Current Limiting Resistor Power Rating

[schoolie]

Hello all,

I want to limit the maximum rectifier plate current, so I'd like to install two 33 Ohm resistors on each leg of the HT secondaries, before the rectifier. This is in a tweed Bassman.

1. I don't think the RMS current should be any higher than 65mA in the HT secondaries, so I think two 5W 33R resistors would do fine. However, I've seen a lot of information saying that 10-20W resistors should be used here (Maybe they are with higher resistance values?). PSU II shows the steady-state, maximum rectifier current is around 300mA and spiky.

2. Is there an difference i resistor requirements, if the resistor is in series with the cathode vs two resistors in series with the anodes? I'm thinking that in series with the anodes, I only need a 5W rating, but would need a 10W resistor if I placed the resistor between the cathode and the reservoir cap?

The reason for doing this is that I want to try out some different rectifier types, like 5R4 and 5V4 which are not as robust as the 5AR4. I also have SS diodes, in series with the rectifier, so maybe that will help too?

Thanks!

Rob

[Alan0354]

I just joint in, don't take me as advice.

Look a 6L6GC, for 450V, AB1, max power is 55W, max signal plate current is 210mA. So back calculate, assume 210 mA is RMS. You have one 33 ohm on one HT wire of the transformer. For bridge rectifier, half the cycle you source current, the other half sinking current. So you basically work 100% of the time. Voltage across the resistor is 0.21A X 33= 6.93Vrms, therefore power is:

P=Vrms^2/R=(6.93X6.93)/33=1.455W.

Seems like you have some margin with a 5W resistor.

[Chuck H]

I want to limit the maximum rectifier plate current, so I'd like to install two 33 Ohm resistors on each leg of the HT secondaries, before the rectifier. This is in a tweed Bassman.

1. I don't think the RMS current should be any higher than 65mA in the HT secondaries, so I think two 5W 33R resistors would do fine.

If you intend to place a cieling on the potential current flow you'll need to get a lot more advanced than a simple resistor. Unless the resistor is rated at a wattage so as to make it act like a fuse. Otherwise all the resistor can do is add current by using watts to impede the current flow demanded by the tubes. Watts are watts whether a resistor or a tube is using them.

Why do you feel you need to place a hard cieling on the potential current?

[schoolie]

If you intend to place a cieling on the potential current flow you'll need to get a lot more advanced than a simple resistor. Unless the resistor is rated at a wattage so as to make it act like a fuse. Otherwise all the resistor can do is add current by using watts to impede the current flow demanded by the tubes. Watts are watts whether a resistor or a tube is using them.

Why do you feel you need to place a hard cieling on the potential current?

Thanks, Alan and Chuck! I want to use a 5V4 or 5R4 rectifier to bring down the voltage a bit for 5881s. I'm concerned about arcing at power up, so I just want to add a resistor to bring the initial current surge down a little.

[tubeswell]

You read Merlin's blurb on limiting resistors here? Two-Phase, Full-Wave Rectifiers

[schoolie]

Thanks, Tubeswell. Yes I did. He wrote that even a 7W resistor would get toasty.

[Alan0354]

Thanks, Tubeswell. Yes I did. He wrote that even a 7W resistor would get toasty.

I don't understand why it get's toasty!!! What did I do wrong in my calculation?

[Enzo]

Toasty doesn't necessarily mean burnt.

He doesn't have a bridge rectifier, he has a full wave tube rectifier with center tapped transformer. I am not into doing the math, but I would imagine your calculation ignored the steady state currents, When signal currents go to zero, the tube is still dissipating from idle conditions.

But as to toasty, 2 watts is a lot of power, it makes somethinjg dissipating it hot. The thing may be perfectly capable of dissipating 10 watts, but that doesn't mean 2 watts won't make it hot.

[Alan0354]

Toasty doesn't necessarily mean burnt.

He doesn't have a bridge rectifier, he has a full wave tube rectifier with center tapped transformer. I am not into doing the math, but I would imagine your calculation ignored the steady state currents, When signal currents go to zero, the tube is still dissipating from idle conditions.

But as to toasty, 2 watts is a lot of power, it makes somethinjg dissipating it hot. The thing may be perfectly capable of dissipating 10 watts, but that doesn't mean 2 watts won't make it hot.

I used the max current spec from the data sheet which is 210mA. That's a lot of current!!! In fact it cannot even be continuous as it is way over the safety limit of 6L6GC. The spec use B+=450V. For 210mA, the power dissipation is 450VX0.21A=94.5W!!! That's way over the pair of tube can give.

In fact the max current can only happen instantaneously, not continuously. I made the very conservative calculation be specifying that I assume it to be RMS instead of peak.

[Alan0354]

You read Merlin's blurb on limiting resistors here? Two-Phase, Full-Wave Rectifiers

I read the article, it said you only need limiting resistor for rectifier tube, not SS diode. It did not explain why. What is the reason only the rectify tube need limiting resistor. Is it because of the peak surge current during the cycle when the sine wave is at the peak?

The solid state rectifiers can take big surge on power up and peak of the sine wave easily. Do you mean the rectifier tubes has problem with peak current?

[Enzo]

Yes, that is why rectifier tubes usually specify the maximum allowed input filter cap size.

[trobbins]

Resistors in that position take a complex current waveform, which as you have appreciated can be quite spikey - the rms current level varies depending on a variety of filter parameters - and can be up to 1.6x the dc level for capacitor input filtering. The 'smoother' the current waveform, the lower the rms level - so valve rectifiers are easier on the resistor than ss diodes. It is not an easy task to accurately calculate the rms level, and much better to be on the very conservative side.

[schoolie]

Thank you for your excellent advice, all. Better to overbuild with 10W resistors, but I don't think they'll fit. Regarding prevention of rectifier arcing, would there be any difference if the resistance were before the anodes versus after the cathode?

[loudthud]

The spec use B+=450V. For 210mA, the power dissipation is 450VX0.21A=94.5W!!! That's way over the pair of tube can give.

You are neglecting power delivered to the load. A 50W amp delivering a rail to rail square wave to a resistive load will pull slightly more than 100W from B+. Under those conditions the tubes have very little dissipation.

PSUD2 (or another modeling program) will calculate the RMS secondary current.

[Alan0354]

You are neglecting power delivered to the load. A 50W amp delivering a rail to rail square wave to a resistive load will pull slightly more than 100W from B+. Under those conditions the tubes have very little dissipation.

PSUD2 (or another modeling program) will calculate the RMS secondary current.

I am doing the worst case calculation, not really worry about the tube output. The point is 210mA of continuous current is much higher than real life situation, so it is a very conservative calculation.

Now the question is the extra RMS current of the ripple after the rectifier and before the filter cap smoothing it out. The cap smooth it out to an average voltage, but there is a kind of a double sine wave after the rectifier that pump current through the resistor.