Old and basic linear power supply

[tony359]

Greetings all,

I have acquired a VERY old signal generator - it's a LEVELL and it's from 1977 or whereabouts I suppose as a note inside the PSU said 1977! It can generate sinewaves and square waves up to 1Khz and up tp 20V p/p.

It was working intermittently and then it stopped working entirely.

It turned out the PSU was gone so I opened it up and found a capacitor and a resistor (R2) gone. I did not have a suitable large 1.2K resistor so I just botched something together for a quick test and replaced both caps: it worked but I could not keep the PSU on for more than a few seconds as the resistor was getting very hot.

I ordered a 2W 1.2Kohm resistor which arrived today. I have soldered it in place in but it still gets very hot. I also found that R1 is also getting very hot, even with no load.

As you can see from the schematics I am attaching, this is a very simple power supply. And probably it would be worth replacing it entirely. However, I'd like to understand what is wrong with it before I just throw it away. The PSU says "33V - 20mA" on the case.

A few notes:
1. The positive rail draws 11mA when the unit is running.
2. if I check the current flowing through the Zener diodes (between A and B, opening the circuit of course), it reads 45mA, which I feel it's quite high considering the whole unit draws 12mA from positive when running. I believe this explains why the 1.2K and 33Ohm resistors are getting so hot - current had to be flowing somwhere!
3. The original capacitors were 100uF or so, I did not have those so I went up in capacitance hoping that that could only make things better?

Across the R2 resistor I read 50V - which is puzzling but it may be just be not fully understanding the circuit? Between 0V and C I read 68V which I suppose it's a byproduct of the fact that half of the circuit is not ground referenced?

As it seems to be such a basic circuit, can anybody help me fixing this thing?

Thank you!

[SilverSolder]

Is there a switch to select between 110V and 220V mains?  What is the AC voltage coming out of the secondary side?  (i.e. from the bottom of the two capacitors to the point where the two diodes meet)

[bob91343]

It does appear that the source voltage is far too high.  Half voltage would be better.  It just has to supply a little more than the zener diode voltage to function.  Run it from 120V and all should be well.  Or make sure the primaries are in series and not parallel.

As it stands it is a poor design.  But not so with half voltage.

[tony359]

SilverSolder

No, there isn't and the box says 200-250V (I am in England). It is also "Made in England".
The voltage I measure at the secondary is 36VAC. Each of the two winding is labelled 15V so that makes sense to me. Though I previously measured 46VAC - probably when the 1.2K resistor and one of the caps were open.

Bob
That's what I thought: voltage too high at the Zener, and they drain way too much current to keep the voltage "regulated" which in turns burns everything.
I am happy to test amending the primary but I did not change it. And They are currently in series, did you mean I should change it to parallel?

Some pics attached - thanks for your help!!

[MarkF]

Why not just build a better supply with simple 3-terminal regulators?   

[tony359]

MarkF

Indeed that’s an option.
But before going that direction I’d love to understand what’s wrong with this one and ‘fix it’.

The only thing is: there is a slight chance that this PSU has always been a poor design (or maybe modified by others) and that I’m chasing ghosts here. Maybe it’s always been running hot as the sun? I do not know to be honest.

[Gyro]

It looks original. It was obviously an add-on PSU option to replace the original battery option of four PP9 batteries.

If it only has to supply 20mA max (in line with PP9 battery life), then you probably have the option of just increasing the resistor value and/or replacing it with a higher wattage one, maybe attached to the Aluminium chassis.

[bdunham7]

What voltage do you measure between the top ends of C1 and C2 (the sides that aren't connected to each other) ? 

How sure are you that it was all connected this way in the beginning?  Can you try connecting both transformers in parallel (watch the polarity--the sides that are connected now will need to be disconnected and crossed) and just putting the 33R resistor in between one side of the transformers and the diode junction?

[tony359]

Yes I suspected it was a replacement for some  batteries indeed.

What resistance value should I choose?  I suppose you mean R2? Would it be connected between the positive rail and ground?

Just increasing the wattage of R2 may not work: R1 is burning too after a few seconds of power being on.

Bdunham
Sorry, you posted while I was replying.
I read 85.6VDC.

I am 100% sure, I did only swap the capacitors and R2. I did not change how the windings are wired.

Let me make a drawing of your suggestion so that I confirm I understand correctly!

[MarkF]

Don't you need a resistor on the negative side as well?
And tie the COM (i.e. 0V) to the center tap to keep it balanced.

[tony359]

MarkF

I was wondering 🙂

The 0V is in fact connected to mains ground. Do you suggest to remove the 33ohm resistor between secondaries and ground it?

[Kleinstein]

The 33 Ohms resistor in between the 2 transformers should not be removed in the original circuit. It reduces the current peaks when charging the filter caps and this way has a little less loss compared to using only the 1.2 K to limit the current. It would be more like making the 33 Ohms larger (e.g. 47 Ohms, maybe 100 Ohms - than with higher power rating).

The power loss in the circuit is quite high as the rectified voltage is so high with the voltage doubling rectifier. When using a normal bridge rectifier and only 1 filter cap the voltage would be about half. So some 40-45 V DC. This would still be just enough to get the 2x16 V from the zeners. The 1.2 K resistor would than need to get a little smaller (e.g. some 300 Ohms), but would get less hot.

The 3 Pin regulators would probably also be OK, though likely with 7815 / 7915 and no need for excessively larger filter caps. However it depends on the circuit, as the current circuit also provides current limiting. So it depends on the rest of the circuit.

[bdunham7]

I read 85.6VDC.

I am 100% sure, I did only swap the capacitors and R2. I did not change how the windings are wired.

Let me make a drawing of your suggestion so that I confirm I understand correctly!

If it is very old, it may have been both marginally designed and designed for lower ranges of line voltage.  The math works--you have ~30VAC which is 42.4V peak, a voltage doubling circuit so 85 volts seems right.  This would give plenty of margin for the capacitors discharge cycle.  If you run the xformers in parallel, you'll have half the voltage, but still more than 10 volts over the sums of the zeners.  You might have to lower the value of R2 as it will drop more than 10 volts at the 12mA you said the device drew.  Try 500R, you'll see 8mA thru the zeners and 12mA to the device.

Or, you can bodge together a better supply as MarkF suggests.

[SilverSolder]

Don't you need a resistor on the negative side as well?
And tie the COM (i.e. 0V) to the center tap to keep it balanced.

You only need one resistor, because the same bias current flows through both zener diode sets.   I.e. with no load connected, the current flows from the positive side of the voltage doubler (point C in the diagram), through R2, then through D6+D4, then up again through D9+D7, and back into the negative side of the voltage doubler.

The zener diodes act as a kind of floating "rail splitter" so COM / 0V should NOT be tied to the transformer center tap.

[SilverSolder]

The circuit appears to work as designed.  Back in the day, using a couple of watts this way was probably not something anyone would lose any sleep over, even though it seems very inefficient by today's standards.  Back then, it was probably often more important to use less components than it was to get high efficiency.  That's why this doesn't use a bridge rectifier - after all, that's two more diodes!

Using a 2W resistor for R2 is marginal.  Probably better to use at least a 3W type, if not a 5 watter, so it can lead a more relaxed life! 

[tony359]

Uhm... Now I am thinking about that I just replaced R2 based on the 20mA figure printed on the case. I went back to the bin and fished out the original resistor and maybe that's a bit more than the 2W I used as a replacement - picture below.

Even assuming that a 5W resistor would do the trick, what about R1 which is getting over 80 degrees seconds after I power up the PSU? Granted, I did not try WITH a load.

[bdunham7]

The thing about power resistors is that even if you get one big enough to easily handle the load, it just cooks everything around it anyway.

[SilverSolder]

Uhm... Now I am thinking about that I just replaced R2 based on the 20mA figure printed on the case. I went back to the bin and fished out the original resistor and maybe that's a bit more than the 2W I used as a replacement - picture below.

Even assuming that a 5W resistor would do the trick, what about R1 which is getting over 80 degrees seconds after I power up the PSU? Granted, I did not try WITH a load.

Basically, these resistors will be loaded with the full 45mA at all times, no matter what load you connect to the supply (the current either flows through the zeners or the load, that's how it regulates the voltage).

So, the 300 ohm resistor will burn about 0.6 watts constantly, and the 1.2K one will burn about 2.5 watts constantly.  Just make sure both are over-rated and they will last a long time.

[SilverSolder]

The thing about power resistors is that even if you get one big enough to easily handle the load, it just cooks everything around it anyway.

In this case, there is no PCB - so as long as there is air flow around the resistor, and it is rated for the heat produced, everything will be fine for another several decades!

[tony359]

SilverSolder

Thank you.
I realised I never checked the currents flowing through those resistors. Indeed I measured 43mA on R2 (Voltage drop of 50V) which makes 2.08W. And on R1 I measured 122mA and a voltage drop of 4V which makes a 0.49W dissipation.

I suppose that when a resistor is rated, say, 2 Watts, it is supposed to be 100 degrees when running at 2 Watts?

I botched together a 4W 1.2Kohm resistor and indeed it gets warm but ok. I suppose I will order a 5 or 6W resistor and be happy.

R1 however is quickly going to 90 degrees after a few seconds. I suppose this is the way it's been running since day one?? I can replace it with a 2W one and it should be ok.

Thanks so far.

Now, part of this whole exercise for me is also to learn something new. So I hope you don't mind if I ask a few very basic questions to understand this circuit.

1. What is the purpose of R1? Current limiting?
2. I measure 50VDC voltage drop across R2 but the transformer is outputting 36V AC - which is 50.76V p/p. If I understand correctly, the diode and the capacitor will rectify the waves and make something close to (36 x 1.41 = 50.76) 50.76VDC - as this is DC, I do not consider the RMS value, that is for alternate current only, correct? So how I do end up with the whole 50V dropped by the 1.2KOhm resistor then? Is that because I had no load attached?
3. How did you come up with the power values from the diagram?

I appreciate your - or anybody's - help on this, thank you!!

[aqibi2000]

Great now that you have it up and running with low wattage resistors, get a 7912 and 7812 regulator like MarkF suggested.

[Kleinstein]

The rectifier is build as a voltage doubling rectifier. One gets nearly the peak value at the positive and negative side. So close to the peak to peak voltage. So 2.8 time 36 V or some 100 V.  Due to R1 and finite capacitor values the voltage at the caps is a little lower (e.g. some 80 V).

Both resistors are there to limit the current. R1 does this before the filter cap and R2 does it behind the filter caps.  R1 is slightly (with the very large caps the difference is not large) more effective (less heat) here, as it takes more of the current when the AC voltage is lower. R1 sees the RMS current and thus more current, so it takes less resistance at R1.
Chances are one could than get a large value for R1 and increase it's power rating correspondingly. One may than get away with the original 1.2 K resistor. With twice the caps twice the value for R1 should be OK.

[SilverSolder]

The circuit consists of two main parts:  a voltage rectifier/doubler part, and a zener regulator part.  Let's look at each individually.




During the positive part of the sine wave from the transformer, C2 is charged via D2.
During the negative part of the sine wave, C1 is charged via D3.

The DC output + and - are made from C1 and C2 in series, hence the voltage is double what it is on either C1 or C2 alone.

You could try measuring the DC value with your voltmeter and see if it hovers around 100V as @Kleinstein predicted.  Use alligator clips or something, keep fingers away, 100V is enough to kill in the "right" circumstances!

[tony359]

Oh don't worry, I am very careful. I do use alligator clips and I always discharge the caps before fiddling. I also have a incandescent bulb current limiter on the line. I appreciate the warning though!

Yes, I think understand that part. I have 36VAC at the transformer which are then rectified by the diodes - D2 will only keep the positive part and D3 the negative part. The capacitors will charge and then smooth the "missing wave" in between each one. I can see that with my scope, there is about 1V of ripple. Hence if 36AC is my RMS value, the actual sinewave is around 36 x 1.41 = 50.76V peak to peak.

Hence my Positive rails becomes around +50VDC and my negative rail becomes around -50VDC - which measure around 100VDC with a DMM. I made a very professional sketch below.

[SilverSolder]

OK so I re-drew the zener part of the circuit to make it more obvious what is happening.




The zener diodes always drop a total of 32 volts (nominally 8 volts each, but maybe a little more in reality).  So the resistor R2 has to drop the remaining voltage,  for example 100V - 32V = 68V

When you connect a load that uses current from the +17 or -17 volt rail, what happens is that the load "steals" current that the zener would otherwise have taken. 

The way this happens is that as the load uses current, the voltage across the zener drops ever so slightly, causing the current through the zener to drop to make up for what the load is using.  The drop in voltage required to change the current is tiny - this is "secret" to how the zener maintains a near constant voltage across it.

This way, the current through R2 is essentially constant no matter what load you put on the supply...  the current either flows through the zeners or through the load, the sum is always a constant (give or take a tiny little bit).