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Paralleling Buck Modules?

Thread starter #1
Hi

This TI note says we can parallel the LM2587.

http://www.ti.com/lit/an/snva552/snva552.pdf

parallel-LM2587.JPG

Is it safe and functional to parallel these off-the-shelf buck modules?

LM2596:
https://www.ebay.com/itm/312344302763

https://www.monolithicpower.com/pub/media/document/MP1584_r1.0.pdf

LM2596.jpg

MP1584:
https://www.ebay.com/itm/282737349512

https://www.monolithicpower.com/pub/media/document/MP1584_r1.0.pdf

LM1584.jpg

i'm wondering, if their voltages are different, then the higher V will try to dump it's extra into the outputs of the lower V, which seems bad. Maybe could prevent with diodes?

thx!
 
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dknguyen

Well-Known Member
Most Helpful Member
#2
Not ideal, but you can do it if you put in a current balancing resistor in series with each output and tie the far end of those resistors together and use that as the output.
 
Thread starter #3
why not ideal? will the difference in voltages between the outputs be dissipated as heat by the resistors? voltage drop across resistors?
 

dknguyen

Well-Known Member
Most Helpful Member
#4
It's not ideal because you need to add current balancing resistors which decreases efficiency.

It's not ideal because the converters are not synchronized to each other so drift and variations in switching frequency will produce beat frequencies.

It's not ideal because the voltage references on the regulators are not synchronzied with each other or don't track each other, therefore they will will always vary a little bit which will cause the regulators to fight each other if current balancing resistors aren't used.

i'm wondering, if their voltages are different, then the higher V will try to dump it's extra into the outputs of the lower V, which seems bad. Maybe could prevent with diodes?

thx!
Diodes, like you suggested won't fix this either. All the diodes will do is stop the regulators from fighting each other but in exchange, only the one with the highest voltage will be supplying the load. The other ones won't participate. They might play hot potato and pass things around but only one supply will ever be supplying the load. That's why you have current balancing resistors. You can size them to be very small so they have less losses than a diode too.

will the difference in voltages between the outputs be dissipated as heat by the resistors? voltage drop across resistors?
Yes, that's their job.
 
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Thread starter #5
assuming the voltages are synchronized, then will there still be efficiency loss?

how can i calculate lost efficiency per volt difference between the bucks? Is this correct?
V = IR
1v = I x 1ohm
I = 1 amp

how to calc heat? correct?
P = VI
P = 1v x 1A
P = 1W

what's the smallest i can make the resistors?
 

dknguyen

Well-Known Member
Most Helpful Member
#6
If the voltages just happen to be synchronized but current balancing resistors are present there will still be a loss, obviously, since current is flowing through the resistors even if the voltage regulators all match. If the regulators track each other to stay synchronized and you don't have current balancing resistors (because you don't need them) then of course there are no losses due to the resistor. The whole point of the resistors is to introduce artificial voltage sag (normally undesirable) so that under a load, the regulator with the most powerful output backs off as it tries to supply more current so that the weaker supplies can take some of the load. So even if the regulators are matching as long as the current balancing resistors are there, there will still be voltage sag caused by the resistors and the resulting losses.

To calculate lost efficiency, you will need to know the efficiency of the original converters, but efficiency varies with load, so unless you have this info you can't calculate it. It's mostly academic anyways.

You can also calculate power dissipated as (I^2)*R which probably makes more sense in this case than P = VI.

You want the resistors to be as small as possible, obviously, so that there is minimal impact on efficiency and minimal voltage sag under load. But the resistors have to be big enough to allow balance out voltage differences between the output. How big or small they are really depends on the regulator and how closely the voltage outputs are. The smaller the the difference between outputs, the smaller the resistors can be. The larger the difference, the larger the resistors need to be. But 1 ohm sounds way too big. I would try something more in the range of 0.25ohms or 0.5ohms, maybe even smaller. In any case, you don't want the resistors so large that they make the voltage droop too much at full load current.
 

ronsimpson

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Most Helpful Member
#7
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.
 
Thread starter #8
To calculate lost efficiency, you will need to know the efficiency of the original converters, but efficiency varies with load, so unless you have this info you can't calculate it. It's mostly academic anyways.
I'm only trying to determine inefficiency due to the resistors, not including the buck's own inefficiency. Are you saying the buck's own inefficiency will affect inefficiency due to the resistors? Academic?

The load is 6A max, can be less. I assume losses are greater with higher load (correct?), so let's say 6A.

Since there's no electrical way to sync these OTS (off the shelf) modules, I can only manually sync them, and hope they don't drift.

LM2596 datasheet gives 4% accuracy. MP1584 doesn't say. So, let's say 4%. At 5V out, ~0.25V max. If one buck drifts up, and the other drifts down, that's 0.5V max difference.

You said to use (I^2)*R. Using the 1A figure from my ohm's law calc above, with a 0.25 ohm R:

watts loss per volt:
(1A^2) * 0.25R
= 0.25W

At 0.5V difference:
0.25W * 0.5V = ~ 1/8W max loss
with two x LM2596 at 5V.

Correct?
Why can't I use a smaller R?
Where do I plug in the 6A?
Thx
 
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dknguyen

Well-Known Member
Most Helpful Member
#9
I'm only trying to determine inefficiency due to the resistors, not including the buck's own inefficiency. Are you saying the buck's own inefficiency will affect inefficiency due to the resistors? Academic?

The load is 6A max, can be less. I assume losses are greater with higher load (correct?), so let's say 6A.

Since there's no electrical way to sync these OTS (off the shelf) modules, I can only manually sync them, and hope they don't drift.

LM2596 datasheet gives 4% accuracy. MP1584 doesn't say. So, let's say 4%. At 5V out, ~0.25V max. If one buck drifts up, and the other drifts down, that's 0.5V max difference.

You said to use (I^2)*R. Using the 1A figure from my ohm's law calc above, with a 0.25 ohm R:

watts loss per volt:
(1A^2) * 0.25R
= 0.25W

At 0.5V difference:
0.25W * 0.5V = ~ 1/8W max loss
with two x LM2596 at 5V.

Correct?
Why can't I use a smaller R?
Where do I plug in the 6A?
Thx
I guess you could calculate the inefficiency of the resistors relative to the raw power output of the converter.

Iout^2*R/(Vout*Iout), per converter

0.25 ohms was just a suggestion. You might be able to go smaller or you might not be able to go to 0.25 ohmd at all.
 

dknguyen

Well-Known Member
Most Helpful Member
#10
I'm only trying to determine inefficiency due to the resistors, not including the buck's own inefficiency. Are you saying the buck's own inefficiency will affect inefficiency due to the resistors? Academic?

The load is 6A max, can be less. I assume losses are greater with higher load (correct?), so let's say 6A.

Since there's no electrical way to sync these OTS (off the shelf) modules, I can only manually sync them, and hope they don't drift.

LM2596 datasheet gives 4% accuracy. MP1584 doesn't say. So, let's say 4%. At 5V out, ~0.25V max. If one buck drifts up, and the other drifts down, that's 0.5V max difference.

You said to use (I^2)*R. Using the 1A figure from my ohm's law calc above, with a 0.25 ohm R:

watts loss per volt:
(1A^2) * 0.25R
= 0.25W

At 0.5V difference:
0.25W * 0.5V = ~ 1/8W max loss
with two x LM2596 at 5V.


Correct?
Why can't I use a smaller R?
Where do I plug in the 6A?
Thx
bolded is wrong. watts x volts is not watts
 

dknguyen

Well-Known Member
Most Helpful Member
#12
"out" means "out per converter"?

6A^2*0.25 / (5V * 6A)
36 * 0.25 / 30
9 / 30
0.3W loss per converter?


how can i calculate min possible resistor?

thx!
correct. Also calculate voltage sag as well since that could be more important than balancing. You won't actually get 5V at the load because of the resistor.

Vsag = IR

That is a hard calculation and plays an important role in your decision. Sagging to 4.9V or 4.8V might be okay for whatever you are powering. Sagging to 4V probably isn't. Current matching doesn't matter if you don't have enough voltage to run your load. At 6A from one converter, 0.25 ohms makes a 1.5V drop which is probably unacceptable. You might need to go a lot lower. You might select R by deciding how much voltage sag you can live with at your load current.

How well current balances isnt' really an issue of hard calculation. Try a value under load and measure the individual currents from each supply to see if they match as much as you want under drift. If they don't increase R, if they match much more than you need and you would rather trade some matching away for higher efficiency then decrease R. It's a compromise and you have to pick what you can live with, as long as you meet your voltage sag requirements.
 
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Pommie

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Most Helpful Member
#15
If you can parallel them as in the application note - I.E use only one error amp - then they should work fine.

Mike.
Edit, posted together. Just lifting two legs of the IC should enable paralleling.
 
Thread starter #16
i modified my comment above before you posted this.
Is that per volts-difference?

Also calculate voltage sag as well since that could be more important than balancing. You won't actually get 5V at the load because of the resistor.
Understood, thx. Let's just focus on power losses plz.

How well current balances isnt' really an issue of hard calculation. Try a value under load and measure the individual currents from each supply to see if they match as much as you want under drift.
Seems it should be calculable, give a max difference of 0.5V between the converters. Any idea what the formula might be?

6A^2*0.25 / (5V * 6A) = 0.3W loss per converter
i made a mistake there, as it's 3A per converter.

thx!
 
Thread starter #18
I do understand the question. I have paralled power supplies. I showed how to do it with out adding resistors. And I showed an alternative. Now I am out.
then i didn't understand your answer. Plz don't be offended.

sounds like you're saying i can disconnect 1 pin on one module, and then send a jumper from same pin to the other module? Correct?
 

dknguyen

Well-Known Member
Most Helpful Member
#19
i modified my comment above before you posted this.
Where does volts-difference go, in the formula?
The volts difference I kind of just ignored because it changes depending on with drift. The formula in post #9 is the average power loss. The loss when drift is present should hover above and below that depending on whether the converter is the stronger or the weaker.

I mainly did that because it is easier and I'm lazy. The change in losses due to voltage differences should be small compared to the regular voltage sag losses unless the regulators are wildly out of match.

It's a bit of a pain. It's not impossible or convoluted, just I'm eating at the moment. I can figure it out after I'm done eating if you want.
 

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