# Op-Amp as a Current Supply

1. Aug 14, 2008

### 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 :)

2. Aug 14, 2008

### 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 [Broken] <Bob Pease article

Last edited by a moderator: May 3, 2017
3. Aug 14, 2008

### dyordyen

thanks very much!!

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

:)

4. Aug 14, 2008

### 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...

:)

5. Aug 15, 2008

### rbj

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

6. Aug 15, 2008

### Staff: Mentor

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

.

7. Aug 15, 2008

### Redbelly98

Staff Emeritus
Looking at Fig. 1a of that article:
http://electronicdesign.com/files/29/18966/fig_01a.gif [Broken]

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".

Last edited by a moderator: May 3, 2017
8. Aug 15, 2008

### Staff: Mentor

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

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

Last edited by a moderator: May 3, 2017
9. Aug 15, 2008

### Redbelly98

Staff Emeritus
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.

10. Aug 15, 2008

### Staff: Mentor

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

11. Aug 16, 2008

### dyordyen

thanks guys,

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

:)