Estimate negative bias voltage

[überfuzz]

How do I calculate the negative voltage rendered if I tap one of the secondary leads of the power transformers 'fender style'. And, how do I estimate how much peak to peak ripple is this negative potential going to have?

[überfuzz]

I assume that the 'size' of the capacitor is going to determine the ripple and that the resistors how fast the idle voltage is going to be reached.

[Justin Thomas]

I'd like to know this, too. I bet I could use Ohm's Law on a few known amps to find out how much current a bias supply draws, then extrapolate a rough guess for a #power tubes etc. I've just always heard a bias supply "draws peanuts for current." If there's an easier way, I'm listening!

Justin

[g1]

Which Fender do you mean? Many of them have a dedicated bias winding, but I think you mean the ones that take a feed off the high voltage winding. You will see there is a resistor in series before the bias rectifier diode and a resistor to ground after the diode.
The resistors form a voltage divider and can be used to adjust the bias voltage.

[Tom Phillips]

Duncan's PSU designer SW application yields data that closely matches actual measurements IF you input the correct parameters. Once you have the parameters set up you can easily experiment with changing one parameter such as the filter capacitor value to see the resulting effect. You can download the free SW at PSUD2

[überfuzz]

Thanks for the input. I still don't know how to calculate it though.

Anyhow, I had a brief look at the duncanamp software. I tend to use other kinds of software when I simulate amplifiers and the magic going on in them. Here for instance is a simulation of the power supply of an amplifier (just something I made up, no particular model).

[R.G.]

The transformer winding can be modeled as an AC voltage source with some internal resistance. For most HV windings, the internal resistance will be in the range of one to a few hundred ohms. This will go through some resistance to a diode, and into a cap. The cap then powers a resistor string.

The resistors that lead to the grids can be considered as going to an open circuit unless you drive the tubes into grid conduction, so ignore the grid leak resistors and grid current.

The cap's effective load is the resistor string that makes the partial voltage that sets bias from the full capacitor voltage, which must be larger (i.e. more negative) than the grid bias voltages. So the capacitor voltage needs to be notably bigger (i.e. more negative) than the bias voltages. I like to use cap voltages about -65 to -75V, but that's a matter of preference.

The current out of the cap is determined by the series/parallel combination of resistors that set the tubes' bias. This is fairly straightforward to calculate. The cap value sets the ripple as Vripple = Idt/C, where I is the resistor current, dt is half your local power line *period* and C is the cap.

You can set the voltage into the cap a couple of ways. One is to use a resistor divider between the winding and the rectifier diode. This has the advantage of being easy to calculate, as it makes an equivalent voltage source into the diode equal to the winding voltage times the resistor divider, through an equivalent resistance equal to the parallel combination of the two resistors in the divider.

Another way is to use a single resistor to the diode. This is harder to calculate, as its action is to choke off much of the input current and spread it out. This is best set by trial and error with some calculation and/or simulation.

Another way is to use a series capacitor to the diode. This gets tricky, as you have to worry about the build up of DC voltage on the series cap as it's just as rectified and filled by the diode as the cap you're trying to fill for the bias voltage. You have to figure out how to empty it on the opposite half cycle.

Each of these has a slightly different method of calculation. The resistor divider is easiest to calculate, of course.

[überfuzz]

Yeah, I used a software called LTspice. I'll attach a *.png of the circuit I used to model the voltages in the power section.

[R.G.]

For that circuit, the 100K in series with the diode and the 47K shunt with the diode form a voltage divider when the diode is on, and a simple 47K load on the cap when the diode is off. The charging voltage is Vsecondary (instantaneous voltage, not RMS) times (47K/147K), or about Vsec/3, and it is charging the cap through a 31.9K resistance equivalent during the time the diode is on. It might be good to put another 100K/diode to the opposite secondary lead, as this would more than double the effectiveness of your capacitance by making it a full wave rectifier. Takes one more diode and one more resistor.

You'll probably have to vary the size of the "100K" resistor down a bit because the time constant of the equivalent 32K charging resistance with the "25uF" cap value is long - I make it about 0.8 seconds, and the power line runs on 8.3mS or 10mS half-cycles. Some tinkering will be needed.

It's possible to simply write the loop and node voltages and come up with an analytic expression for the cap voltage as a function of time, but it's VERY tedious. I spent way too much of my life doing that in college. And in fact, writing and solving the node and loop equations is what simulators do.

[überfuzz]

Quite right! Eh, well ,almost right. It's actually even lower than that. It's an old Fender power supply circuit. I'm not sure but I think the high, or big, resistors was used to even out the voltage. I guess your advice goes along nicely with some beefy capacitors.

[überfuzz]

This is slightly off topic, but now when I started to talk about the size of capacitors in the power supply section. How does the size of capacitors affect the amplifier. My point being, why now just used some BIG capacitor after the rectifier so that the supply voltage is super smooth?

[R.G.]

There's a subtle problem hidden here.

If the bias takes too long to come up (er, down?) then the power tubes may be running without bias until it does. This turns it into a horse race between the heaters and the bias supply. You want the bias supply to be stable before the heaters bring the power tube emissions up. Very long cap time constants will keep bias too small for a while and let the tubes run low- or no-biased at each power on.

You're probably OK, as tube warm up is about 10-15 seconds usually, and even with a 0.8Sec time constant, five time constants is 4 seconds; but it's something to think about and check.

[überfuzz]

I thought about that. I'm not sure a sec or two is enough to bust a power tube. I don't have the balls to test it either.

[Dave H]

Before I had LTSpice I used to guesstimate the bias voltage by assuming the cap will eventually charge up to the average value of the half wave rectified sinewave and then divide that by the potential divider ratio to get the bias voltage. For the circuit in post #8 that would be 366V*0.65 /2 (for the half wave rec.) then divide that by 3 (for the 100k, 47k divider) which gives 39V. To guesstimate the p-p ripple divide 366V by the ratio of the 100k resistor to the impedance of the cap at the ripple frequency (25 or 30Hz) which is 266/(100k/250) = 0.67V p-p

[überfuzz]

I did calculate it like that a voltage divider, but figure I was to far off to be satisfied with the result. Ah well, I guess it's close enough to use as a start and iterate until the right components are established. Kind of what I do now... :-|