To obtain a negative voltage from a transformer winding, is it nesessary for it to be centre tapped?
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Originally posted by bunyip6 View PostTo obtain a negative voltage from a transformer winding, is it nesessary for it to be centre tapped?
If you're trying to get a negative voltage from a non-centre tapped winding with bridge rectifier then you'll need an extra cap+resistor to feed the bias diode with AC.
I'm sure some Marshall amps do it often, although this is the only example I can find at the moment, and it's a bit of a confusing schem. http://schematicheaven.com/marshalla...30w_jtm310.pdf
You can see C127 and R123 supplying the bias diode D107, from the HV winding on the transformer, which isn't centre tapped. The rest of the supply can be a standard Fender style bias supply.
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Originally posted by Merlinb View PostNo it isn't necessary.
If you're trying to get a negative voltage from a non-centre tapped winding with bridge rectifier then you'll need an extra cap+resistor to feed the bias diode with AC.
I'm sure some Marshall amps do it often, although this is the only example I can find at the moment, and it's a bit of a confusing schem. http://schematicheaven.com/marshalla...30w_jtm310.pdf
You can see C127 and R123 supplying the bias diode D107, from the HV winding on the transformer, which isn't centre tapped. The rest of the supply can be a standard Fender style bias supply.
Can you or anyone else comment on the reasoning behind the bias circuit shown in the linked scheme? I've seen this in other Marshall circuits before and wondered about it.
The path from the HV winding is 4K7-47nF-56K (R107, C127, R133) to ground, with the bias supply diode at the top of the 56K.
Why is that network there? While there is a low-pass shelf filtering effect, there is hardly any attenuation.
Then, the series of two 15V and one 9V Zener to ground after the bias supply diode fixes the half-wave bias voltage at the junction of the 220K bias feed resistors at -39V, and it is going to be essentially a square wave, no? The 10uF from there to ground is for filtering, but what is the 100K for?
Thanks,
MPM
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Ok, firstly you need to appreciate that with a centre-tapped winding on the PT, each "leg" of that winding swings positive and negative. The positive swings feed the B+ rectifier. The negative swings can feed the bias rectifier.
BUT! With a non-centre tapped PT with bridge rectifier each leg only swings positive! The negative swing is clamped to -0.7V by the bridge rectifier, so there is no negative voltage to feed a "Fender style" bias supply. So how do you get bias?
You use a coupling capacitor to feed the bias supply- the "positive only" pulses get turned into AC again on the other side of the cap. After that the bias supply is perfectly ordinary.
In the PDF I linked to, there are a few other bits to distract you:
The 4.7k resistor (R107) is just a current limiter, which I'll come back to.
C127/R133 are the coupling cap and ground-reference (it won't work if R133 is removed).
The string of zeners prevent the bias supply from ever exceeding -39.1V. Obviously that's the bias voltage they were aiming for. Now, normally you need a resistor to feed zeners, so where is it?
Answer: It's the 4.7k resistor (R107). Ok, they chose to put it right at the start of the circuit, but so what? Still does the same job.
The 100k (R132) looks like a bleeder to me.
However, the other possibility is that R133 and R132 together set the bias voltage like any potential divider, and the zeners are there just to protect the smoothing cap from over-voltage at start-up or during standby. But the fact that they are set to an odd value of 39V makes this unlikely to me.Last edited by Merlinb; 08-23-2009, 02:10 PM.
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The 10uF cap is in parallel with the zeners, so it will be smoothed DC- no tops at all really!
Here's a sim.
The brown line is the voltage on the "leg" of the PT.
The green is the voltage right after the coupling cap, across R133.
The light-green line is the bias voltage, except I used much small smoothing cap (100n) so you can see what's going on better. It's clamped at 39V by the zener, and the cap fills in the gaps.
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Wow, awesome! The end result is pretty much what I imagined, with relatively narrow pulses (time at 0V longer than time at -39V). I'm puzzled by the shape of the brown line, though. That's the voltage after the bridge, yes? I'd thought that that would just be a FW rectified shape. You've been more than patient, but is there a quick answer to the reason for the two flats at ~360 and ~40V? I guess that is the Zeners conducting, but I can't quite make sense of it.
MPM
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I haven't figured that out yet. It is to do with the presence of the coupling cap network (the flats aren't present without it), but exactly what the interaction is I'm not sure yet. I suggest you download a sim program and have a play with it yourself.
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