Rating for resistors in power supply?

[bluegti]

I think I understand how to determine the power rating for resistors in a power supply, but I want to be sure. The following image is just for example:



According to PSUD, the RMS voltages at C1, C2 and C3 are 456, 413 and 164 respectively.

To determine the power rating needed for resistor R1:

* Calculate voltage drop: 43V = 456 - 413
* Calculate current across resistor: 0.043 = 43 / 1000
* Calculate power dissipation using: I * I * V or ~0.08 = 0.043 * 0.043 * 43
* Choose a rating 2 to 3 times the amount dissipated: 0.24 = 0.08 * 3

Therefore, a 1/4 Watt resistor would be a good choice for R1.

To determine the power rating needed for resistor R2:

* 279 = 413 - 164
* ~0.04 = 279 / 7000
* ~0.45 = 0.04 * 0.04 * 279
* 0.9 = 0.45 * 2

Therfore R2 will require at least a 1 Watt resistor but go with 2 Watts for extra reliability.

While I'm at it... how do I calculate the rating for the plate load resistor in a Class A Grounded Cathode Amplifier?

Using the circuit defined in the 6SN7 data sheet. Assuming a load of 5.1K and a B+ of 180 Volts. Do I calculate the same as above?

~0.035 = 180 / 5100
~0.22 = 0.035 * 0.035 * 180

So the 5.1K resistor would need to have a rating of at least 1/2 watt, but 1 watt for extra safety.

Please let me know if I am doing this correctly. Thanks!

[rotaspec]

Not quite. Power dissipated in a resistor is I*I*R, not I*I*V. So your R1 needs to be minimum rated at 1.85W, and a 5W would do.
If you are given the voltages, you don't even need to calculate the I, just use V*V/R where V is the voltage difference across the resistor.

Gary

[Richard Ellis]

If you 'check' the boxes for current of R1...& R2.. R1 pops up to 230mA briefly then settles down to 65mA....that makes it 52.9 Watts briefly then settles to 4.22 Watts.
The second ...R2 runs 110 mA to 65mA making for 12.1 Watts to a settled 4.22 Watts.
A twenty Watt wirewound will work Ok for R1 & R2. R1 might handle the current "burst"...but you never know, perhaps a fifty Watt.
Try using the different functions, just don't check more than a few of the "boxes" as it gets hard to read.

__________________________________________________ Rick....

[bluegti]

Good thing I asked!

So, it appears I could use V*V/R to determine the rating if there were no large peaks in current (where V is the voltage dropped).

However, since there are large peaks during the initial startup I need to use I*I*R to determine the power rating (where I is the Max current for the resistor).

Here is the actual schematic I am looking at:



According to PSUD:
I(R1-max) = ~119mA
V(R1-max) = ~56V

I(R2-max) = ~23mA
V(R2-max) = ~127V

Max watts dissipated for R1 = ~6.8W = 0.12 * 0.12 * 470
Max watts dissipated for R2 = ~2.9W = 0.023 * 0.023 * 5600

So if I go with 2 or 3 times rule I will need a 20W resistor for R1 and a 10W resistor for R2.

Do I really need this high a power rating for a peak that around 10ms? Doesn't seem like most of the schematics I see have resistors rated this high. What am I missing?

[AndrewT]

I would go for two to three times continuous duty power dissipation.
You'll find that gets close to taking the short term demand with an overload factor that is slightly greater than 1. Hopefully not greater than 2.

[bluegti]

According to PSUD,

V(R1-rms) = 35V
V(R2-rms) = 60V

Therefore power continuous dissipation for R1 = 2.6 watts = 35*35/470
Therefore power continuous dissipation for R2 = 0.64 watts = 60*60/5600

So I could use a 10 watt and 2 watt resistor for R1 and R2 respectively.

If I change R1 to 330R, then V(R1-rms) = 25V so

Power rating would be 1.9 = 25*25/330 so I could use a 5 watt resistor. This keeps me in the ballpark of where I want the B+ to be.

Please let me know if this is correct (and safe). Thanks.

[chrish]

My thoughts:

The power rating of a resistor is an indication of how much heat it can dissipate. There will be a little thermal mass in the resistor, so short startup current peaks should not be an issue unless they are outrageous.

You know the current passing through the circuit, as you have defined it in the model. Just use Ohm's Law power = current^2 * resistance.

Your first resistor of 470R is passing 64+10 mA so 2.57 watts. You want to add a safety factor, typically double, so a 5 watt resistor should be OK.

10 mA through 5.6k is 0.56 watts. You may want a 2 watt resistor.

[SGregory]

For a PSU, especially a DIY system, I feel it is often best to use a minimum factor of 3x power, usually more depending on what the end application is. In my case I play around with the bias quite often. If the supply feeds both channels and I wish to increase the bias current, it doesn't take much to eat up the safety factor since energy is a function of current to the second power.
In a psu, a cheap wire wound is generally more than acceptable. The price difference between 3, 5 and 10 watts is only about $1.00 USD at the extreme. By going bigger, it is nice knowing that there isn't a mini igniter in the chasis...

[chrish]

Point taken about the safety factor, I guess my point was you don't have to get as esoteric working out the actual power dissipation.

Why such a large value for the first resistor? are you looking to drop some voltage? If the power dissipation factors are too high for the parts you can easily get, you can parallel resistors, or you may want to use half the value and series them with another RC network to further reduce ripple. See Morgan Jones power supply section for discussion on this.

Chris

[indianajo]

1/4 watt resistors made in the 70's were about 1/2" long and were fine for tube circuits. (This thread is posted on a tube forum, although the circuit doesn't refer to the load). 1/4 watt resistors being sold now are rated at roughtly 300 Volts max, probably because they are so small. Resistors for tubes should be rated 500 volts. I'm buying 1 and 2 watt resistors even for transistor circuits, because anything in my bins might go on tubes. For the voltage rating, see the resistor selectors on Newark(US) or Farnell (UK) 's website newark.com