# Resistance calculation

Discussion in 'General Electronics Discussion' started by selva, May 6, 2010.

1. ### selva

25
0
Apr 26, 2010
Hi ,

Pls don't mind for this simple question. i got confused need some help.

The output from battery is 3.3V and 4250mA, i have to reduce it to 2.5V and 11uA which is a input to an IC. I have to connect some resistor to reduce the voltage and current from the battery.
can anyone plz help in calculating the resistor value to be connected.

Warm regards,
Selva

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
unless the IC has a fixed and rather accurate 11 uA current requirement, a simple resistor won't do it.

You would probably be better off with a 2.5V voltage regulator. You could use an adjustable voltage regulator similar to a LM317 (preferably in a small package) but its quiescent current will exceed the 11uA many times over.

For such a low current, you may be able to get away with a voltage reference chip such as an LM285-2.5

3. ### selva

25
0
Apr 26, 2010
Its not that Steve, i am not gonna implement in physical, i just want to know how the calculations are made to get in to an particular resistor in specific point of a circuit,i don't want to be accurate for that current lets make it around by 11uA to 20uA or 25uA.

If i have misunderstood something wrong excuse me and inform me that.

thanks

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
In that case, you simply need the appropriate resistor to drop 0.8V at 11uA. See here.

5. ### NickS

367
0
Apr 6, 2010
Vout = R2 / (R1+R2) * Vin

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6. ### 55pilot

434
3
Feb 23, 2010
That assumes I-Out is zero. That is not the case here.

The full equation is:

Vout = R2 / (R1+R2) * Vin - Iout*(R1*R2)/(R1+R2)

---55p

Last edited: May 6, 2010
7. ### NickS

367
0
Apr 6, 2010
Right you are if he taps off at Vout 55p.

I was assuming that R2 was his device, like I thought Steve had directed. So I was thinking he would approximate his IC as R2 (by knowing it requires 11uA @2.5V) and then he could solve for just the in line resistor R1.

But looking at it now that was not so obvious. I should have been more specific

8. ### 55pilot

434
3
Feb 23, 2010
I agree with you there. It is obvious that it is not obvious what is being asked, especially to the OP.

Did I make myself obscure enough?

---55p

9. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
I am certainly not certain that I am in uncertain agreement on that point.