Not sure you can use a resistor because of ohms law. V=3DI/R
Since the current drawn isn't constant, the voltage won't be either. I
suppose that's why we have voltage regulators :)
I have a PSU that does something similar to your requirements.
To drop the 40v to 15v (which is used as input to the 12v regulator) I
use a resistor and zenner diode between 40v + and - rails. Their
junction connects to the base of a 2N3055. Collector goes to the 40v and
emitter gives the 15v ish...
Depending on the current you need, you can vary the transistor and mount
it on a heat sink.


-----Original Message-----
From: Rex Byrns [mailto:rexb@TESTENGEER.COM]
Sent: 15 May 2002 18:31
To: PICLIST@MITVMA.MIT.EDU
Subject: [EE] Power supply ignorance


I have a problem that has got to be a simple fix, and I am already
feeling
stupid.

I am trying to build power supply that will give me 40v, 12v, and 5v.
I have the 40v part working fine. My 12v regulator will only take 35v
max,
so I got a resistor, put it between the +40 output and my meter, and it
shows about 30v.  I installed the regulator with this same resistor
btween
the input and the +40v. Ground to the -40v.

It does not work.  The meter only shows .638 volts at the input of the
regulator. Output is 0.

I planned to then put a 5v regulator off of the 12 volts.

Do I need to put the resistor at ground instead of + ?

Maybe the regulator is bad?

Thanks

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email listserv@mitvma.mit.edu with SET PICList DIGEST in the body

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email listserv@mitvma.mit.edu with SET PICList DIGEST in the body