from 'votes' so far. Hard to know what to conclude from this :-). \$\endgroup\$
    – Russell McMahon
    Jun 24 '14 at 9:16
  • \$\begingroup\$ Oh man, can't wait to read this. I actually managed to figure out the part about how it varies the voltage from 0-30 by myself on the weekend, I was pretty happy with myself, I'm impressed you got the value like that though, right on the money haha. You clearly do know your way around this device \$\endgroup\$
    – MJXS
    Jun 25 '14 at 23:47
  • \$\begingroup\$ So how do the transistors function? I have a basic grasp on their operation, getting more familiar with small signal characteristics, etc, but besides knowing the 3055 is the pass transistor and power transistor, that's about it. What went into the design of that portion, especially with the feedback transistor? Why use that instead of the current sense/limit? \$\endgroup\$
    – MJXS
    Jun 25 '14 at 23:55
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The LM723 is one of the first variable voltage linear regulator ICs produced. It was first introduced in the early 1970's. It is still available today and used in some designs, though there are better choices now available, depending on the specific application.

The LM723's primary intended function is to "regulate" a DC voltage. "Regulate" in this context means that it can process a varying, or "variable", higher voltage into a "constant" lower voltage. These terms seem generic and innocent enough, but have a special meaning within the context of voltage regulators. It's the terminology that often confuses beginners.

If you don't have an LM723 data sheet at hand, you should download one from DigiKey or Mouser. Then study the example application circuits given in it. You may find there are several data sheets available from different manufacturers. Get each one as they usually contain the same information presented in different ways. This can be helpful if you are a beginner trying to decipher the terminology.

"Variable" in this context means "any voltage within the specified range of the LM723". There are two aspects to this. First, the LM723 has a maximum allowable input voltage of 40 volts. Go above this and kaput! -- gone forever. Second, the input voltage must be at least 3.0 volts higher than the configued output voltage. ("Configured" means how you set the resistor ratio of the feedback resistors. In your case these would be R3 & R4.) If you break this 3.0 volt specification, the output voltage will not be held so constant as when the voltage difference is higher than 3.0 volts.

"Constant" means the output voltage will maintain one solid value regardless of how much current you draw from the LM723's output and how much the input voltage varies within the legitimate input voltage range of the LM723.

So "regulate" means: maintain the configured output voltage no matter how much the input voltage may "vary", and how much the attached load (what you are driving on J5 & J8 terminals of your circuit) is made to vary. All understood to mean "within the legitimate bounds of the LM723's capabilities".

When you study the data sheets, you will see that you can use the LM723 "on its own", though with a few necessary external resistors and capacitors. However, it will only regulate a maximum output current of about 150 milliamps. To get more output voltage you need to use "booster" transistors. In your case these are Q1, Q2 & Q3/4. The latter being a very curious designation. Does it imply there are actually two transistors connected in parallel? With booster transistors you can increase the LM723's useful output current range to several amps. But again, all with certain limitations.

Most transistors "explode" in experimental situations because they are connected wrong. In non-experimental contexts they explode because a reverse voltage is inadvertantly applied across its terminals, or a very large current passes thru its terminals.

I don't see a "zener diode" in the schematic. Perhaps you are confused by the circular symbol with the enclosed "V" tied across the output terminals ( J5 & J8 )? This is the conventional schematic symbol for a voltmeter.

If you are only trying to build a 0-30 volt 2 amp power supply, there are better ways to do it in 2014. "Better" meaning fewer components and fewer fatal traps for the unwary.

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  • \$\begingroup\$ Hi there! Indeed I've been looking through several datasheets, the latest ones from TI being the most robust (got them from Jameco). My question is more to do with how the circuit itself operates because I'm not sure. I'm especially unclear exactly how the transistors operate, as their connections aren't clear to me. Is Q1 an amplifying stage? Is Q2 providing feedback? Is R8 a bias resistor? That sort of thing. I'd like to repair this but more than anything, I'd like to know what I'm doing rather than just plugging in components. (Hang on, out of space) \$\endgroup\$
    – MJXS
    Jun 23 '14 at 12:44
  • \$\begingroup\$ In which case, since the 723 outputs "2-37V," how do they get it from 0-30? Is that due to pin 9 being grounded? If so, why? I can't seem to replicate this behaviour so I suspect the interaction with the transistors is super important. Q3/Q4 specifies one of two power transistors: Q4 being a 3055. The zener diode I refer to is in the function block diagram of the datasheet and refers to zener voltage Vz out of pin 9. Thank you for your response! \$\endgroup\$
    – MJXS
    Jun 23 '14 at 12:44
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The output of the '723 is a pass transistor with the zener Vz (pin 9) connected to ground, so that it pulls Vc (pin 11) down in order to increase the output voltage. If the zener is not grounded it won't work properly.

That increases the base current through Q1, which pulls up the base of the parallel(?) main output pass transistors Q3 and Q4.

Q2 is for short-circuit protection- it detects the voltage across R10.

If Q1 is failing, that implies that Q3 and/or Q4 are dead (so the tiny A1015 Q1 tries to supply the entire load current), so unplug them and test them to start with.

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  • \$\begingroup\$ Thank you very much for your reply, this explains a lot in only a few lines. I'm going to spend some time pouring over the circuit with this info. Only one of Q3 or Q4 are used, in my case being Q4, a 3055 or similar MJ1004 I think. I think the 3055 is okay because I did swap it with another with no difference. I know for sure the 723 is okay (and funnily enough discovered 3 out of 4 new ones I bought are nonfunctional). None the less, I'm going to look up the 3055 datasheet and give her a test. Thanks again for your time. \$\endgroup\$
    – MJXS
    Jun 23 '14 at 12:49
  • \$\begingroup\$ Frankly, assuming there are no shorts, the only thing that can kill Q1 is if Q4 is dead, and then only if you have a relatively low resistance load on the power supply. I suspect you'll find a collector-base short on Q4. \$\endgroup\$ Jun 23 '14 at 17:44
  • \$\begingroup\$ That makes a lot of sense. I tested it with a 5-ohm power resistor. (With the voltage turned right down of course). I'll run a test on these today when I get out of the lab. \$\endgroup\$
    – MJXS
    Jun 23 '14 at 17:46
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About 15 years ago I worked for a HV power supply company. A 723 was used to regulate the HV output (15KV). Using a new batch of 723's the HV output started going high & unregulated.

The problem was traced to the 723. The old one had 2 of the pins open while which were being used for voltage feedback sensing. The new 723 replacement had these two pins shorted internally.

Just wanted to mention this for something to think about while troubleshooting. Sorry I can't remember which pins. But if your power supply is very old, and you have replaced the IC. Something to verify.

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