Injecting a voltage onto the feedback node of a regulator

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I am having trouble understanding how to calculate the output voltage of a regulator when a voltage is injected into the feedback node. (See pic below) This technique is an easy way to create a digitally controlled regulator using, for example, a filtered PWM waveform. Surely the calculation is simple and I just can't see the forest for the trees. Any hints?

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This type regulator will try to always maintain 1.25 volts between OUT and ADJ. If you "push" ADJ higher or lower from another source, the voltage at OUT will follow. If ADJ is forced to 4.5 volts, OUT will be 5.75. You could for example use an opamp as a pwm integrator for the V2 source and remove R2. The value of R1 can be increased to reduce the current the opamp must to handle. Vout will be 1.25 volts above
the integrator output (V2).

Tom Pappano
Tulsa, Oklahoma

Tom Pappano
Tulsa, Oklahoma

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You CANNOT inject a voltge. Inject a current which is derived from the voltage and a resistor.

Think of the regultor as an op-amp. The non-inverting "input" is hidden inside as Vref. The normal feedback net consists of R1 and R2. The feedack node is the summing junction.The "output" is the node "out". Normal op-amp relationships hold here.

Jim

 

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I think it will indeed work as I described. The STK-500 power supply is done this way. I looked further at the LT1083 data sheet though, and it does need a minimum 10ma load. This is why the resistors shown above were such low values.

Tom Pappano
Tulsa, Oklahoma

Tom Pappano
Tulsa, Oklahoma