# Transistor based voltage regulator with normal diode

On another thread I was discussing about Power Supply for an equalizer that originally takes a 9VAC as input, then I found some interesting things on the input circuit of the equalizer that I did not quite understand it:

First of all, I realized the original transformer that comes with it, actually converts 230VAC to 12VAC and not to 9VAC as expected, but fine so far.

Then, as far as I could understand the circuit (which I drew on the notepad on first picture), it seems the upper cycle of the AC input will be rectified and stored on the upper right capacitor, producing the upper voltage rail and the same will happen below for the lower cycle of the AC. However, the power switch is on the big board (located in series with the resistors on the diagram), which means, once turning it on, current will flow out of the emitters lighting up all LEDs and ICs on the board, resulting on a voltage of approx. +7.5V on the upper rail and -7.5V on the lower rail.

The problem is, I would expect that both diodes (in parallel with the respective capacitors) should be Zener diodes, to work as a simple voltage regulator together with the transistors. However, when analyzing further to be able to buy a replacement for one of the diodes that burned, I realized they are apparently normal diodes, because if I turn off the red LEDs of the big board while the device is turned on, voltage on both rails jumps to +8.30V and -8.30V respectively. And if I disconnect the big board and short circuit the pins that represent the power switch, I get about +16V and -16V respectively on top of these diodes (which means they cannot be Zener). But funny enough, if I connect a music source and sound system to the output and then play with the Gain (making it clearly louder or quieter), it does not affect the voltage levels at all, even though I would expect some changes since the "voltage regulator" does not indeed regulate, since it is draining more power. Am I missing something here? Is this a typical way of adjusting input voltage levels of a board? If so, how come it does not change voltage levels while changing the gains of the OpAmps?

EDIT:

The small input board also has 2 OpAmp ICs which as presumably being fed also by this input circuitry. So with power connected, I get the following readings:

• +16V/-16V on the big electrolytic capacitors when the device is OFF
• +12.5V/-12.5V on the big electrolytic capacitors (with ripple) when the device is ON
• +7.5V/-7.5V on the Emitters when device is ON

The resistors are 5.1kOhm

• Measure the voltage across D4 (the good, likely Zener) during normal operation. If you see about 0.7 V, then it's a regular diode. If you see ~7 V, then it is a 7 V Zener. Commented Nov 14, 2022 at 20:43
• But during normal operation the diodes are in reverse direction, it won't and should not have 0.7V across it. What I did was try to remove almost all the load after the emitter and seeing if I get a fixed voltage across it, but it just raises to 16V (which is almost the voltage from the big capacitors). That's exactly what made me wonder that they are not Zeners. Commented Nov 14, 2022 at 20:48
• Right, a Zener works by "breaking down" in the reverse direction. A normal diode does not "break down" until some high voltage, like 60 V or 100 V or something. But a Zener is designed to breakdown at a specific voltage, such as 7 V. This is why they are sometimes used in voltage regulator circuitry. It would make sense that D3 and D4 are 7 V Zeners in this circuit. Commented Nov 14, 2022 at 21:32
• Yes, it would make total sense to me as well. To be honest, I was actually sure they were Zeners and I just performed the tests to find out it's value (i.e. confirm that they were indeed around 7V Zeners). But then, when removing most of the load on the emitter, it raised to almos 16V. The same on both rails. How could that be possible if it was a Zener? Commented Nov 14, 2022 at 22:16
• I think this is just leakage through the transistors. i.e., there must be some small load (perhaps 10k) to make them operate as intended. Commented Nov 15, 2022 at 13:35