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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:

Circuit1 Circuit2 Circuit3

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

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  • \$\begingroup\$ 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. \$\endgroup\$
    – rdtsc
    Nov 14, 2022 at 20:43
  • \$\begingroup\$ 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. \$\endgroup\$ Nov 14, 2022 at 20:48
  • \$\begingroup\$ 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. \$\endgroup\$
    – rdtsc
    Nov 14, 2022 at 21:32
  • \$\begingroup\$ 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? \$\endgroup\$ Nov 14, 2022 at 22:16
  • \$\begingroup\$ 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. \$\endgroup\$
    – rdtsc
    Nov 15, 2022 at 13:35

1 Answer 1

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Replacing the zener diode with a regular diode means the regulator will no longer regulate, and will output close to the input voltage (with some capacitor multiplier action so the resulting hum won't be as bad as it might be). The regulated output voltage (assuming sufficient input voltage and at least a light load) should be the zener voltage minus about 0.7V. With a regular diode (or nothing at all, which is just the same in this case) in there, there will be no regulation.

With no load at all (or just a multimeter) there's probably enough leakage through the transistors to get a get a misleading reading. Check with a load of at least 1mA or so (a 4.7kΩ resistor will suffice). If you have a lower input voltage than the circuit was designed for, or if the transformer cannot supply enough current, then the regulators may not have enough head room to regulate.

Minor nit: you should draw the big electrolytic capacitors on the left, just after the diodes so the circuit flows from left to right.

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  • \$\begingroup\$ Of course the diodes at the bases must be zener diodes. The resistors must have a low enough resistance to power the zener diodes plus the maximum base current. \$\endgroup\$
    – Audioguru
    Nov 14, 2022 at 20:43
  • \$\begingroup\$ The term "no load" written on the paper is a bit misleading so I've edited the question. It's still weird for me that it shows a big voltage drop when turning on the device, but no changes whatsoever when changing the gain. So I was wondering, if I indeed use a 9VAC power supply as it should be (instead of the 12VAC that came with it - disguised as 9VAC), will I see also less than +7.5V/-7.5V on the emitters when fully operating (due to less input voltage) or will it keep these levels somehow? \$\endgroup\$ Nov 14, 2022 at 20:47
  • \$\begingroup\$ 9VAC should yield about +/-12VDC at the large input filter capacitors (far right in your schematic). Wall warts are rated for voltage under full load- it's not unusual for them to output 20% or more higher than rated voltage with no load. The transistor base should be about 0.7V higher than the output voltage, so for 7.5V out you'd expect maybe an 8.2V zener. \$\endgroup\$ Nov 14, 2022 at 20:50
  • \$\begingroup\$ I measured the transformer without any load and it outputs indeed 12.2Vrms with something like 38Vpp. I'm quite sure it is a 12V transformer with a 9V case somehow \$\endgroup\$ Nov 14, 2022 at 20:52
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    \$\begingroup\$ Sounds like a 9V transformer to me. \$\endgroup\$ Nov 14, 2022 at 20:52

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