4 volts from 6.3 volts - UK Vintage Radio Repair and Restoration Discussion Forum

michael99 · 6th Oct 2011, 1:30 pm

If you half-wave rectify a 6.3V sinusoidal supply, and assume 0.6V drop across the diode, you get within a few millivolts of 4.0V RMS - useful if you need, but don't have, a 4V winding. This can be proved by doing the integration from first principles (1st year A-level stuff), bearing in mind that the peak voltage is 6.3xroot2 minus 0.6, and the diode is conducting for slightly less than 180 degrees. If you're running more than one valve, point the diodes in opposite directions to avoid DC in the transformer. Easier than removing turns, and cooler-running than using resistors. The technique was common in the days of series-connected TV heater chains, but I've never seen it spelt out that, here, the numbers work out particularly conveniently.

ppppenguin · 6th Oct 2011, 1:50 pm

Looked at in a simpler way for those whose integral calculus is rather rusty. I know mine is, because I just tried the integral and was a factor of 2 out for reasons I haven't quite worked out yet.

The power is on for half the time, so it's half the power. Hence the RMS voltage is divided by SQRT(2), about 4.5V. Subtract the 0.6V diode drop and it's near enough 4V RMS. Actually I should subtract the 0.6V first which gets even closer to 4V.

I like the idea of using 2 diodes to balance the load.

G8HQP Dave · 6th Oct 2011, 1:54 pm

You can do the calc roughly without using integration. 6.3V rms for half the time is 6.3x0.707=4.45V rms. Then take off the diode drop, which will be approximately 0.6x0.707=0.42V rms. One snag, will it create buzz because the heater supply for any one valve will now include mains harmonics, so will jump any capacitive link more easily?

6th Oct 2011, 1:30 pm	#1
michael99 Retired Dormant Member Join Date: Jul 2011 Location: Hastings Posts: 2	4 volts from 6.3 volts If you half-wave rectify a 6.3V sinusoidal supply, and assume 0.6V drop across the diode, you get within a few millivolts of 4.0V RMS - useful if you need, but don't have, a 4V winding. This can be proved by doing the integration from first principles (1st year A-level stuff), bearing in mind that the peak voltage is 6.3xroot2 minus 0.6, and the diode is conducting for slightly less than 180 degrees. If you're running more than one valve, point the diodes in opposite directions to avoid DC in the transformer. Easier than removing turns, and cooler-running than using resistors. The technique was common in the days of series-connected TV heater chains, but I've never seen it spelt out that, here, the numbers work out particularly conveniently.

6th Oct 2011, 1:50 pm	#2
ppppenguin Retired Dormant Member Join Date: Dec 2003 Location: North London, UK. Posts: 6,168	Re: 4 volts from 6.3 volts Looked at in a simpler way for those whose integral calculus is rather rusty. I know mine is, because I just tried the integral and was a factor of 2 out for reasons I haven't quite worked out yet. The power is on for half the time, so it's half the power. Hence the RMS voltage is divided by SQRT(2), about 4.5V. Subtract the 0.6V diode drop and it's near enough 4V RMS. Actually I should subtract the 0.6V first which gets even closer to 4V. I like the idea of using 2 diodes to balance the load.

6th Oct 2011, 1:54 pm	#3
G8HQP Dave Rest in Peace Join Date: Sep 2008 Location: Solihull, West Midlands, UK. Posts: 4,872	Re: 4 volts from 6.3 volts You can do the calc roughly without using integration. 6.3V rms for half the time is 6.3x0.707=4.45V rms. Then take off the diode drop, which will be approximately 0.6x0.707=0.42V rms. One snag, will it create buzz because the heater supply for any one valve will now include mains harmonics, so will jump any capacitive link more easily?