LM10011: Adjusting Output Voltage of an isolated DC-DC converter module with LM10011

ghasem aghdam

Part Number: LM10011

Hi,

I would like to use LM10011 to adjust the output voltage of an isolated DC-DC module dynamically. The module has a trim pin called "Vadj" and remote sense pins (+Sense, & -Sense), please see below. The initial output voltage is 12 V and it can be adjusted from 9.6 V to 13.2 by populating a single resistor between the Vadj pin & +Sense pin or the Vadj pin & -Sense pin. I'm just wondering if LM10011 can be used to adjust the output voltage of the module dynamically.

Thanks & BR,

Ghasem

over 3 years ago

0 JohnTucker over 3 years ago

TI__Guru**** 191445 points

Ghasem,

The answer is "maybe". Can you share the part number and / or datasheet of your module?

0 ghasem aghdam over 3 years ago in reply to JohnTucker

Intellectual 455 points

Hi John,

Thank you so much for the prompt reply.

Here is the link to datasheet of my module. This module has no remote sense pins.

https://flexpowermodules.com/documents/fpm-techspec-pku4113d-pku4213d

BR,

Ghasem

0 JohnTucker over 3 years ago in reply to ghasem aghdam

TI__Guru**** 191445 points

Ghasem,

The module scales the voltage up by injecting current into the Vadj pin and scales the voltage down by sinking current from the Vadj pin. They do not disclose much information on the details. LM10011 can only source current, so it can only adjust up directly. The possible work around would be to set the Radj resistor for the minimum output voltage them use LM10011to adjust it upwards from there. There is not enough information in the module datasheet to suggest details for the design of such a circuit though. Can you ask them for details about the adjustment circuitry?

0 ghasem aghdam over 3 years ago in reply to JohnTucker

Intellectual 455 points

H John,

Many thanks for your support.

I will check, but I would like to know what particular information/details you need regarding the adjustment circuitry.

Best Regards,

Ghasem

0 JohnTucker over 3 years ago in reply to ghasem aghdam

TI__Guru**** 191445 points

Ghasem,

How much do you now about voltage margining? I need to know the details of what is connected to the Vadj pin of that module. Is it a fixed voltage? Is there an internal voltage divider? How much current is required to change the voltage?

0 ghasem aghdam over 3 years ago in reply to JohnTucker

Intellectual 455 points

Hi John,

Sorry for the late reply.

Reference is in the range of 1.225 - 1.25 V. The point between two resistors internally of 5.1 kOhm goes to the module reference, but it is 5.1 kOhm when calculating current. It seems that 2.45 uA per % trim is required.

Looking forward to hearing from you,

Thanks & Best Regards,

Ghasem

0 JohnTucker over 3 years ago in reply to ghasem aghdam

TI__Guru**** 191445 points

Ghasem,

I am not clear about the connection. Can you sketch it out?

0 ghasem aghdam over 3 years ago in reply to JohnTucker

Intellectual 455 points

Hi John,

Please see the attached image. I was talking about resistors R229 and R215 in my message above. Both are 5.1 kOhm resistors.

Thanks & Best Regards,

Ghasem

0 JohnTucker over 3 years ago in reply to ghasem aghdam

TI__Guru**** 191445 points

Ghasem,

That is a rather unusual way to adjust the voltage for voltage mode control. If you set the resistor to ground to 15.33k for a 9.6 V output, then directly injecting current from the LM10011 will allow you to adjust up to 11.34 V, so you have about 1.77 V of adjustment range, or 12 V +/- 0.88 V. For a larger range you would need a current amplifier.

0 ghasem aghdam over 3 years ago in reply to JohnTucker

Intellectual 455 points

Hi John,

I'm so sorry for the late reply and many thanks for your response.

How did you get 11.34 V? How did you get 1.77 of adjustment range? Could you please expand on this?

BTW, could you please let me know how I can use a current amplifier? Do you have a reference design?

Looking forward to hearing from you.

Best Regards,

Ghasem

0 JohnTucker over 3 years ago in reply to ghasem aghdam

TI__Guru**** 191445 points

Ghasem,

Quite simple. I selected the resistor pull down to select the minimum voltage of 9.6 V. I believe it was 245 k. Then I calculated the node voltages with an additional 60 uA injected from the LM10011. 60 uA is the maximum output current. You may want to contact your module manufacturer to see what they suggest for external adjustment.

0 ghasem aghdam over 3 years ago in reply to JohnTucker

Intellectual 455 points

Hi John,

Thanks for your message.

15.33 kOhm should be selected to be able to adjust the output voltage to 9.6 V and then current from the LM10011 is directly injected to allow me to adjust up. But I'm still confused about how you calculated 11.34 V - sorry.

Could you please write the formula you used to calculate 11.34 V?

Best Regards,

Ghasem

0 JohnTucker over 3 years ago in reply to ghasem aghdam

TI__Guru**** 191445 points

I have a couple urgent projects pending. I will try to look at this later in the week. You should be able to simply derive them yourself from the supplied schematic. Unfortunately, I did my calculations literally on the "back of a napkin" at lunch the other day, so I will have to redo it when I have some spare time.

0 ghasem aghdam over 3 years ago in reply to JohnTucker

Intellectual 455 points

Hi John,

Thanks for your response.

Please share the details of calculations with me when available. Much appreciated.

Best Regards,

Ghasem

0 ghasem aghdam over 3 years ago in reply to ghasem aghdam

Intellectual 455 points

Hi John,

Have you had a chance to look at this? If so, please share the details of calculations with me.

Thanks & BR,

Ghasem

0 JohnTucker over 3 years ago in reply to ghasem aghdam

TI__Guru**** 191445 points

OK starting from the beginning. LM10011 can only source current so it will margin the voltage up. Seeting the output to the minimum of 9.6 V:

Radj (kohm) = 5.11 * ((100/20)-2) = 15.33 k

And the internal voltage at the non-inverting error amp input is:

1.25 / ((5100/(5100+15330) +1) = 1.0003

Back calculating the output voltage:

Vout = 9.6 * 1.0003 = 9.6 V

Now tha maximum upward adjustment occurs with Idac = 60 uA and:

(1.25 - Vadj) / (5100 + 5100) + 60E-6 = Vadj / 15330

Solving for Vadj = 1.11808

And the non-inverting input voltage is"

((1.25 - 1.11808) / 2) = 1.18404 V

The resulting output voltage is:

Vout = 9.9 *1.18404 = 11.3668 V

Sorry, I do not have any reference designs. If it were me, I would control the output using a digital potentiometer for both the upwards adjusting and downwards adjusting resistors.

0 ghasem aghdam over 3 years ago in reply to JohnTucker

Intellectual 455 points

Hi John,

Thank you so much for the calculations. It's now crystal clear.

Best Regards,

Ghasem

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LM10011: Adjusting Output Voltage of an isolated DC-DC converter module with LM10011