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I am finding it hard to find a suitable buck

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king.oslo:
Thanks guys. I got enough datasheets to read up on for a while now :) M

king.oslo:

--- Quote from: amspire on April 29, 2012, 12:06:38 am ---As far as varying the voltage, most of the IC's have an internal voltage reference of 1.25V or similar connected to one input of the comparator internally. Lets say your device is 1.25V. You then have a voltage divider on the the other input to set the output voltage. If instead of taking the lower resistor of the divider to 0V, you take it to a variable 0 - 1.25V source, then as the voltage of this source goes from 0 - 1.25V, then the output should in theory go from maximum to zero.

Richard.

--- End quote ---

Just wanted to ask (and learn) how you arrived at the conclusion that the error amplifiers does not compensate the voltage relative to the 1.25V reference. Say at V_FB = 1.25 = V_REF the output is uncorrected, and if V_FB increases, V_OUT is lowered to again put F_FB = V_REF? Like an inverting amplifier?

Thanks :) M

amspire:

--- Quote from: king.oslo on May 06, 2012, 10:16:00 pm ---
--- Quote from: amspire on April 29, 2012, 12:06:38 am ---As far as varying the voltage, most of the IC's have an internal voltage reference of 1.25V or similar connected to one input of the comparator internally. Lets say your device is 1.25V. You then have a voltage divider on the the other input to set the output voltage. If instead of taking the lower resistor of the divider to 0V, you take it to a variable 0 - 1.25V source, then as the voltage of this source goes from 0 - 1.25V, then the output should in theory go from maximum to zero.

Richard.

--- End quote ---

Just wanted to ask (and learn) how you arrived at the conclusion that the error amplifiers does not compensate the voltage relative to the 1.25V reference. Say at V_FB = 1.25 = V_REF the output is uncorrected, and if V_FB increases, V_OUT is lowered to again put F_FB = V_REF? Like an inverting amplifier?

Thanks :) M

--- End quote ---
I am not completely sure what you are asking, but if you are asking if there is negative feedback to ensure that the feed back voltage ends up equaling the internal reference voltage, then the answer is yes. So if the voltage on the feedback input is too high, the regulator will reduce the output voltage till it the feedback input is matching the internal reference.

Richard

IanB:

--- Quote from: king.oslo on May 06, 2012, 10:16:00 pm ---Just wanted to ask (and learn) how you arrived at the conclusion that the error amplifiers does not compensate the voltage relative to the 1.25V reference. Say at V_FB = 1.25 = V_REF the output is uncorrected, and if V_FB increases, V_OUT is lowered to again put F_FB = V_REF? Like an inverting amplifier?
--- End quote ---

Like Richard, I am also not quite sure what you are asking?

But another key point to bear in mind is that voltages are always defined as a difference between two points. There is no such thing as an absolute voltage of "1.25 V". For practical purposes you cannot have such a reference. So the regulator is controlling the voltage it sees to be "1.25 + X volts" where X is some external voltage offset under your control as the circuit designer. In most cases X is zero, but if you adjust X to be a negative voltage like -1.25 V then the regulator can regulate the output down to zero.

king.oslo:
I understand that Richard says 0v at FB-pin = lowest duty cycle, 1.25v = highest duty cycle.

If I feed in 1.15V from a DAC to VFB, how do I know that the regulator does not drop the duty cycle to minimum, in an attempt to get VREF = VFB?

Does that make my question clearer? :) M

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