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vingenz 12th November 2012 02:36 PM

Big B+ voltage drop using diode rectifier.
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Hello folks!
I've built an amp with 2x6l6gc P-P AB1 in the power section. There is problem that I can't solve: after full wave rectifier, B+ is 460VDC (372VAC on secondary) at idle, and 395VDC (350VAC on sec.) at full power.
Power transformer specs are: 250mA; 372-0-372VAC.
6l6 are biased (fixed bias) at 35mA each. Diodes are 1n5408 (1000v, 3A)
Well, i need more clean headroom and I can't understand why it's sagging too much . Suggestion? Tricks?

DF96 12th November 2012 03:31 PM

Remember that transformer specs assume a resistive load. A rectifier/cap load will impose a peaky load so voltage drop will be greater.

The figures you give don't seem quite right. A 372V AC secondary should give an off-load DC of around 520V, but you are getting about 10% less. It may be that the AC waveform has the peaks flattened so the ratio between peak and RMS is not the 1.414 for a pure sine wave. Take the 350VAC under load, knock off 10% then multiply by 1.4 to get 441V. If the DC current is 200mA then the ripple will be about 33V so we are down to 408V.

It is a feature of power supplies that they usually droop more than expected.

vingenz 12th November 2012 05:11 PM


Originally Posted by DF96 (
It may be that the AC waveform has the peaks flattened


DF96 12th November 2012 05:13 PM

A high resistance joint in your supply wiring, or neighbours drawing heavy current into conventional PSUs can do this.

Yvesm 12th November 2012 06:04 PM

Could you measure the resistance of the primary and secondary windings of your transfo ?
If it was designed for use with tube rectifier, secondary resistance if probably too high.


tomchr 12th November 2012 07:23 PM

I'm guessing the sag is caused by the increased supply ripple. I suggest increasing the reservoir cap. A quick calculation estimates the ripple voltage to be about 40 V at max load; 5.8 V at idle. Increasing the reservoir cap to, say, 470 uF would give you 5.3 V ripple at full load...

For a better understanding, I suggest running a sim in LTspice or PSUD II.

From physics: C = Q/V; Q = i*t --> V = (i*t)/C; t = 1/(2*f) for a full-wave rectifier. --> V = i/(C*2f)
C = capacitance in Farad; Q = charge in Coulomb; V = peak-peak ripple voltage in Volt; t = time in seconds; f = line frequency in Hertz; i = load current in Ampere.

Above calculation assumes zero resistive losses, neglects the diode forward drop, and assumes a zero conduction angle. So take it as a worst case ripple estimate. For a more exact value, run a sim.


Mickeystan 12th November 2012 09:10 PM

I agree with Tom's suggestion of excessive sag due to too small a reservoir cap. Try Tom's suggestion and you will see a marked increase in the B+ Mickeystan

kevinkr 13th November 2012 12:27 AM

Excessive winding DCR?

vingenz 13th November 2012 08:30 AM

But if i use a bigger capacitor value, more current is need to recharge the cap
[V = i/(C*2f)] and so more voltage drop. Sag is not on secondary winding (only few volts), but after the diodes/filter cap.

bayermar 13th November 2012 09:28 AM


Originally Posted by vingenz (
Sag is not on secondary winding (only few volts), but after the diodes/filter cap.

That is exactly what you would expect if the reservoir cap is too small.
It's quite commonly used in guitar amps where some designers use that on purpose to create PS sag.

What is coming after the first cap? Choke or resistor?
I can't imagine this is directly connecting to the OPT center tap?!?
Try to sim it in PSUDII once you know the winding resistance of the PT (as tomchr suggested), it's an eye-opener.


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