Increase filters on Rectifier tubes?

[Raybob]

I understand from RCA tube manuals the ratings for filter capacitance on rectifier tubes, but can't understand at all why? What harm could it do to, say a GZ34 if the filter capacitance was increased to something like 200 mF. Could this cause it to overheat? I really can't understand why, and am about to try it, just to see what happens. Hate to kill a good amp in trying it, but if there's a reason I shouldn't, someone here prolly knows or has been there and done it...

[Dave H]

Increasing the capacitor above the recommended value could exceed the peak current rating of the tube. A bigger capacitor results in narrower but higher peak current spikes.

[Raybob]

Increasing the capacitor above the recommended value could exceed the peak current rating of the tube...

Sounds like a good thing to me. The recto will only produce the current it gets from the transformer. Am I wrong?

[Dave H]

The recto will only produce the current it gets from the transformer. Am I wrong?

It's the charge that's constant not the current. Each half cycle the recto has to replace the charge removed by the power amp. If the cap is larger the voltage will sag less meaning the recto will turn on later in the cycle but it still has to replace the charge removed but has less time to do it so the peak current has to be greater.

[eschertron]

At initial power on, the recto has to 'fill' the cap to full potential. A bigger cap needs more coulombs (joules? whatever) to fill it. The recto works harder. SS diodes don't sweat it, but a vacuum tube recto will over heat and die given enough cycles (power on) that stress the tube. Note that it's the tube, not the "amp" that will fail, and it may survive the first couple power on cycles, and the big inrush current is only at power on. You wanna see what new production rectifier tubes can stand? Go for it. Please report back with your findings.

[R.G.]

All good answers. Here's another take of the same info:
1. Tube rectifiers have a peak current limit. Exceeding this peak current damages the cathode so it can't provide as much current later. How fast the damage accumulates depends on how much you exceed the peak current. But each overcurrent instant adds up.

2. You're pretty certain to exceed the peak current at start up, when all the caps are empty. Tube makers back in the Golden Age figured out how many power-ons was "good enough" to give perceptibly long life for their products, and advised accordingly.

3. Capacitor-input DC filter caps have a bit of mathematical funniness. The caps only get charged up near the peak of the incoming rectified AC half-cycle, but they supply DC for a full cycle. So between peaks, the rectifier diodes are turned off by the cap voltage and what the incoming AC voltage is doing, but the caps supply DC and their voltage runs down. This is the sawtooth ripple voltage you see on your B+. It gets bigger with more load, because with higher current, the cap runs down further.

Since the cap has run down further with higher loads, the diodes turn on earlier in the AC half-cycle and start charging the cap a bit sooner in the AC half-cycle. So low loads give little ripple, high loads give bigger ripple. But you can also make ripple smaller by using a bigger capacitor. It gets charged to the same peak voltage (roughly), but runs down less quickly, because a bigger cap stores more charge. This lower ripple is no doubt what you're after.

But Mother Nature says that there ain't no such thing as a free lunch. A bigger capacitor holds the ripple voltage to be smaller, but because of that, the charge portion of the AC cycle from the diodes gets shorter. So to get the same charge into the cap in a shorter time, the size of the current being shoved in has to get much bigger.

For the same load current, ripple goes down about linearly with capacitor size. But peak current into the cap goes up disproportionately quickly as the cap gets much bigger.

It's those repeated peak currents that can wear out a rectifier much more quickly than expected. Surge at turn-on is one thing, but having EVERY AC half cycle exceed the peak rectifier current is another.

There are sneaky circuit tricks to get around this, of course. But the advice on rectifier tubes and max capacitance amounts to the tube manufacturer saying "if you go above this capacitance, you're going to be buying a lot more recitifier tubes."

[eschertron]

It's those repeated peak currents that can wear out a rectifier much more quickly than expected. Surge at turn-on is one thing, but having EVERY AC half cycle exceed the peak rectifier current is another.

Good to know. I consider myself sufficiently spanked

[Chuck H]

Since the inrush current as the caps are trying to charge is going to be much greater than anything drawn by the circuit for amplification all you really need to do is limit that charging current. R.G. already alluded that there were circuits for doing this. Perhaps the easiest one is the pi filter. From the rectifier tube you filter with a cap that meets the max uf spec for the tube. Then a resistor of sufficient value to limit current to a higher value cap and then to the first HV node. I would love to give a formula for figuring a suitable resistor value for a given larger cap value, but I don't know formulas well enough. On a recent single ended build with a rect. tube I used:

rect. tube/22uf cap/100r resistor/80uf first HV node filter.

I'm only guessing that a larger first node filter may require a larger pi filter resistor to slow the charging time proportionately. Maybe R.G. can clarify that for me.

[DRH1958]

If I remember correctly and have interpreted the data correctly, this max uF rating you speak of is only if the amp draws the full load current the rectifier tube is rated for. If you draw less than that max, then that uf rating can be increased. Since the max current draw is lower than rated, this decreases the current draw that eventually destroys the tube. So a 5AR4 tube that supplies an amp that draws ~125mA like an 18 watt P-P amp, it can have a reservoir cap larger than rated since that tube can supply twice that current.

[g1]

Note that it's the tube, not the "amp" that will fail

If that rectifier tube failure is a short, the fuse may not save the power transformer, so that is something more serious to consider.

all you really need to do is limit that charging current. R.G. already alluded that there were circuits for doing this. Perhaps the easiest one is the pi filter. From the rectifier tube you filter with a cap that meets the max uf spec for the tube. Then a resistor of sufficient value to limit current to a higher value cap and then to the first HV node.

Choke input filter also gets around the max. capacitance limit.

[R.G.]

Good to know. I consider myself sufficiently spanked

No spanking intended or implied. There is a huge inrush at power on, and it does often exceed what a rectifier can do. It's only tempered by the rectifier filaments heating up slowly and acting like a current limiter because the cold cathode can't produce enough electrons. Power-ons are brutal.

Even worse: flipping an amp on and off every second or so, or into/out of standby for circuits that let the caps run down. The heaters are still hot, and the tubes ran down the caps quite a bit, so you get surges limited only by the ability of the rectifiers to conduct. AC30s are BAD for going into/out of standby, to the point where some techs like to disable standby entirely.

[Enzo]

An uncharged cap is a short circuit for the first moment when power is applied. Hit zero ohms with 400v and tell me what current the circuit WANTS to draw.

This is a separate issue from what the amp draws during operation.

[Old Tele man]

Circuit Theory 101:

• at ƒ(0) L is open-circuit; C is short-circuit.
• at ƒ(∞) L is short-circuit; C is open-circuit.

[R.G.]

Choke input filter also gets around the max. capacitance limit.

It does, and does a GREAT job of reducing ripple, too.

Problem is that the minimum current and inductance values are hard to calculate, and when the current rises to where the inductor is doing its best job, the output voltage falls to about 63% of the peak input voltage, unlike a capacitor filter where the voltage stays near the peak voltage until the capacitor is way overloaded (i.e. ripple voltage is high). At zero or nearly so output current, an inductor input filter reduces to a capacitor input filter and the output voltage is near 100% of peak. This falls as output current increases to the level that makes the inductor conduct all the time.

Probably the best of these situations is to have a small-to-moderate first filter cap, then some impedance, like maybe an inductor or a resistor, then a second and larger filter cap. That describes what happens in most amps, but the inductor is after the big current user, the power stage, so it's vastly smaller and cheaper than it would be if it had to supply the output stage currents.

[R.G.]

Circuit Theory 101:
• at ƒ(0) L is open-circuit; C is short-circuit.
• at ƒ(∞) L is short-circuit; C is open-circuit.

??
At f(0) L is a short cirucuit, C is an open circuit.
At F(∞) L is an open circuit, C is a short circuit.