What range pot do i need to reduce 1.25v, 200 mA. to a range from .2v to.5v?

Question

This will be used to adjust the signal sent from the O2 sensor to the control module in an automobile. I got 20Ma when I tested the output voltage of the sensor. I need to raise the voltage sent to the control module, if the sensor is sending .3v I need to module to get .4, etc. The sensor sends a signal that varies from .2v to .9v. I made a typo in my first post, the .5v should be .9v.

When I checked the output signal from the sensor, it produced a pretty steady 20mA.

Answer 1

The fact that you gave a current rating suggests that you want to use this circuit to power something. Using a pot as a series resistor only works if the current draw is constant, which I think is very unlikely.

I would use a NPN transistor in emitter-follower configuration. You need a medium power transistor like a BC327, a 100ohm pot and a 120 ohm resistor. Make these connections:

1.25V to top of pot.

resistor to bottom of pot

other end of resistor to 0V

wiper of pot to base of transistor

collector of transistor to 1.25V

emitter of transistor to output

That will give you the output range you need, reasonably well regulated.

Edit:

Your additional info is a lot more helpful. Forget about the 20mA, it is irrelevent, this looks like a voltage output - voltage input connection. If I understand correctly you either want a circuit where

Vout = Vin * K (a non-inverting amplifier), or

Vout = Vin + Vk (an adder)

Where K is a number > 1 and Vk is a constant voltage.

You can do either of these with a simple op-amp circuit, but they are too complicated to describe here. First you need to understand op-amp basics, so start by googling basic op-amp circuits.

It will need a power supply of some kind, eg. a battery.

Answer 2

need a more detail for a definitive answer...

But assuming you just want a series resistor to reduce the voltage, and that the 200 ma is fixed (othewise it won't work).

Then Rhigh = (1.25 - .20 ) / .2 = 5.25 ohms

....power = .21 watts

Rlow = (1.25 - .50 ) / .2 = 3.75 ohms

...power = .15 watts

So you need to cover a range of 3.75 ohms to 5.25 ohms, so say 3 to 6 ohms.

A 10 ohm pot will do, they are usually 1 watt, so that is ok.

A 5 ohm would be better if you can find one, with a 2-3 ohm resistor in series.

A 3 ohm pot would be best, with a 3 ohm series resistor