baby_1
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I think that's not possible....Vout=2.5(1+12000/100)=302.5
Yes i'm agree with you about this ! i think he confused with some definitions !Vout=2.5(1+12000/100)=302.5
This formula is only correct if there is one output voltage, connected to the reference pin with one resistor.
Yes, but in fact you calculated it correctly the first time! Remember you said this:if we now assume that o single input and omit the other fraction ((Vout2 - 2.5) / R2 + (Vout3 - 2.5) / R3) we have
(Vout1 - 2.5) / R1 = 2.5 / Rx
example one that R1=47.5 and Rx=2.7 , Vout1=20
(20-2.5)/47.5=.36
2.5/2.7=.92
they aren't the same.
That is quite right. If you take away the first two output voltages,then the third one will go up from 20V to 46.48V.Vout=2.5(1+47.5/2.7)=46.48 :shock:
Why? There is a resistor in series with the LED to limit the current. Surely that is all that is needed?A notation : you should use a resistor in parallel with the LED of your opto coupler . don't forget it .
The missing feedback makes the diference.what is my fault?is it not the similier circuit to SMPS feedback output?
But besides involving a more complicated calculation, what's the purpose of the multi-feedback circuit? Did you think about how it works? Does it bring a feature that's helpful for your application?In this case the calculation is more complicated.
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