Paralleling Buck Modules?

So I can put a quarter watt resistor in series with a 10,000 amp load, no probs?

(Not talking about parallel)

johnyradio said:

So I can put a quarter watt resistor in series with a 10,000 amp load, no probs?

Click to expand...

Sigh. Let me ask you this: Do you think you need a series resistor when you are only using a single, unparalleled regulator?

Answer: You don't. So you would never put that quarter there to begin with. You just wouldn't have a resistor there because you don't need a series resistor with a single regulator, because the resistor is only there for current balancing but current balancing isn't needed if you only have one regulator, ergo no resistor.

You asked this:

johnyradio said:

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?

Click to expand...

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.

Becaise you only have a single regulator, therefore no need to current share, therefore no resistor.

That is the question I have been answering these last few posts, and as far as I can tell you are asking the same question over and over again but not understanding the answer I am giving.

Sure, if you wanted to put a resistor there then yes, it would need to be 10.8W, but there are no benefits, and only detriments to such a resistor if you only have a single, unparallel regulator, so just don't put one there. Use a piece of wire instead.

dknguyen said:

If you wanted to put a resistor there then yes, it would need to be 10.8W

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Thx for finally answering the question.

You didn't have to keep repeating "you don't need it". I kept repeating "i got that".

So, if drop affects resistor rating with a single reg, why doesn't drop affect resistor rating with parallel regs?

The

johnyradio said:

Thx 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?

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If you are asking, what I think you're asking, it does. It's just hidden away in the equations

Power = VI
V = IR

That means that Power = V^2/R = I^2*R.

It just depends whether you want to (or whether it makes more sense for your specific problem) to use V or I, since between V, I, R, if you define two of them you automatically imply the third.

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

johnyradio said:

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

Click to expand...

(V^2) / R, but seems more like you just mistyped it since your answer is correct.

Also, the datasheet said +/-4% initial accuracy and 4% over temperature. So worst case is that one regulator is 4% high and another is 4% low out of the box and then temperature overlapped with that (though it's not clear how from the datasheet). But that doesn't as much as you might think because in this case your regulators are manually adjustable so you can adjust them to be really close using the same multimeter and then (hopefully) they all drift in the same direction with temperature. With manual tuning you would hope that 4% is as bad as it can get and not 8%.

I would think you would have better results since you are manually tuning and the devices should drift in the same direction with temperature (and they should be similar temperature too if they are load sharing properly). So you can probably reduce the R in reality from what you calculated but you won't know by how much unless you test it out.

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

they all drift in the same direction with temperature

Click to expand...

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.

ronsimpson said:

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.
______________
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.

Click to expand...

To Ron's excellent suggestion, I would suggest a minor circuit improvement by using a dual diode package for the rectifiers. That way the forward voltage drop change, due self heating will closely track between both diodes.... A minor improvement.

dknguyen said:

your answer is correct.

Click to expand...

Thx. But then I don't understand your previous comment

So I can select resistors with adequate power rating.

Click to expand...

You only need Wdif for that. Not Wreg since the power is dissipated in the resistor.

Click to expand...

Drift: do the regs drift more in parallel than singly?

johnyradio said:

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.

Click to expand...

I must have been crossed eye. I looked at that three times.

Ehhhh, most of the time things are listed in the datasheet because they're either:
1. so bad that they don't list it because the badness is not important for the expected application
2. they just didn't bother spending the time or money to test it too carefully, sometimes (many times?) because it's difficult or not too important for most applications

johnyradio said:

Thx. But then I don't understand your previous comment



Drift: do the regs drift more in parallel than singly?

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Ignore that comment. I thought you meant something different by Wreg at the time.

Regulators drift whatever they will drift. It doesn't really matter if they're in parallel or not.

You can kinda see why now paralleling converters that aren't meant to be paralleled using current balancing resistors is non-ideal. I'd just go with a bigger one or some that can be synchronized together 95% of the time. Ron's suggestion is the the most ideal modification you can make. The main downside is you have to lift pins and run wires between the boards.

In my scenario, it's about size and cost. The two 3A modules in my OP are significantly smaller and cheaper for two than any single 6A module I can find.

I don't understand why you don't use the method in the datasheet as Ron suggested.

If instead you use 1Ω 3W resistors then at full load your output voltage will be 3V lower than at no load. If you use 1/3Ω 1W resistors then it's better - a 5V no load output will drop to 4V on full load.

Mike.

Pommie said:

I don't understand why you don't use the method in the datasheet as Ron suggested.

Click to expand...

Thx Mike. For my use, modding the modules may not be possible, especially the MP board.