Paralleling Buck Modules?

dknguyen · Dec 11, 2018

johnyradio said:
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?

More or less. It basically makes it so all the modules use the feedback loop of just one of the modules, synching their voltages together so they aren't each doing their own thing anymore. You need to lift the FB pin from all but one of the modules, as pommie, said to disable the module from trying to run its control loop independently. The one module then becomes the master and connecting its COMP pins to the other modules, as Ron said, spreads the master control signal and slaves them.

johnyradio · Dec 11, 2018

dknguyen said:
The volts difference I kind of just ignored because it changes depending on with drift.

At 4% drift (according to datasheet), that's 0.2v drift at 5v out. So, assume max 0.4 volts difference, if one converter drifts up and the other drifts down.

I'm eating at the moment. I can figure it out after I'm done eating if you want.

gosh, don't let me interrupt your dinner!

appreciate your help!

The change in losses due to voltage differences should be small compared to the regular voltage sag losses

"regular"? you mean drop across the resistors? that's why i want to make the R as small as possible.

"change in losses"? The "losses" are reduction in amperage due to V difference, dissipated as heat, correct? Need to calc max loss at 0.4v difference

dknguyen · Dec 11, 2018

johnyradio said:
At 4% drift (according to datasheet), that's 0.2v drift at 5v out. So, assume max 0.4 volts difference, if one converter drifts up and the other drifts down.

gosh, don't let me interrupt your dinner!

appreciate your help!

"regular"? you mean drop across the resistors? that's why i want to make the R as small as possible.

by regular i mean the voltage sag that normally occurs due to the presence of the resistors, even if the voltages were matched.

johnyradio · Dec 11, 2018

I'm guessing the R needs to be big enough so that so it's own V drop is bigger than the V difference between the converters?

dknguyen · Dec 11, 2018

johnyradio said:
I'm guessing the R needs to be big enough so that so it's own V drop is bigger than the V difference between the converters?

Yeah. There's a certain amount of extra current that will be pushed out be the weaker supply that will develop a voltage drop across the resistor. This voltage drop adds onto the voltage of the weaker supply so that so the far end of the resistor (where all the supplies meet) is equal amongst all supplies. The larger this resistor is, the less extra current is required to match output voltages. OTOH, the larger the resistor is, the more losses you have at all times, even if there happens to be no imbalance at the time.

Pommie · Dec 11, 2018

If you do use resistors to balance them make sure you get their voltages as close as possible before paralleling.

Mike.

johnyradio · Dec 12, 2018

dknguyen said:
R needs to be big enough so that so it's own V drop is bigger than the V difference

Click to expand...

Yeah.

But then my math shows it can be as small as I want:

Max Vdifference: 0.4V

3A^2*0.25R / (5V * 6A)
0.075W max loss per converter.
Vsag = 0.75V.
0.75V < 0.4V

3A^2*0.001R / (5V * 6A)
0.0003 W max loss per converter.
Vsag = .003
0.003V < 0.4V

The change in losses due to voltage differences should be small compared to the regular voltage sag losses

Unclear. One is watts. The other is volts. How to compare?

johnyradio · Dec 12, 2018

Ok, this article explains

https://www.electronicdesign.com/power/double-your-output-current-parallel-voltage-regulators

Rshare is chosen so the voltage dropped across it at full load is several times larger than the maximum difference voltage between the regulators.

In my case, I tried 3x:
Vdif = 0.4V
Vmargin = 0.4V x 3 = 1.2V
Rshare = V/I = 1.2V/3A = 0.4R
Where REG1 output's 3A.

Full-load difference current =
(VREG1 - VREG2)/RSHARE .

Me:
0.4V / 0.4R
1Adif

I don't know how to claculate heat. Should
include Wreg + Wdif?

This part I don't understand:

By choosing a full load difference current of IOUT × 10%, defining Tolerance = T, and setting the maximum (VREG1 - VREG2) value to 2 × T × VREG, we get:

RSHARE = (VREG1 - VREG2)/(IOUT × 10%) = 20 × T × VREG/IOUT.

Me:
(VREG1 - VREG2)/(IOUT × 10%)
0.4/(3A x 10%)
1.3R
(Not sure if Your is per regulator or both regulators)

I guess that's the recommended R. My 0.4R, above, is too small.

Difference current:
0.4V / 1.3R
0.3A

How to get heat?

Thx

dknguyen · Dec 12, 2018

johnyradio said:
But then my math shows it can be as small as I want:

Max Vdifference: 0.4V

3A^2*0.25R / (5V * 6A)
0.075W max loss per converter.
Vsag = 0.75V.
0.75V < 0.4V

3A^2*0.001R / (5V * 6A)
0.0003 W max loss per converter.
Vsag = .003
0.003V < 0.4V

I don't understand what you are trying to do here. This seems to be the second time (at least) you have swapped voltage ans watts. They are not the same. Where is is 0.075W and 0.75V (or 0.0003W and 0.003V) coming from?

Vsag is not THE criteria. It is a limiting criteria that prevents R from being too large, while current matching is a limiting criteria that prevents R from being too small. You pick one that satisfies both. If you just use Vsag, then yes smaller is always better ad infinitum. Likewise with current matching and R being large to infinite. But you would never do that because you have two competing criteria, not one.

johnyradio · Dec 12, 2018

dknguyen said:
I don't understand what you are trying to do here.

Plz see my newer reply

dknguyen · Dec 12, 2018

johnyradio said:
I don't know how to claculate heat. Should
include Wreg + Wdif?

Yes, but why do you care about total power dissipated? You can't do anything about the module which is already built. Your R neds to be physically large enough to handle the power though.

johnyradio said:
This part I don't understand:

It is setting the resistor so that a 10% difference in current will be enough to cause a voltage drop across the resistors to balance out the maximum voltage difference expected. In other words, the currents will match by 90% in the worst case, or better. You can choose any percentage you are willing to live with. Lower current matching means lower R which is more efficient. There is a tradeoff.

Also make sure to check that your chosen R meets your voltage sag requirements. Again lower R means less matching but less sag. Tradeoff.

The more current matching you have the closer you can use full current across all supplies, at the expense of voltage sag. And vice versa. Tradeoff.

You can see now why its not ideal. Synching converters so their voltages track each other has none of these issues but iusn't always possible. Synching their clocks is a separate issue but not doing so can also cause issues sometimes

johnyradio · Dec 12, 2018

dknguyen said:
why do you care about total power dissipated?

So I can select resistors with adequate power rating.

dknguyen said:
Lower matching means lower R which is more efficient

Doesn't lower matching = more heat due to difference current, which is less efficient?

johnyradio · Dec 12, 2018

dknguyen said:
why do you care about total power dissipated?

So I can select resistors with adequate power rating.

dknguyen said:
Lower matching means lower R which is more efficient

Doesn't lower matching = more heat due to difference current, which is less efficient?

dknguyen said:
make sure to check that your chosen R meets your voltage sag requirements

I can raise the regulator V to compensate, right?

dknguyen · Dec 12, 2018

johnyradio said:
So I can select resistors with adequate power rating.

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

Doesn't lower matching = more heat due to difference current, which is less efficient?[/QUOTE]

Lower current matching as in how tightly do you wamt output currents between modules to match.

johnyradio said:
I can raise the regulator V to compensate, right?

Only somewhat. Your load current can vary. Vsag is highest at high currents and lowest at low currents. Fully compensating for it at high current might make the voltage too high at low currents.

johnyradio · Dec 12, 2018

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

Say I've just one regulator, with a series resistor on output. If my load is pulling 3A at 5V, don't I need a 15W resistor?

Lower current matching as in how tightly do you wamt output currents between modules to match.

I understand. Doesn't lower matching = more heat due to difference current?

Thx

dknguyen · Dec 12, 2018

johnyradio said:
So I can select resistors with adequate power rating.

Doesn't lower matching = more heat due to difference current, which is less efficient?

I can raise the regulator V to compensate, right?

if you just have one regulator then you dont need that resistor because you don't need to current balance it with anything.

No, lower current matching means that one supply will do more work than the others so you will be farther away from achieving the full current that all supplies could output together. The current won't be shared as evenly so not all the supplies can be used to their full current potential. The weaker supplies just supply less current and the stronger supplies supply more. The R is smaller iin this case so more efficient, but less overall maximum current capability.

johnyradio · Dec 12, 2018

dknguyen said:
if you just have one regulator then you dont need that resistor

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?

dknguyen said:
load current can vary. Vsag is highest at high currents and lowest at low currents

You mean, even if a single regulator is rock solid by itself, a series R on it's output will make it vary as current varies? (I mean without a parallel reg)

dknguyen · Dec 12, 2018

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?

You mean, e

I understood what you were asking so my response doesn't change. 3.9V drop actually from your numbers. You don't need the resistor because your load is taking all the power. Th resistors were only ever there to allow the parallel modules to current share with uneven output voltages. No parallel modules, no resistors.

johnyradio said:
You mean, even if a single regulator is rock solid by itself, a series R on it's output will make it vary as current varies? (I mean without a parallel reg)

yes. this is voltage sag and an undesirable trait. All supplies have this due to unavoidable internal resistance in series with the output. but for current sharing between unsynchronized converters, you need more because it relies on that mechanism to workso you add more, but there is a price to be paid.

the higher voltage regulator outputs more current, but the R makes the voltage sag more at higher currents which lets the other regulators with lower voltage participate. It's gimping the strongest members so that they must work with the group because the group can get more work done as a whole.

johnyradio · Dec 12, 2018

You don't need the resistor because your load is taking all the power. Th resistors were only ever there to allow the parallel modules to current share with uneven output voltages. No parallel modules, no resistors.

Yes I understand that. You're still not getting my question.

I know I don't need the resistor with only one regulator. I put it there anyway. Won't there be heat due to v drop?

yes. this is voltage sag and an undesirable trait. All supplies have this due to unavoidable internal resistance in series with the output. but for current sharing between unsynchronized converters, you need more so you add more, but there is a price to be paid.

I think. you misunderstood this question too. Let's assume this regulator has 0% drift. Let's also assume they are synchronized. Will V still vary with load?

dknguyen · Dec 12, 2018

I understood. You were talking about just one regulator with no parallel. Those are the questions I was answering. Then after answering the question for a single unparallel regulator I mention current sharing in parallel modules to tie in why they are there for parallel, but not for singular.

Paralleling Buck Modules?

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