Question about buck converter feedback.

Thread Starter

thedoc8

Joined Nov 28, 2012
155
I have been studying buck converters and I have a question about a simple one. At the output of the buck I see people making a voltage divider and sending this voltage back to the drive circuit to control the pwm. How do you determine what to send back to the control ic, or what resistors to use for the divider. Does the control circuit look at whatever feedback is there and just try to maintain whatever it was. Thanks for any help.
 

Papabravo

Joined Feb 24, 2006
17,032
I have been studying buck converters and I have a question about a simple one. At the output of the buck I see people making a voltage divider and sending this voltage back to the drive circuit to control the pwm. How do you determine what to send back to the control ic, or what resistors to use for the divider. Does the control circuit look at whatever feedback is there and just try to maintain whatever it was. Thanks for any help.
The IC has a feedback input that is compared to a reference voltage. To compute the resistors in the divider you must know the reference voltage, and the output voltage. You may have to size the divider to draw a certain amount of current from the output.
For example:
  1. Suppose the divider must draw 10 mA and the output voltage is +5VDC. This means the total resistance of the divider is 500 Ω = 5 / .010
  2. Let us say the reference voltage is +1.25 VDC, then the resistance R of the lower resistor in the divider needs to satisfy: 1.25 VDC = (R/500)5VDC which implies R=125Ω
  3. This means the other resistor must satisfy Ru = 500 Ω - 125 Ω = 375 Ω
 
Last edited:

Thread Starter

thedoc8

Joined Nov 28, 2012
155
The IC has a feedback input that is compared to a reference voltage. To compute the resistors in the divider you must know the reference voltage, and the output voltage. You may have to size the divider to draw a certain amount of current from the output.
For example:
  1. Suppose the divider must draw 10 mA and the output voltage is +5VDC. This means the total resistance of the divider is 500 Ω = 5 / .010
  2. Let us say the reference voltage is +1.25 VDC, then the resistance R of the lower resistor in the divider needs to satisfy: 1.25 VDC = (R/500)5VDC which implies R=125Ω
  3. This means the other resistor must satisfy Ru = 500 Ω - 125 Ω = 375 Ω
Thanks very much, this I can understand and will help a bunch. I like examples
 
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