Bias circuit.what.s the clue?

[catalin gramada]

Hy. Just read few articles how important full wave rectifiers and perfect filtering are into a fixed bias circuit to avoid any trace of psu modulation into signal path. Again guitar amp builders was acused for poor implementation using half wave rectifier and poor filtering. But looking over more contemporany hi.fi amps from their glory era like mc.intosh. bogen. harman kardon. marantz seems they did not understood importance of this advice as well.
Gentlemans what is the clue? Why half wave rectifier ?maybe ripple frequency reason or so Please?Thanks.

[frus]

Maybe because it's..... good enough?

[catalin gramada]

Have not sense.I don.t talk about cheap chinese amps where each fraction of cent is matter .Full wave means charging twice as fast with almost no cost talking about high fidelity amps.This guys overengineered almost everything in respect with performances as I know. The only reason I can imagine is: as time the ripple cannot be total rejected with small-moderate capacitor values-which ensure acceptable charging time-it is preferable to keep it at lowest frequency...best compromise I think...but could be wrong...maybe experts knows better...

[g1]

How about the self-cancelling of same phase signals in a push-pull amp?
This will depend on how well the push and the pull are matched.

[The Dude]

^^^^^^ Yes, that. It's not necessary for the bias supply to be as clean as a 5v supply for a microprocessor. Any slight hum is cancelled by the PP amp. A lot of the old Fender amps had a "tubes matching" or "hum balance" adjustment to make sure this was so- in case tubes were not well matched.

[R.G.]

For a simple rectifier circuit with a capacitor filter, fed with a sine wave like the AC mains waveform, the capacitor only charges until the peak of the AC waveform is reached. When the AC input waveform start to drop towards zero again, the rectifier diode is reverse biased, so the capacitor supplies all the current to the load, and its voltage decreases at a rate of dV/dt = I/C where I is the load current and C is the capacitance. dt is the time until the incoming AC voltage once again turns on the diodes, and may be assumed to be only slightly less than the AC cycle (for half wave rectification) or AC half-cycle (for full wave rectification).

If dt is fixed, and you know your load current, you can reduce the ripple voltage to as small as you want it to be by changing the value of C. Double the capacitance, half the ripple voltage. It's a little more complicated than that, but this is a good first order approximation.

However, it is cheaper to add another few diodes to do full wave rectification than to double the capacitance. This is why full wave rectification is so universally liked - it makes better use of the capacitance because it halves dt.

In a bias circuit, though, the load is indeterminately small. Grid leak resistors are usually in the 100K to 470K range, and that is in series with a tube grid, so the loading on a bias voltage is very, very, very small. In fact, the grid leaks can usually be neglected, and the load thought of as only the resistors that set the actual grid voltage. These are usually in the 47K to 100K range.

Making a guess at bias as 50V, and with a 47K bias setting network, the current pulled from the bias voltage is 1ma. If we want to pick a capacitor that holds ripple down to, say, 2% of a 50V bias, we want C = I*dt/dv = 1ma * 16mS/(50*0.02) = 16uF. So a 22uF cap would hold ripple well under 2% for a full wave rectifier.

The actual numbers will vary, but the message is clear: the load current is so low that ripple on a grid bias supply can be suppressed to any degree with reasonable capacitor sizes even in a half wave rectifier. There's no need to spend the money for full wave.

[66 Kicks]

FW doesn't charge up twice as fast as HW when the resistors and capacitors are the same in both cases.

[R.G.]

I'm a little confused by that.

It's correct, or close - in that FW and HW charge up at the same speed, limited by the AC signal's source impedance, the diode's forward resistance and the capacitor's ESR, not whether the rectification is FW or HW. The charge-up time (at power on?? on each half or full cycle??) is about the same. The charge down speed is the same, governed by the capacitor's voltage and capacitance and the load current. The increased ripple on a HW is purely due to the capacitor having to hold up the load current for twice as long as in the FW case.

So the statement is true, but I have this feeling that I don't understand what you were trying to say.

[66 Kicks]

I'm a little confused by that.

It's correct, or close - in that FW and HW charge up at the same speed, limited by the AC signal's source impedance, the diode's forward resistance and the capacitor's ESR, not whether the rectification is FW or HW. The charge-up time (at power on?? on each half or full cycle??) is about the same. The charge down speed is the same, governed by the capacitor's voltage and capacitance and the load current. The increased ripple on a HW is purely due to the capacitor having to hold up the load current for twice as long as in the FW case.

So the statement is true, but I have this feeling that I don't understand what you were trying to say.

In Reply #3, it was stated that FW means charging twice as fast, which I took to mean at power on. There seems to be a lot of concern that the bias will not reach its full potential before the power tube heaters heat up and the current will run wild. I've never lost any sleep over it myself.

Your question about charge-up time on each half or full cycle kind of caught my attention. If half cycle means what happens with HW rectification and full cycle means what happens with FW rectification and charge-up time means the time that the diode is forward biased and the capacitor is charging, then I don't think the times are the same for HW and FW. In HW, the discharge time is longer than in FW before the rising sine wave reaches the voltage that forward biases the diode. It also means that the forward bias point will be at a lower voltage on the sine wave for HW. This means that the time the sine wave spends charging the capacitor before it reaches its maximum voltage is longer for HW than for FW. I might lose a little sleep on this one if you don't straighten me out. I just noticed that they were talking about negative portions of the sine wave and I gave an example with positive portions of the sine wave. The same thing happens, just turn the paper upside-down.

[Enzo]

No, what they meant, I believe is that a FW rectifier charges twice as often as a HW. 120 times a second rather than 60. Twice as fast is improper terminology for twice as often.

[catalin gramada]

Lots of valuable informations here.Thank You

[robrob]

Maybe because it's..... good enough?

Yep, simple, reliable, inexpensive and good enough.

[frus]

This guys overengineered almost everything in respect with performances as I know.

Probably in those times marketing didn't work like today
"Your amp has FULL WAVE BIAS SUPPLY!!! It cost us like 50 cents more, but we'll charge you $1000 for bragging rights, because it's UNIQUE in this regard! Even the legends like (aforementioned) didn't have one! Not that it's any better, and it's not like many people even understand what the bias supply is, but 'full wave' sounds better than 'half wave', right?"

[robrob]

F*^k that half wave s*%t.

[big_teee]

So is it as simple as just adding the second diode?
Using the same filtering caps, or is there other components that have to be added?
I haven't read the whole thread.
T