johnyradio
Member
So I can put a quarter watt resistor in series with a 10,000 amp load, no probs?
(Not talking about parallel)
(Not talking about parallel)
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Sigh. Let me ask you this: Do you think you need a series resistor when you are only using a single, unparalleled regulator?So I can put a quarter watt resistor in series with a 10,000 amp load, no probs?
As I have been saying the entire time these last few posts, the answer is NO. You do not need a 10.8W resistor because YOU DO NOT NEED A RESISTOR AT ALL. When I say you don't need a resistor, I am not saying you don't need a 10.8W resistor and can use a smaller one. I am saying you do not need a resistor at all. You don't need a 1/4W resistor, you don't need a 10.8W resistor, you don't need a resistor. You use a piece of wire.You misunderstood. Forget the parallel regulator for a minute. Different scenario. Let's say I've just one regulator, with a 1.3 ohm series resistor on output. The R serves no purpose. If my load is pulling 3A, that's a 3.6V drop, right?
3A x 1.3R = 3.6V drop.
3.6V x 3A = 10.8W
In that scenario, don't I therefor need a 10.8W resistor?
Thx for finally answering the question.If you wanted to put a resistor there then yes, it would need to be 10.8W
If you are asking, what I think you're asking, it does. It's just hidden away in the equationsThx for finally answering the question.
You didn't have to keep repeating "you don't need it". I got that.
So, why doesn't drop affect resistor rating with parallel regs?
(V^2) / R, but seems more like you just mistyped it since your answer is correct.Ok. So getting back to my earlier question, to calculate heat, I Should
include Wreg + Wdif. Not just Wdif.
Wreg:
3A x 1.3R = 3.6V drop.
3.6V x 3A = 10.8W
Difference current:
0.4V / 1.3R
0.3A
Wdif:
0.4V x 0.3A = 0.12W
total heat:
10.8W + 0.12W = 12W .
Need 12W or higher resistors.
Correct?
Thx
Ah really? Then, assuming they drift equally, the only difference is how well i sync them manually. Which could be a lot less than 0.4v or even 0.2v.they all drift in the same direction with temperature
If you parallel PWM with different error amplifiers one supply will pick up the load first and will go into current limit before the next will start working. If you supply will work at full power this is OK.
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Look at your first circuit. The COMP pins are wired together. Only one error amp is being used. (Check to see if the IC you are using can do this)
This forces all the ICs to have the same duty cycle, thus almost the same current.
Many error amplifiers pull down hard but have very little pull up power. This is done so an outside signal can over ride the amplifier.
Thx. But then I don't understand your previous commentyour answer is correct.
You only need Wdif for that. Not Wreg since the power is dissipated in the resistor.So I can select resistors with adequate power rating.
I must have been crossed eye. I looked at that three times.I didn't mistype. I used ohm's law and power
I = V/R
P = VI
Difference current:
0.4V / 1.3R= 0.3A
Difference power:
0.4V x 0.3A = 0.12W
Ah really? Then, assuming they drift equally, the only difference is how well i sync them manually. Which could be a lot less than 0.4v or even 0.2v.
Maybe datasheet doesn't list drift because it's really really tiny.
Ignore that comment. I thought you meant something different by Wreg at the time.Thx. But then I don't understand your previous comment
Drift: do the regs drift more in parallel than singly?
I don't understand why you don't use the method in the datasheet as Ron suggested.