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The output voltage should give 0-5 V with the maximum current of 500mA. should use 1 Op-Amp and 1 MOSFET (no commercial regulator chip).
the Voltage is varied using a potentiometer

I tried to do using circuit maker but the voltage step if too big when the potentiometer is varied

2007-05-13 03:21:15 · 4 answers · asked by Anonymous in Science & Mathematics Engineering

4 answers

how is the voltage varied in your setup? try using a micro adjustment in your voltage selector if you are using resistance to vary the output.

2007-05-13 03:30:31 · answer #1 · answered by lnfrared Loaf 6 · 0 0

You have at least two choices, both assume that the MOSFET is rated for well more than 500 mA at 5 V:
1. Use the FET as a variable resistance, with gain determined by the op-amp circuit as a gate driver. Not a really good choice, but it could be made to work.
2. Make a buck converter, with the FET as the main switcher and the op-amp as the gate driver. In this circuit, I would assume that the supply voltage is either a rectified ac (from a transformer wall wart) or directly from a battery. Create an astable multivibrator (oscillator) using the op amp with the potentiometer as the fine adjustment. Output of this controlled oscillator drives the FET gate. Be sure to use a suffciently large capacitor on the output of the FET, and be sure to ramp up the output on start-up, so you don't blow the FET while charging the cap initially.

A quick search on buck converters should yield a few example circuits. If you want to be really advanced, you could even make your own very low power single-phase DC-to-AC inverter. Good luck!

2007-05-13 05:46:05 · answer #2 · answered by Steve W 5 · 0 0

I guess there are many ways. I have used a simple Wall adapter and potentiometer playing around.
I not sure this link what you want, but, may give some ideas.
http://www.seattlerobotics.org/encoder/200112/variable.html

2007-05-13 03:31:43 · answer #3 · answered by Snaglefritz 7 · 0 0

I recoment a wheatstone bridge.

2007-05-13 03:42:08 · answer #4 · answered by eric l 6 · 0 0

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