Hello,

I'm trying to replace the wall-wart transformer of my Moog Rogue synthesizer with a built-in transformer.

The original transformer specs according to the schematics are 24V AC, which is then rectified by an 7812 and 7912 to provide +12 and -12V.
The Moog's power is rated at 6W in the service notes and needs, according to the same notes, 200 mA. The actual wall wart original Moog transformer is rated 20V, 500 mA, so it supplies 10VA and since the Moog is rated at 6W, it wil draw some 300 mA max (20V*300mA = 6W), correct?

Anyway, my main question is this: I happen to have a 12VAC, 1A transformer; would this be enough?
My reasoning: 12VAC RMS = 16.97 VAC peak. 0.7V drops over the 1N4004 diode (in positive halve of the cyle), so 16.27Vpeak is fed into the 7812. The 7812 has according to the datasheet a typical dropout voltage of 2.5V, so it needs at least 14.5V (peak, I suppose?) to deliver a stable 12V at the output. If I feed it with 16.27V, that should be ok, no?
Same for the negative half of the cycle; 0.7V drops over the other 1N4004 diode, so I'm feeding -16.27V in the 7912.

As for the amperage, my transformer would be able to supply 1A and since the Moog only needs 300 (or, by a safer margin, 500), that should also suffice? 12V*500mA = 6W?

Is this correct or am I confusing things or overlooking something?

 

Peak input voltage: 12*1.414 = 16.968V
Subtracting from the the diode drop (assumed 0.7 but may be greater) and the minimum regulator input voltage (2.5V) and the regulator output voltage leaves just 1.768V. There will be some ripple and that voltage will be the peak of the ripple. The smoothing capacitors are 1000uF and by my calculations the current drawn when the ripple is 1.768V would be about 88mA. This is the maximum current that would ensure sufficient input voltage to the regulators. Further none of that takes into account what happens at times of low mains voltage, nor the actual output voltage of the transformer under load.

 

Peak input voltage: 12*1.414 = 16.968V
Subtracting from the the diode drop (assumed 0.7 but may be greater) and the minimum regulator input voltage (2.5V) and the regulator output voltage leaves just 1.768V. There will be some ripple and that voltage will be the peak of the ripple. The smoothing capacitors are 1000uF and by my calculations the current drawn when the ripple is 1.768V would be about 88mA. This is the maximum current that would ensure sufficient input voltage to the regulators. Further none of that takes into account what happens at times of low mains voltage, nor the actual output voltage of the transformer under load.

Thanks for your thoughts! Although you're going a bit too fast for me...
- if 0.7 (or greater) is dropped over the diode, doesn't that mean that the remaining 16.268V is supplied to both the smoothing cap AND the 7812 (since they're in parallel)? So 16.268V is the voltage that gets "smoothed" by the 1000 uF cap (with a certain amount of ripple) and is then also presented on the input of the 7812?
- how do you calculate that ripple voltage and the maximum current without a specific load resistance?
- in the end you are basically saying: don't do it, it will not suffice, since in the "real world" you are forgetting that mains voltage might be a bit lower as might output voltage under load?

 

Peak input voltage: 12*1.414 = 16.968V

but the schematic shows 24VAC in

 

The peak of the ripple voltage at the input to the regulator is 16.268V. The minimum voltage that the regulator needs, at its input, is its output voltage plus the dropout voltage - 12V + 2.5V = 14.5V, so the bottom of the ripple must be at least 14.5V for the regulator to work properly. Therefore the maximum peak to peak ripple voltage is 16.268V - 14.5V = 1.768V.

The smoothing capacitor is charged to the peak input voltage each cycle of the mains cycle and then the output current discharges the capacitor during the rest of the cycle - 20mS for 50Hz or 16.7mS for 60Hz.

Now we can calculate the output current which will make the ripple voltage equal to that maximum allowed ripple voltage:
Q=CV
It = CV
I = CV/t
I = 0.001F * 1.768V / 0.02S
I ≈ 0.088A
Any current in excess of that will cause the input to the regulator to fall below the minimum required for part of the cycle. As you need more current than that it will not work properly. You must either increase the input voltage or increase the size of the smoothing capacitor. I am not clear from your description just what output current you require but let's assume it is 200mA. Then we can calculate the capacitor value needed to keep the ripple below 1.768V at 200mA:

C = It / V
C = 0.2A * 0.02S / 1.768V
C = 0.0022F = 2200uF

These calculations are for 50Hz but if you have 60Hz mains then the ripple will be a bit lower and your transformer will probably provide more than 12V at 200mA as it is rated for 1A. You will have measure that and do the calculations for your actual values. Remember that you will need to allow some margin for low mains voltages.

 

Use diodes 1N5819.
See

 

The peak of the ripple voltage at the input to the regulator is 16.268V. The minimum voltage that the regulator needs, at its input, is its output voltage plus the dropout voltage - 12V + 2.5V = 14.5V, so the bottom of the ripple must be at least 14.5V for the regulator to work properly. Therefore the maximum peak to peak ripple voltage is 16.268V - 14.5V = 1.768V.

The smoothing capacitor is charged to the peak input voltage each cycle of the mains cycle and then the output current discharges the capacitor during the rest of the cycle - 20mS for 50Hz or 16.7mS for 60Hz.

Now we can calculate the output current which will make the ripple voltage equal to that maximum allowed ripple voltage:
Q=CV
It = CV
I = CV/t
I = 0.001F * 1.768V / 0.02S
I ≈ 0.088A
Any current in excess of that will cause the input to the regulator to fall below the minimum required for part of the cycle. As you need more current than that it will not work properly. You must either increase the input voltage or increase the size of the smoothing capacitor. I am not clear from your description just what output current you require but let's assume it is 200mA. Then we can calculate the capacitor value needed to keep the ripple below 1.768V at 200mA:

C = It / V
C = 0.2A * 0.02S / 1.768V
C = 0.0022F = 2200uF

These calculations are for 50Hz but if you have 60Hz mains then the ripple will be a bit lower and your transformer will probably provide more than 12V at 200mA as it is rated for 1A. You will have measure that and do the calculations for your actual values. Remember that you will need to allow some margin for low mains voltages.

Genius! Thanks a lot for that detailed explanation! I learned the theory behind all this not so long ago, but to actually apply it in a practical world was a bit over my head. Just wondering: you state (correctly) that the cycle of 50 Hz takes 20 ms, but don't I just need half a cycle here? One cap charges and discharges through one half of the cycle (upper half), so that's 10 ms and the other charges and discharges on the other half, so that's another 10 ms?
That would give me, in both half cycles +/- 176 mA. Still too low of course, but already a bit better. Or am I wrong again?

"Some margin for low mains voltage"; how much are we talking about then? I read somewhere that + or - 5% is sort of a general rule and since I don't live near big power plants or other crazy inductive loads, can I apply that rule of thumb? (which would mean 230V mains could be as low as 219V mains, which would provide only 11.43V RMS secondary instead of the full 12)

 

don't I just need half a cycle here? One cap charges and discharges through one half of the cycle (upper half), so that's 10 ms and the other charges and discharges on the other half, so that's another 10 ms?

No. The positive rectified output capacitor is charged on every positive cycle of the input. Positive cycles happen every 20mS. Look at the red and green waveforms in the simulation above.

 

Again, how are you hoping to get 24V from a 12V wall wart? You need a different transformer!

I'd look for a modern SMPS power supply to replace the wallwart approach.

 

No. The positive rectified output capacitor is charged on every positive cycle of the input. Positive cycles happen every 20mS. Look at the red and green waveforms in the simulation above.

True, of course, should've known better. Positive half of the cycle is only 10 ms, but the cap discharges throughout the whole cycle before the next half starts (that's the reason it's there in the first place). So 20 ms!
If I take into account that V(RMS) could swing to 11.4 if mains voltage fluctuates 5%, I get:

V(peak) = 11.4*1,414 = 16.1196V
Vripple(peak) = 16.1196V - 14.5V = 1.6196V

Now take some margin on the amperage (schematic shows 200 mA, but let's take 300 mA to be sure):

0.3A * 0.02s / 1.6196 = 3700 uF.
I'll see if I have some lying around here and might just give it a shot.

The only think I haven't taken into account from your previous post is "what happens to the transformer's output voltage under load", but I guess that's something I should test in practice, since I don't know the transformer regulation percentage of this particular transformer.

Again, how are you hoping to get 24V from a 12V wall wart? You need a different transformer!

I'd look for a modern SMPS power supply to replace the wallwart approach.

Wayneh, thanks for your comment: I've bought a 12V transformer to build in my Moog (so not a wall-wart transformer). I don't specifically NEED 24V, I need +12V and -12V rails (supplied by the 7812 and 7912). You are correct that the schematic states I need a 24V AC transformer in order to get that +12V and -12V, and although that will certainly suffice, it doesn't HAVE to be the full 24V AC (for example, my current, original Moog wall-wart only supplies 20V AC and my Moog works perfectly).
So my question was: if it works with 20V AC, as it does now, will it work with the 12V AC transformer I have lying here? Seems it will if I change the caps, although there's not a lot of margin...

 

I think the answer is no, but of course it is possible it will operate off a much lower voltage than it is designed for.

Note that a transformer rated at 20V AC, at some current level, gives a much higher voltage when rectified and filtered with a large capacitor for low ripple. In other words, it gives at least 28.2V, less 2 diode drops, or about 27V DC. If the 20V is specified under a load of, say, 1A then at current loads less than that, the voltage will be even a little higher.

Instead of a 12V transformer, you might look at a laptop power brick. You can often find these free at the local electronics recycling center. They have plenty of capacity and usually output ~19V DC. You could split that into ±8V or so, depending on the regulators you can find.

Look for "low drop out" LDO regulators so that you can get as much voltage as possible.

 

As your transformer is rated to give 12V at 1A, it will probably give more than that at the currents you need. Only measurement can get an answer to that.

 

I think the answer is no, but of course it is possible it will operate off a much lower voltage than it is designed for.

But I won't let it operate off a much lower voltage. The 7812 and 7912 supply +12V and -12V and all the opamps and other IC's in the circuit need that +12 and -12. The Moog needs 200 mA according to the service manual. And it's rated 6W.
If you'd feed it 24V, 250 mA, the 6W makes sense, so that's a safe value to start from

Using AlbertHall's formula's:
The wallwart power transformer that came with my Moog is rated at 20V, 500 mA (that's also "lower than it is designed for", since it's designed for 24V...)
20V RMS = 28V peak.
28V peak - 0.7V over diode = 27.3. I need at least 14.5V on the input of the 7812 to supply a stable 12V output. So the ripple voltage can be as large as 27.3-14.5 = 12.8V
If I take that as maximum ripple, with the 1000 uF cap supplied in my Moog, the current I get is: (0.001F*12.8V)/0.02s = 640 mA max. Which is a higher than the 250 mA it's designed for, but then again: that's a maximum value, so to me it seems to make sense...

That also means that if I lower the input AC voltage even more but raise the caps' value, I should get the same stable +12V and -12V and sufficient mA's... I might just try it out as AlbertHall suggests, especially since the transformer will probably provide more than 12V if I don't use the full 1A... But I thanks to all your help I also realise it might be too tight, so maybe I'll just get me a 20V transformer and avoid unnecessary difficulties

 

You're not hearing it. You cannot get 24V DC (same as ±12V) out of a 12V AC transformer. Best you can hope for is about 12 x 1.41 = 16.9V peak. Subtract 2 diode drops (not one) and you're at 15.5VDC. You cannot easily get to ±12V from that without some relatively tricky circuitry. It is possible (as in #6 above), but not terribly practical when you have other options.

Maybe #6 done with LDO regulators is an option.

 

You're not hearing it. You cannot get 24V DC (same as ±12V) out of a 12V AC transformer. Best you can hope for is about 12 x 1.41 = 16.9V peak. Subtract 2 diode drops (not one) and you're at 15.5VDC. You cannot easily get to ±12V

You can because the Moog power circuit uses a voltage doubling rectifier so that 15.5V becomes ±15.5V. See the circuit in the first post.

 

You're not hearing it. You cannot get 24V DC (same as ±12V) out of a 12V AC transformer. Best you can hope for is about 12 x 1.41 = 16.9V peak. Subtract 2 diode drops (not one) and you're at 15.5VDC. You cannot easily get to ±12V from that without some relatively tricky circuitry. It is possible (as in #6 above), but not terribly practical when you have other options.

Maybe #6 done with LDO regulators is an option.

Oh, but I'm hearing it, I'm simply not understanding it, I'm afraid. Does that mean that AlbertHall's calculations and suggestions are wrong? I was thinking that a 12V AC transformer means 12 V AC RMS, means 16.9V positive peak which is just sufficient to be half wave rectified to +12V and -16.9V negative peak which will be half wave rectified to -12V. I know that's close and margins are small, but you're telling me that reasoning is plain wrong then? Have you looked at the schematic in my first post? From what I understand, the Moog rectifier circuit is designed as a "voltage multiplier", as described here: http://www.electronics-tutorials.ws/blog/voltage-multiplier-circuit.html, so shouldn't I be able to get 2 times VACout from VAC in? I'm not contesting it, you obviously have a better understanding of this than I have, but I'm a bit lost (just when I thought I got it).

 

That makes two of us. I wasn't aware of the voltage multiplying circuitry already present. I suppose there is a good chance it will work fine. Maybe more importantly, I don't see any downside to testing it.

 

If the peak rectified voltage from the 12VAC transformer provides sufficient headroom for the 7812 and 7912 regulators to work, then you are okay.

If needed, you might try some large Schottky rectifiers in place of the junction devices as they will give you a few more tenths of a volt output.

 

Go on ebay and BUY a cheap +- 12 volt supply. They are under 20 bucks free shipping.

 

Go on ebay and BUY a cheap +- 12 volt supply. They are under 20 bucks free shipping.

He already has the transformer - zero cost