View Single Post 18th Oct 2019, 5:56 pm #128 cdm1christopher Heptode   Join Date: Apr 2019 Location: Birchington, Kent, UK. Posts: 528 Re: Philips BX281U20 valve radio Okay yes I understand how those calculations are done it's just remebering ohms law E over I×R. E voltage, I current amps and R. So just moving around depending on what your looking for in this case voltage drop. This part Lawrence has written I do not fully understand and at present I'm still trying to interpret abbreviations used on valve spec sheets as well as trying to understand the graphs. Both the heater current and the rectifier current will flow through the resistor, use the heater current given in the valve data sheets and add that to the rectifier current, for a ball park figure for the rectifier current assume that it's 2.2 times greater than the DC load current. The heater current given in the valve data sheets is 0.1 amps, the DC load current can be worked out by adding up the anode and screen grid currents given in the table in the manual or alternatively it can be found using Ohms law by dividing the voltage given across C75 in the manual by the resistance of R75 which equates to 8.2 (volts) divided by 121 (ohms) which equals approx. 0.068 (amps) Multiplying that by 2.2 equals approx. 0.150 amps, adding that to a heater current of 0.1 amps equals 0.25 amps... Bit lost on DC current side but sure repetitive reading will help work that out. Now I was lost on the R75 121 ohms now I assume since R75 is 2 resistors in parallel that would mean R1xR2 ÷ R1+R2 so 59400 ohms ÷ 490 ohms = 121 ohms. Is that correct? Now since putting in the new resistor in series C75 has dropped to 7 volts bang on. So 0.05 amps or I x 2.2 = 0.12 + 0.1 = 0.22 amps. So think I understand that math. Cheers guys really appreciate your time and the 100 or so posts I will need to print off the important ones so makes it easier to reference. I've noticed a drop of 45v by adding in a 180 ohm resistor. Not checked heater currents will do that bit later. Cheers guys 