Op-Amp as a Current Supply

[dyordyen]

guys, i need some help...

is it possible to construct a variable-output current supply using op-amp(s)?

if so, can anyone help me about how to construct one?

thanks :)

[Zapitgood]

There are tons of circuits on the web, search for current source, not supply.

Here are a few schematics.. You can beef up the current with a transistor. Also check out Bob Pease article on current sources.

http://www.edn.com/contents/images/6309110f1.pdf

http://electronicdesign.com/Articles/ArticleID/18966/18966.html <Bob Pease article

[dyordyen]

thanks very much!!

so that's why, i used "supply" for searching... should have used "source"...

:)

[dyordyen]

oh, and one more thing..

how about a variable op-amp current source?

when i search the web i only find constant current sources using op-amps...

or is there any of the kind?

sorry for the "newbie-ness" of mine...

:)

[rbj]

by "variable", do you mean voltage-controlled? if no, your current source is determined by the component values. those can be changed.

[berkeman]

As rjb says, you could use a potentiometer for the current setting control. Or, if you want it to be voltage controlled (like from a microcontroller), you could use a "digital potentiometer" device, or use a MOSFET as a voltage controlled resistance (but there are nonlinearities in doing this that need to be accounted for).

Digital pots and uCs: http://www.maxim-ic.com/appnotes.cfm/an_pk/408


.

[Redbelly98]

http://electronicdesign.com/Articles/ArticleID/18966/18966.html <Bob Pease article

Looking at Fig. 1a of that article:
http://electronicdesign.com/files/29/18966/fig_01a.gif

Shouldn't the current be


I = \alpha \ (V_s - V_{in}) \ / \ R


???

Then to get a variable current, you could use a potentiometer for the resistor appearing between the words "Current" and "V_in".

[berkeman]

Looking at Fig. 1a of that article:
http://electronicdesign.com/files/29/18966/fig_01a.gif

Shouldn't the current be


I = \alpha \ (V_s - V_{in}) \ / \ R


???

Then to get a variable current, you could use a potentiometer for the resistor appearing between the words "Current" and "V_in".

Think the current I is just I = \frac{ (V_s - V_{in}) }{ R }

Don't see where alpha would enter in...

[Redbelly98]

Don't see where alpha would enter in...

I'm guessing alpha accounts for the difference in collector and emitter currents for the transistor, and is very close to 1. So your equation would work just fine too.

[berkeman]

I'm guessing alpha accounts for the difference in collector and emitter currents for the transistor, and is very close to 1. So your equation would work just fine too.

Oh, I see the alpha factor now. Thanks Redbelly.

[dyordyen]

thanks guys,

this forum works like magic... well, strictly-physics speaking, magic don't exist... but not in this case...

:)