Re: How to lower the output of this FET regulator

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Sunsun22

Re: How to lower the output of this FET regulator

I bought this all FET power regulator from diyzone.net. It can provide + - 10V to + - 20V with 700 mA current. However, I wish to know is there a way to lower the output voltage to + - 5V?

I tried to ask diyzone.net but their answer is "post the question on the forum". I did but there is no reply.

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guitvinny

Voltage divider ?

Frank Berry

Reduce the value of R12 and R13.

Sunsun22

By how much I should reduce the value of these 2 resistors? Would the circuit becomes unstable?

Sunsun22

It looks like I have an opposite advise on the values of R12 and R13. When I reduced the value from 500 ohms to 220 ohms, the output turns out to be 11.8V to 24.5 V. When I chaned the value to 3.9K, the output turns out to be 6.5V to 18V.

I believe I can achieve 6V eventually by increasing the value of R12 and R13 but will it casue the circuit into unstable?

Ipanema

Hi,

You can short resistor R11 and R14 to get +/-5V. If it oscillate, then increase C11 or C13.

Regards.

jan.didden

Paid Member
Ipanema said:
Hi,

You can short resistor R11 and R14 to get +/-5V. If it oscillate, then increase C11 or C13.

Regards.

I agree. The reference is 5V at VCR. To get 5V out you need also 5V at the feedback point VCF. To do that just short R1 (R14) as noted. There's not too much gain in this circuit so it may be stable as-is, especially since C13 is already shorting R11 for hi freqs. But if it oscillates, increase C11 and C12 (not C13) as noted.

Jan Didden

Sunsun22

Thank you. I'll try this way tonight.

What IF

I want a + - 6V output? Should I keep on increasing the value of R12 and R13 or should I reduce the value of R11 and R14, let say, to 500 ohms?

jan.didden

Paid Member
Sunsun22 said:
Thank you. I'll try this way tonight.

What IF

I want a + - 6V output? Should I keep on increasing the value of R12 and R13 or should I reduce the value of R11 and R14, let say, to 500 ohms?

Just play with R11. Assume that VCR = VCF = 5V (that's the reference value from D1 IIRC). You can caluculate the ratio of R11 to the resistor below it to get the Vout. Example: if R11 is 1k, and the resistor below it is 5k, you know that the voltage across that resistor below it is 5V. Therefore, V across R11 = 1 V, so Vout = 6V. Real simply, really.

Jan Didden

Sunsun22

I love your conceptual explanation. I'll try it tonight.

Thank you very much.

jan.didden

Paid Member
Sunsun22 said:
I love your conceptual explanation. I'll try it tonight.

Thank you very much.

There's nothing so practical as a really good theory

Jan Didden

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