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