i've been trying to figure out how this PSU in my amplifier is working by studying the schematic, and would like to check that my assumptions are correct, but also, as per the title, i'd like to understand how this circuit is sourcing its negative rail from a diode bridge.

So far as i can see, the diode bridge is acting as a full-wave rectifier, but rather than the voltage passing back into the neutral side of the AC connection through the diodes D209 & D210 respectively (depending on the phase of the AC) as in a typical bridge rectifier, it's being sunk to ground through a center tap on the transformer. The anodes of these same two diodes are then being tapped to provide the -45V negative rail. It seems that in effect these two diodes are in fact not acting as rectifiers, but are instead reverse-biased and blocking positive voltage from going into the negative rail, and that in fact only D207 & D208 are doing the rectification. Is this correct?

I have probed the diodes and confirmed this assumption, the anodes are indeed sitting at -45V in reference to the common ground terminal.

How then can current pass into the negative rail from the AC side through the diodes D209 & D210 which appear to be reverse biased and thus effectively open circuit..?

It would appear to be quite a fundamental rule of electronics that i'm missing.

Thanks for any help !

Cambridge Audio A500 Power Amp PSU Schematic

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    \$\begingroup\$ Maybe this will give you a hint electronics.stackexchange.com/questions/392010/… \$\endgroup\$ – G36 Feb 4 at 15:05
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    \$\begingroup\$ I think the fundamental rule you're missing is that what "positive" and "negative" are depends entirely on your (arbitrary!) choice of ground reference. \$\endgroup\$ – Hearth Feb 4 at 15:06
  • \$\begingroup\$ @d4afft6j Is this a Cambridge Audio amplifier PSU? \$\endgroup\$ – G36 Feb 4 at 19:40
  • \$\begingroup\$ So the answer here, people, with reference to my post title, is that the negative voltage is not in fact only present at the anodes of the diodes D209 & D210. The winding of the transformer uses a center tap to set a 0v point from which separate voltage rails are derived. These two rails, positive and negative, are then used to power the rest of the circuit. The voltage arriving at the diodes is measured as negative in relation to this 0v winding, which is tied to ground, and is the reason why my probing showed it to be negative, as i had the common probe clamped to the ground lug. \$\endgroup\$ – user211895 Feb 4 at 21:37

Here you can see how the current will flow in your circuit when the load resistance is symmetrical (\$R_1 = R_2 \$). For positive and negative half.

enter image description here

For asymmetrical load, the example situation is shown here when (\$R_1 < R_2\$):

enter image description here

Play with this:

enter link description here

And the DC equivalent circuit

enter image description here

Because the charged capacitor behavior very similar to the "battery".

  • \$\begingroup\$ The diode conducts the current only in one direction do you agree? And at the secondary voltage at the transformer output is a sine wave (a DC that is changing continuously) from let us say 0V to 10V and from 10V to 0V and change the polarity and change again form 0V to -10V and from -10V 0V. i.stack.imgur.com/hj7qO.png Is that clear to you so far? \$\endgroup\$ – G36 Feb 4 at 19:59
  • \$\begingroup\$ And the resistors in my diagram represent the "Load" current. The current that is "consumed" by the amplifier circuit. \$\endgroup\$ – G36 Feb 4 at 20:03
  • \$\begingroup\$ One question. Do you understand why the voltage is negative in this simple circuit that only contains two voltage sources connected in series? i.stack.imgur.com/GDbQF.png \$\endgroup\$ – G36 Feb 4 at 20:51
  • \$\begingroup\$ one answer: yes. ground is relative. like morality \$\endgroup\$ – user211895 Feb 4 at 21:07
  • \$\begingroup\$ @d4afft6j So I don't get you. Because in AC circuit we have exactly the same situation and the diode bridge "make" the current to flow only in one direction (DC). Hence we have two "DC voltage sources" connected in series (c1 and c2). And because of the way are measuring the voltage in this circuit (with respect to the reference point) we get a positive and negative voltage. \$\endgroup\$ – G36 Feb 4 at 21:15

The four diodes do work as a full bridge rectifier. This is a quite common circuit and used in many circuits where a symmetrical supply (positive and negative supply rail) is needed.

Can you understand how this circuit works?


Instead of using a full bridge rectifier (4 diodes) it uses a transformer with a center tap. Note that this gives a positive output voltage. (But I could reverse the diodes and then I would get a negative voltage.)

Now you might notice that there will only be a current flowing through the transformer during half of the sinewave. Either D1 conducts a current and no current flows through D2 or the other way round (no current through D1, current through D2). So half of the time no current is flowing though a diode and the transformer tap it is connected to.

Can we use those transformer taps more and also use it for the current in the other direction? Yes we can, see below.


simulate this circuit – Schematic created using CircuitLab

This is exactly the circuit at the right side of your schematic.

Here the "previously unused" part of the sinewave is now used to make a negative voltage -V.

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    \$\begingroup\$ @d4afft6j No it does not work like that. The diodes do not "source" their current. Also V+ and V- will be DC voltages as they are smoothed by C1 and C2 (in my schematic but many capacitors in your circuit). DC voltages do not have a phase. I suggest that you do some research as to how a (bridge) rectifier works because you seem to lack some basic understanding. There are some explanations on Youtube as well. It goes too far to explain all that here. \$\endgroup\$ – Bimpelrekkie Feb 4 at 18:32
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    \$\begingroup\$ Be careful with how you read emotion in text. I do not think Bimpelrekkie was being snide or condescending. He is well known around here for being patient and helpful with newbies. It might help if you could more explicitly state the particular part of the explanation that is eluding you. Remember that voltage is always relative. \$\endgroup\$ – evildemonic Feb 4 at 20:01

It's a center-tapped transformer. The orange wire is declared to be ground, or 0V. At any given time, red will be positive, and yellow negative, OR red will be negative and yellow positive. So either D207 and D209 will be conducting, OR D208 and D210 will be conducting.

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    \$\begingroup\$ Half the time, the red wire is positive compared to the yellow wire. The current flows through one diode, then your circuit, then returns through the other diode. The other half the time the yellow wire is positive compared to the red. The current then uses the other pair of diodes to do the same thing. All of the current returns to the source. 'Ground' is just a reference point. \$\endgroup\$ – evildemonic Feb 4 at 21:01
  • \$\begingroup\$ @d4afft6j The thing they haven't actually drawn on the diagram is the load - in this case the amplifier. That's what is actually letting the current through. But the electrolytic capacitors will also leak a little current. \$\endgroup\$ – Simon B Feb 4 at 23:40

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