Linear change of resistor value

[badflyer]

Hi all, i require a circuit that will linearly change a resistor value from 5K to 500R in a 10mS time frame. Is it possible to do perhaps with a FET? its for a power supply projetct im doing that requires me to drop up to 10v in the timeframe. The PSU as the availability to use an external resistor to set the volts.
Cheers

 

[Russlk]

The FET is a variable current sink, not a resistor, but it could do the job. How linear does it have to be? The FET is not very linear at low current. Have you looked into digital potentiometers?

 

[badflyer]

RE:

Thanks for the reply! Thinking about it, all i need is to be able to bring the resistance up or down in 10mS. i just need to find a simple method of doing it. Cheers

 

[Exo]

with a digital pot?
how much current goes trough this resistor (at lowest resistance) ?

 

[badflyer]

RE:

I'm not too sure how much current will pass, but a 1/4 watt resistor is specified if i require a fixed reference voltage. I do know that the programming coefficient is 500R per Volt. Cheers

 

[Exo]

do you have a scematic or something you can put here ?

 

[badflyer]

RE:

Sorry i dont have one my friend! i'm trying to design a circuit that will do the job. All i know is that to give me a 1v o/p on the psu, i need a fixed 500R across the o/p reference. If i wanted 10v o/p i would need to put a fixed resistor of 5k.

I just need a circuit that will do this swing in a 10ms time frame. A digital pot seems a possble solution? Cheers

 

[Exo]

Yes, a digital pot will do the job, but the maximum current and power dissipation a digital pot can handle is very low.
Therefore it's important to know where this resistor will be connected to...
if it, for example, drives a transistor or something, then it can probabely handle it. But if it needs to drive the load directly then it can't...

 

[Roff]

The tolerance on a digital pot used as a rheostat is pretty sloppy. If you control the pot digitally, how much resolution do you need (in ohms)?

 

[badflyer]

re

Hello! i'm not too woried about the resolution for the time being, just a quick voltage jump or down is sufficient. Thanks

 

[Russlk]

It would appear that all you need do is put 5K across the terminals to set the 10 volts and in parallel with that, an FET with 500 ohms in series with the drain. When the FET is turned on, the voltage will jump down to 1 about 1 volt (a little less because of the 5K in parallel).

 

[Oznog]

It is likely- very likely- that this resistor is just used as a voltage divider to create a DC voltage. You'll do better to find which terminal and just supply the DC voltage, frankly.

Find what voltage is across that resistor. I think a digital potentiometer is usually limited to 5V.

You can create a 0-5v reference voltage directly from a PIC's PWM and a lowpass filter. If this voltage is not high enough, or the output impedance of the filter is too high, use an op amp to buffer it.

It seems somewhat unlikely you will be able to get the output to follow a 10mS change in the reference. There may be a low pass filter in the power supply's feedback loop preventing changes around this speed. And why would you need to do this? If you want to amplify a signal, the right way to do it is a power amplifier circuit powered by the supply.