MAX8792 buck AC coupled feedback. What does this do to V_FB?

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king.oslo

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Hello there,

I set MAX8792 buck controller by connecting a voltage divider between REF-pin (2V output) and REFIN-pin. Regulated V[SUB]Out[/SUB] = V[SUB]REFIN[/SUB]. But If I want higher than 2V output, I must use a voltage divider that reduces V[SUB]FB[/SUB]. V[SUB]OUT[/SUB] will be rise directly proportional with this reduction. Fig. 4 on p. 17 of the datasheet has a very effective illustration: https://www.electro-tech-online.com/custompdfs/2012/01/MAX8792-2.pdf

I opted for ceramic caps. According to datasheet I must use AC or DC coupled current sense compensation if I wishes to do this (p. 23 of the datasheet.) I chose AC-coupling.

The question is what happens to V[SUB]FB[/SUB] when I use this compensation? Surely R[SUB]COMP[/SUB] (67Ohm) must alter V[SUB]FB[/SUB]? Without knowing I cannot determine which V[SUB]OUT[/SUB] (and a bunch of other things) I will get.

What do you think?

Thank you for your time.

Kind regards,
Marius
 
V[SUB]FB[/SUB] has no significant effect on the output voltage in that circuit since the FB input is high impedance. To get a higher output voltage you need to add a resistor to ground at the FB input to give your desired output voltage.

Since R[SUB]COMP[/SUB] is part of the compensation network you should make the equivalent parallel value of R[SUB]COMP[/SUB] and the added resistor to ground equal to the desired compensation value (670 ohms per your calculation). This will keep the time-constants of the network the same.
 
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Thank you for taking the time to look at my problem Carl.

I drew up this schematic. Tell me if it is exactly what you suggested: R8 and R4 form the voltage divider. I adjust R9 to get 0-14.4 V[SUB]OUT[/SUB].

In addition, I get the ac-coupling and consequently the stability required to use ceramic output caps.

Is this correct?

Thank you

Kind regards,
Marius
 
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If you want the equivalent impedance to be 640 ohms, then the parallel value of R4 and R8 should be equal to 640 ohms. The formula for the parallel resistance is Rp = (R4*R8)/(R4+R8). For equal value resistors, each resistor would be 1280 ohms.
 
crutschow said:
If you want the equivalent impedance to be 640 ohms, then the parallel value of R4 and R8 should be equal to 640 ohms. The formula for the parallel resistance is Rp = (R4*R8)/(R4+R8). For equal value resistors, each resistor would be 1280 ohms.
Click to expand...



Dear Carl,

You are assuming I am more skilled than I am.

Why are we talking about parallel impedance? I know what parallell impedance is, and the formula, but to me, R4 and R8 appear to be in series and form a voltage divider. To me they dont appear to be in parallel.

Please explain to me in simpler terms. Why are walking about parallel resistances here?

Thanks.M
 
We are talking about the impedance at the junction of R4 and R8 since that is the impedance seen by the compensation network. The AC impedance at that point consists of R4 going to the output (which is a low impedance) and ground. Thus the equivalent impedance at the junction of R4 and R8 is their parallel value.

Does that make sense?
 

Thanks for that Carl.

The impedance seen by the compensation network? Seen by FB you mean?
Are R4 and R8 in parallel at the junction of FB?

I am sorry, as you can tell, I am still confused

Kind regards,
Marius
 
Yes, from an AC signal point of view at the FB input, R4 and R8 are in parallel.
 
Oh, will the voltages at FB be swing from positive and negative, relative to ground?

Thanks.M
 
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In normal operation the voltage at FB will be DC, other then perhaps a little ripple from the output. Why would you think it swings from positive to negative?
 
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Hello Carl,

Great to hear from you again!

crutschow said:
Yes, from an AC signal point of view at the FB input, R4 and R8 are in parallel.
Click to expand...

For the last week I have struggled to understand this. I am not giving up. But I still need some help.

If the voltages was swinging from positive to negative, I would have understood how the above could be true. But if the signal is DC, I cant see how R4 and R8 are in parallel, or that the signals are viewed from an AC point of view

Can you please explain this to me in laymans terms?

Kind regards,
Marius
 
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The FB signal may appear to be DC but it actually has an AC component (any change in output due to load change, temperature drift, noise, etc.) that is part of the negative feedback to correct for output errors and perturbations. This AC component must be compensated so that the feedback loop, which typically has an AC frequency response of a couple kHz or so, is stable and does not oscillate. That is the purpose of the RC feedback compensastion components.

Making more sense?
 
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Dear Carl,

Finally! That make sense. If I want Rcomp of 1K2, then R8 and R4 must be 2k4.

So R4 must be equal to R8. But I want to adjust the Vout ceiling to ~15V. It appears to me that the only way to lower the voltage sensed at FB is to make R8 != R4. Because if R4 = R8, then Vout of the R4-R8-volage-divider will always be half of Vin.

But that is not possible, because R8 must = R4. Or not?

So how to determine *the correct value of R4 and R8 be to get a Vout ceiling of 15V and not alter Rcomp?

Thanks.

Kind regards,
Marius
 
King,
You picked a very complicated project. There are at least three things happening here.
DC: In the DC model all the capacitors are open and you only have to think about R4 & R8.
AC: To make the power supply respond faster, there is typically a C or RC across R4 to send any ripple on the output back the the FB pin.
Slope compensation: This is really complicated. Simply said the supply works better if there is a triangle wave added to the FB pin. So R7 and C21,22 make a filter that changes the square wave on the coil into a small triangle wave and sends it back to the FB pin. C21,C22 have dual functions.

"So how to determine *the correct value of R4 and R8 be to get a Vout ceiling of 15V and not alter Rcomp?"
My first idea is to increase R4 and R7 by the same ratio, and to reduce C21,22 by the same ration.
 
Ron, good to read your response!

I am just working this out, then I will have a prototype made up. Really exciting

Yes, I feel it is tough. I am only a beginner Could it be correct that the datasheet for the max8792 doesn't give instructions on rising the ceiling whilst keeping the compensation the same?

Kind regards,
Marius
 
R4 does not need to be equal to R8 for their parallel value to equal the desired resistance. For example the parallel resistance value of 3k0 and 2k0 is also 1k2.

So first you determine the ratio of R4 to R8 that gives you the desired output voltage. Then you calculate the value of R4 and R8 using the parallel resistance equation I mentioned in my previous post (or use this calculator) that will give you the desired equivalent resistance at the FB node.
 
Thanks guys,

The resulting schematic is attached to this message.

Can any of you see anything else about the schematic that should be addressed before I get this prototype-board sent off to be manufactured?

Thank you very much for your time with this challenging task!

I am learning a lot, which is very valuable!M
 
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I suggest you breadboard it before you send it to be manufactured (or assume that the first manufactured board is the prototype breadboard and not likely the final product).
 
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