# Circuit to subtract two currents

Discussion in 'Electronic Design' started by owais, May 17, 2007.

1. ### owaisGuest

Hi,
I need some guidance or basic starting point on this. I need to
generate a current that is the difference of two currents. One of the
currents is constant at ~ 25mA and the other (input current) can vary
from 0 to 1000mA. My original idea is to do a I-V conversion on the
two inputs, do a voltage subtraction of the two using a basic opamp
difference circuit and then an V-I circuit to convert back to
current...but there has to be a cleaner way.
Owais

3. ### Jim ThompsonGuest

Compliance? That is, what supply voltages do you have available, and
what must the difference current drive, impedance-wise?

...Jim Thompson

4. ### John LarkinGuest

Currents in parallel add! So connect the 0-1 amp thing to the same
load as a -25 mA current source.

John

5. ### Mike MonettGuest

He used "~' in front of the 25mA, not a minus sign, so you don't know the
polarity of the 25mA current. If it's positive, it will add instead of
subtracting.

Also, connecting the two current sources together means the load voltage
will appear at both outputs.

Since you don't know the load resistance, you don't know the max voltage,
and you don't know if the 25mA current source has sufficient compliance to
handle the max voltage.

If the currents are the same polarity, probably the easiest solution is to
invert the 25mA current, then add it to the 1A as you suggested. The
conversion circuit can then address any compliance issue with the 1A
source.

In any event, it would probably be best to leave the 1A current as is.
Doing I-V and V-I conversions on a 1A current would dissipate quite a bit
of power.

Regards,

Mike Monett

6. ### John LarkinGuest

OK, I wouldn't hire you.

John

7. ### owaisGuest

Hi,

The power supply driving the whole solution is rated at 12V, 1A
continuous, 3.5A PWM; the load is a string of high-power LEDs. The
difference current output would ideally be 12V at up to
(InputCurrent-25mA); where InputCurrent is 0->1A

THanks

[...]

Regards,

Mike Monett

9. ### owaisGuest

Thanks to everyone for helping. I now understand how to do this. I
am going to implement an current sink using a current mirror connected
to the same point as the 1A supplying the load (which on the led is
the anode) Seeing how simple the solution is, I'm kind of embarrassed
at my original question, but really, thanks!

Owais

10. ### Mike MonettGuest

Nice of you to let us know you found your solution. Many people ask for
help, then simply disappear.

What kind of current mirror do you plan to use - Wilson?

Regards,

Mike Monett

11. ### Fred BloggsGuest

Mmmmmmm....maybe even simpler than that depending on what you're doing.
If the LED forward drop can be considered constant, then a simple
parallel resistor can be used to divert the 25mA.

12. ### owaisGuest

Actually, I'm using a Widlar and because I don't have bjts on hand,
I'm going to end up using nmos.
Thanks!

Bye.
Jasen