In the attached circuit I can't figure out how to calculate reference voltage. Can someone help with this?
Thanks.

 

That's interesting. Usually, 7805's are used as intended. The output will try to be some fixed voltage above apparent ground. The currents through the fixed resistors may tell you something.

Next time, use an LM317.

 

I see no reference to reference voltage on the referred schematic. What reference voltage are you referring to? Just for reference.

 

I'm using Vout =Vref (1+r2/r1) to figure out Vout. I have r2 and r1. So would Vref be 5.0 V?

 

Apparently, Fairchild did sanction this kind of arrangement for 78XX: http://www.fairchildsemi.com/ds/LM/LM7805.pdf (see P.23).

You will see that they give a formula for Vout: VO = VXX(1 + R2 / R1) + IQR2 , provided that I(R1) ≥ 5 IQ


Vref is indeed 5V for 7805 but you also have to allow for the quiescent current IQ. This current is about 5 to 6mA, so you end up wasting an extra 25mA or more to get an adjustable output.

As beenthere suggests, if you want an adjustable regulator, why not use something designed for the job?

 

Apparently, Fairchild did sanction this kind of arrangement for 78XX: http://www.fairchildsemi.com/ds/LM/LM7805.pdf (see P.23).

You will see that they give a formula for Vout: VO = VXX(1 + R2 / R1) + IQR2 , provided that I(R1) ≥ 5 IQ


Vref is indeed 5V for 7805 but you also have to allow for the quiescent current IQ. This current is about 5 to 6mA, so you end up wasting an extra 25mA or more to get an adjustable output.

As beenthere suggests, if you want an adjustable regulator, why not use something designed for the job?

Thank you Adjuster. Really what i was looking for was to verify if I was using the correct reference voltage. This is not a circuit I am using, it's for a lab question.