Reservoir capacitor - UK Vintage Radio Repair and Restoration Discussion Forum

TrevorG3VLF · 7th Dec 2016, 11:21 pm

What sort of peak to peak voltage should be expected with a 32uF reservoir capacitor in a normal valve superhet radio and what should be expected after further smoothing?

Trevor

Herald1360 · 8th Dec 2016, 12:46 am

Very roughly, at 50mA load with FW rectification, about 10V p-p on the reservoir. Add 1K series R and another 32uF and it falls to about 300mV p-p

If you don't have a simulator to "cheat" with, then roughly calculate the voltage drop on the reservoir capacitor in around 8mS (about the time between peak volts and the next half cycle catching the fall and heading back up to peak volts) using V=iT/C (use amps, seconds and farads to get the right order of magnitude). This would give V= 0.05*0.008/0.000032 = 12.5V p-p (a bit pessimistic).

The next stage is a bit more difficult, but again treat the 1K 32uF as a simple potentiometer with 1K and 1/(2*pi*f*C) or 1K and 1/(2*3.142*100*0.00032)

= 1K and about 20R, so residual ripple = 12.5*20/1020 = 245mV (now a bit optimistic)

Close enough to be useful anyway.

7th Dec 2016, 11:21 pm	#1
TrevorG3VLF Rest in Peace Join Date: Apr 2011 Location: Matlock, Derbyshire, UK. Posts: 1,378	Reservoir capacitor What sort of peak to peak voltage should be expected with a 32uF reservoir capacitor in a normal valve superhet radio and what should be expected after further smoothing? Trevor

8th Dec 2016, 12:46 am	#2
Herald1360 Dekatron Join Date: Feb 2007 Location: Leominster, Herefordshire, UK. Posts: 16,535	Re: Reservoir capacitor Very roughly, at 50mA load with FW rectification, about 10V p-p on the reservoir. Add 1K series R and another 32uF and it falls to about 300mV p-p If you don't have a simulator to "cheat" with, then roughly calculate the voltage drop on the reservoir capacitor in around 8mS (about the time between peak volts and the next half cycle catching the fall and heading back up to peak volts) using V=iT/C (use amps, seconds and farads to get the right order of magnitude). This would give V= 0.050.008/0.000032 = 12.5V p-p (a bit pessimistic). The next stage is a bit more difficult, but again treat the 1K 32uF as a simple potentiometer with 1K and 1/(2pifC) or 1K and 1/(23.1421000.00032) = 1K and about 20R, so residual ripple = 12.520/1020 = 245mV (now a bit optimistic) Close enough to be useful anyway. __________________ ....__________ ....\|____\|\|__\|__\_____ .=.\| _---\__\|__\|_---_\|. .........O..Chris....O Last edited by Herald1360; 8th Dec 2016 at 1:01 am.