Ignorant Question #637 - Cathode Resistors

[solderstain]

This place is having both a positive effect (I'm learning a hell of a lot), and negative effect (I feel more stupid every day) on me.

Question about Cathode Resistors in a cathode-biased output stage:

I've done cathode-bias builds before, but have never dug into the thinking and theory behind some of it all. All of my cathode-bias style builds to date have been 2xEL84 based. Since 'everyone else does it', I dutifully use a minimum of a 5w resistor for the cathode bias resistor.

And I don't understand WHY it needs to be such a big, honkin' resistor. Doing basic power calculations based on voltages, etc. from my builds, it SEEMS that even a 1w resistor is overkill by at least 100%.

So I'm now clear that I'm missing something REALLY BIG in my math calculations. From my current project, (2xEL84), I have the EL84s sharing a 100 ohm (at the moment) cathode resistor. The voltage drop across the cathode resistor is 9.15 volts. The plate voltage is 304 volts.

Someone show me the error in my thinking that I COULD get by with a nice, 1w Dale resistor for my cathode resistor instead of the 5w resistors I have been using.

[tubeswell]

100R at 9.15V = 9.15/100 = .0915A (or 91.5mA)

.0915A x 9.15V = .84W. Double that for a safety margin = 2W minimum

You could use a 1W, but you are sprayin' close to the wind

But 5W won't do any harm

[tubeswell]

Dang double posts

[Phil_S]

9v is at idle. I don't have any idea what it rises to in use, but I wouldn't be surprised if it doubles. If 2W is the safe size at idle. 3W is probably still not beefy enough. You don't want this thing to fail while the circuit is hot. Use the 5W.

The concept of "derating" is what were getting at here.

[Structo]

All you gotta do is feel that resistor after playing for a while.

I use 10 watt resistors there and it gets pretty warm.

[solderstain]

Okay, guys... thanks.

So far, my math isn't off. But maybe some of my operational concepts are.

I know that the 91.5ma is at idle, and I know it goes up, but I didn't think it would go up much. I didn't think at all that it doubles. I've been operating under the assumption that an output stage of this configuration is already idling at about 90%, and won't rise 'much'. Do I have that wrong?

And I don't disagree about doubling the safety margin, and I'm not saying I won't continue using the 5w resistors I've been using... just think of this question as coming from your five-year-old kid going through their "Why? Why? Why? Why? Why?" stage of life (before they get to the age where they think you don't know anything... )

[Structo]

I think you are right about the cathode current staying relatively stable on a cathode biased amp.

[Kregg]

So ... should we think of it as a heat sink which helps to prolong the life of the tube?

[Structo]

Well it is dissipating a good deal of current and the common rule is to go double what it calls for.

In my mind, I would rather have a 10 watt resistor getting warm verses a 2 or 5 watt resistor getting hot.

And don't forget that heat can cause a resistor to change in resistance.

[tonestack]

So ... should we think of it as a heat sink which helps to prolong the life of the tube?

No, increasing the power rating merely protects the cathode resistor from failure under heavy load. A 3W resistor is more than large enough for a 2xEL84 amp.

If one measures the voltage across the cathode resistor at idle and at peak drive voltage on a cathode-biased EL84 amp, one finds that it rises around two volts. The cathode-biased topology is self-limiting because the bias voltage becomes more negative as cathode current increases. At some point, the amp will transition from Class-A to Class-AB operation; however, the cathode current cannot rise indefinitely because it will eventually rise to a point where the negative bias voltage developed across the cathode resistor increases to a point where it limits cathode current.

[Andy Le Blanc]

The cathode resistor one the power side can be the one single component
that must handle all of the power going thru the power tubes. With a pair of
EL84, thats in the neighborhood of 28 Watts plate dissipation. Now this
statement is misleading, the reality of the math works out differently, but it
makes a fairly safe assumption. All manufactures at some point have used
the least expensive parts they can get away with, with resistors that means
the lowest watt rating you still sell to customer without them experiencing
reliability issues. So if you do the math, you can get away with 3w to 5w
that you can source for as low as 25 cents per unit, you can get up to a 25w
for less than a dollar per but then you run into a mechanical reliability issue
around the size of the part and how best to mount it. Most older amps of any
quality would have something like a Ohmite 200 or 270 series tied to a
component board, but if reliability was the absolute goal then the resistor
would be mounted to the side of the chassis with the appropriate hardware
and an oversized watt rating. Many current makers choose aluminum housed
wirewound power resistors, they solve both issues of power handling and
mechanical reliability. Id get a good handful of different value 10w cement
resistors and 1%1ohm to measure current across, so you can noodle
to find the operation point and bias that best suits your ear,
then make your choice of bias resistor and watt rating, even the aluminum aren't too bad
for $$$ the cost per unit doesn't jump until you get to 75w, in the current
Mouser pricing the 25w are less than 5$.

[tubeswell]

I think you are right about the cathode current staying relatively stable on a cathode biased amp.

Yeah you're right - the cathode voltage hardly changes at all - especially if the cathode resistor is bypassed - it stays almost the same (and that is one of the main reasons why we bypass them after all). If it is un-bypassed, it will go up and down a little bit following the voltage swing at the anode, thereby providing internal feedback and reduced gain from the stage, but the cathode voltage doesn't double.

[Andy Le Blanc]

Its not the voltage you have to worry about. The amount of heat that is
generated depends upon the square of the current, the resistance and time,
and must not be confused with the temperature to which it will eventually rise.
Its expressed in power units, E2/R = I2R = EI = P. P is the power lost in the
resistance as heat, over time it really adds up. You lose the heat at the same
time to convection but do really want to exercise a 3W component in that manner.

[Andy Le Blanc]

getting back to the stated numbers, and assume that the EL84 are biased
near maximum plate dissipation, 90 ma. at 304v gives 27.36w tot. plate
dissipation for two tubes, and the stated 9.15 v at the cathode. So P = EI
9.15 x .09 gives .82 which looks like you can get away with a 1w resistor.
But it is a power unit, this means the rate of doing work. The same total
amount of work may be done in different amounts of time. If all the work is
done instantaneously more energy will be converted to heat more than if the total
amount of work is done over a longer time. The watt means the rate at which
electrical energy is changed into another form of energy, such as heat or light.
When power is used in a resistance, the rate at which electricity is changed to
heat increases and the temperature of the resistance rises. The larger the wattage rating,
the more easily the device will absorb and give up heat.

I ran across an old army surplus projection set a while back all original,
the tranny's were dated 53', one bad bathtub in the power side. The
pair of 6l6 were cathode biased with a 32w ohmite. If it died in the field
the ONLY thing that need be replaced are tubes. Going large here pays in the end.

[tonestack]

The power level at which a resistor must be rated is based on the average power it must dissipate, which for a cathode resistor is the product of the square of the current flowing through the resistor times the value of the resistor in Ohms. The average power a cathode resistor on a 2xEL84 amp must dissipate is ~1W. Doubling this value provides a large enough safety margin for normal operation, and we want this resistor to open under abnormal operation. A resistor costs less than a dollar. A transformer costs serious money.

A 2W Xicon metal oxide resistor is rated for -55C to +235C ambient temperature operation. The power derating curve starts at 70 degrees C. However, at 155C, a 2W Xicon power resistor can still dissipate 1W of power. Other components will start to fail if we allow the inside of our chassis to reach 155C.