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
Google "current mirror"
Compliance? That is, what supply voltages do you have available, and what must the difference current drive, impedance-wise?
...Jim Thompson
-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et |
Currents in parallel add! So connect the 0-1 amp thing to the same load as a -25 mA current source.
John
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
OK, I wouldn't hire you.
John
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
LOL - too bad. We could have had so much fun! Regards,
Mike Monett
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
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
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.
Actually, I'm using a Widlar and because I don't have bjts on hand, I'm going to end up using nmos. Thanks!
add -25mA to your input current. Bye. Jasen
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