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I'm just experimenting with rectifying a 120v to 12v dual winding transformer and smoothing it out two a dual rail power supply while I work towards attempting to build a clean power supply. I realized the 12v outputs were too low for my intended purpose, so I decided to just test it out with what I had on hand while I wait for a more appropriate transformer to come and other parts.

I was trying to somewhat emulate the PSU circuit from here: https://www.diyaudio.com/forums/pass-labs/317803-whammy-pass-diy-headphone-amp-guide.html

Whammy PSU Schematic

This is what I made with the parts on hand: Test PS Schematic

With no load, the positive rail measures at ~+11.95v after the regulator. Okay, sounds good. The problem is, when I put any load on it more than a few mA, it immediately sags down to a measured ~+10.8v and varies a lot around that area.

With just an LED and resistor set for 10mA current on the LED, the rail held as if nothing changed. But anything beyond that and it immediately drops out of regulation. The pre-regulator positive rail measures well over 14v (I believe almost 15v) and the regulator has a 2v dropout voltage, so I thought it'd be okay.

What am I doing wrong or not understanding about this? I know to seasoned EE's (I'm just a hobbyist), the capacitor and resistor values probably seem random and not ideal (it was what I had on hand), so please note that I do know the values are probably not ideal. But, I thought it'd be plenty of smoothing through the CRCRC and work similar to the referenced circuit.

Thanks!

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    \$\begingroup\$ You don't need the 10 ohm resistors, short them out. \$\endgroup\$ – Kevin White Oct 23 at 18:54
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The resistors between your filter capacitors are counterproductive. They cause extra voltage drop due to load current, and prevent the later capacitors from charging to peak rectified voltage.

Here's what happens to the unregulated output when a load of 100mA is applied (simulated in LTspice):-

enter image description here

Voltage is very smooth but has dropped to 14V, which is right on the dropout voltage of a 7812.

Your regulators remove the ripple, so the filter capacitors just have keep the unregulated voltage above the dropout voltage of the regulators. The circuit will work better if you simply join all the filter capacitors together to make one larger capacitor.

Without resistors and the 2200uF capacitors paralleled to make 6600uF the unregulated output looks like this:-

enter image description here

Now the voltage is staying above 16V so the regulator has plenty of headroom. There is a small amount of ripple, but the regulator will remove it.

The pre-regulator positive rail measures well over 14v (I believe almost 15v) and the regulator has a 2v dropout voltage, so I thought it'd be okay.

If the transformer is really putting out 12VAC then you should be getting over 16V with no load. Under load it will drop, but should still be good for up to ~100mA. Transformers usually put out higher than rated voltage when unloaded to ensure that they meet their spec at rated load, so in practice it should be even higher. Lower than expected unloaded voltage suggests either incorrect transformer winding or low mains voltage.

If the unregulated voltage is over 14V under load then the regulator should have enough to work on (at least at currents well below 1A). A meter only reads average voltage so it won't tell you the lower ripple 'trough' voltage, but your circuit should have very low ripple if the capacitors are anywhere near the correct values. So if the loaded regulator input voltage is almost 15V then something else is causing the low output voltage.

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What am I doing wrong or not understanding about this?

  • You're using too low a voltage of transformer. If the transformer ratings are typical, then each output winding is 12VRMS when the AC input is 120V. The output is pulsed DC that does go to around 16V when it is unloaded, but the capacitors have to hold up the voltage between pulses. Even keeping low dropout regulators happy for a 12V output will require very short current pulses from the transformer. You probably want a transformer rated for 15 or 16V per output winding (there's a rule of thumb out there somewhere, but I've forgotten it).
  • You're using high dropout regulators. This just exacerbates your problems.
  • You're using filter resistors. This is great for smoothing out the AC, but terrible for causing voltage drop -- and you're causing voltage drop on top of the already inadequate AC voltage.
    • Unless you need a super-quiet supply, ditch the resistors.
    • If you do need a super-quiet supply, it's probably just for a portion of the circuit -- so tap off of the rectifier output (or use separate rectifiers) for a separate, super-small super-quiet supply.
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  • \$\begingroup\$ Thank you for the response, much appreciated. \$\endgroup\$ – crabbyone Oct 23 at 20:12
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I assume your transformer outputs 12 V a.c. This means that after rectification you will get 12 multiplied by square root of 2 minus diode voltage drop. An approximate of 15 V assuming about 1 V drop per diode. This is the peak voltage and in theory it is enough to supply the linear 7812/7912 linear regulators and allow them to function properly (3 volts above output).

When a load is inserted in the circuit, the condition to have the input voltage at least 2 V higher than the output voltage of the regulator all of the time is no longer met. Two reasons why this happens:

  • filter capacitors discharge faster;
  • voltage drop across 10 ohm resistors increases;

Solutions:

  • use a transformer with higher output voltage and/or
  • use low voltage dropout regulators
  • remove those 10 ohm resistors (or use transformer whose output voltage can compensate the drop across resistors).
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  • \$\begingroup\$ Thank you for the explanation and options. Excuse my ignorance, but how do I calculate the voltage drops across the resistors in question if this were connected to a two stage op amp headphone circuit powering headphones of difference impedance such as 32ohm-600ohm? I know the basic ohm's law calculations, but I'm not sure how I account for the circuit past this power supply in calculating the voltage drop. \$\endgroup\$ – crabbyone Oct 23 at 20:11
  • \$\begingroup\$ @crabbyone determine the maximum current required by your circuit (practical measurement, theoretical calculation) and use Ohm's law to find voltage drop across resistors. Without knowing the circuit, I can't tell what current it does require. The total current through resistors is the current drawn by regulator (very small) plus the current drawn by the load circuit. \$\endgroup\$ – Cornelius Oct 23 at 20:16

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