Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: mxrshiver on January 02, 2026, 02:50:26 am
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any clue why someone would do this instead of just not connecting the CT?
voltages are supposed based on the transformer and circuit, not measured - this thing's got a lot of issues needing attention first. fairly typical 4x EL34 Marshall JCM style power section overall.
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any clue why someone would do this instead of just not connecting the CT? ...
Connecting the CT to what?
The resistor has no apparent function / purpose.
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sorry, i meant as opposed to just not using the CT, as with a normal full wave bridge.
i'm scratching my head about what the purpose might be, too... or what the effect might be, and if it might be problematic.
the transformer is 165-0-165. so this resistor would allow about 6uA to bypass one of the two conducting diodes and one half of the winding, on each AC half cycle.
if the resistor was lowered in value all the way to a wire, then it'd become a two-diode two-phase instead, current would no longer flow in the two diodes between the transformer windings and ground, and the rectified DC output would be halved compared to the bridge.
...seems like this could be a quick and easy way to slightly bring down the B+? without resorting to a resistor in series with the B+ after the rectifier, it avoids any significant sag, or the necessity for a large power resistor dissipating a lot of heat.
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sorry, i meant as opposed to just not using the CT, as with a normal full wave bridge. ...
Are you getting a full wave bridge rectifier mixed up with 2 phase full wave?
The latter must have a transformer winding with a grounded CT.
Grounding the CT in post 1 image will draw fault current.
https://www.valvewizard.co.uk/bridge.html
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if the resistor was lowered in value all the way to a wire, then it'd become a two-diode two-phase instead, current would no longer flow in the two diodes between the transformer windings and ground, and the rectified DC output would be halved compared to the bridge.
That's not quite right. The first thing to happen will be the two diodes that connect to ground will be destroyed. If they should happen to fail open, then yes, the FWB circuit becomes a conventional full wave rectifier with half the dc voltage output. BUT, if the diodes fail shorted, then your PT will try to smoke. Hopefully your fuse will do it's job!
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i understand a two diode two phase rectifier would have a grounded center tap, and a bridge wouldn't. that's the confusion with this particular circuit, because it's a bridge with the center tap grounded through that resistor. in trying to figure out why this resistor was included, i was trying to imagine what would happen if the resistance value was reduced, all the way down to a wire. apologies for phrasing that poorly.
hah, i was totally missing the fact that that would connect the CT directly to either side of the winding on each half cycle, resulting in magic smoke...
my best guess with that resistor is still that it was an attempt to slightly lower the B+. think i'm just gonna clip it out. if anyone has any further ideas about why this resistor was included or what effect it would have, though, i'm very curious. thanks folks!
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... my best guess with that resistor is still that it was an attempt to slightly lower the B+ ...
If so, it was completely wrong headed, that resistor wouldn't make a blind bit of difference on a beefer transformer suitable for a typical 4xEL34 amp.
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I tested this circuit using a Hammond 290DAZ on a rheostat to give me 430VDC off the bridge using a 10uF 450 Cap and 100K 3W load [resistor]. The bridge does not short with the CT grounded through a 270K 2W resistor. The output voltage does not change either with the CT connected or disconnected. What I did find interesting is the AC feeding the bridge does change. With the CT disconnected, I measured 151VAC to one side of the bridge while 166VAC on the other. With the CT connected to ground through the 270K [
was], I measured 173VAC on both sides of the bridge (again with no change in output VDC). This was only one test but it looks like this arrangement balances the AC to either side of the bridge. Whether this was the intended purpose and what advantage it would ultimate provide, I am unsure.
[Edited]
[I did a second test increasing the load using a 10K 40W load resistor. Although I believe I saw a couple 10ths of a voltage drop on the output with the CT connected, it is pretty much meaningless. Voltages across the bridge were 166VAC and 173VAC with the CT disconnected and 181VDC both sides with the CT connected.]
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I measured the DC current from either side bridge. I measured an average difference in current of 0.55mA without the CT connected, and 0.05mA with the CT connected.