RCA 117

6/10/2014 3:55:49 PMfrank goddard

Hi Folks

I need to replace the tapped resistor dropper to this radio. On the schematic it is R17 - 18 - 19 - 20. What wattage rating should I go for.?
Any help appreciated.

6/10/2014 4:44:23 PMCV

R17 = 2 watts
R18 = 5 watts
R19 = 2 watts
R20 = 3 watts

The above values have already been derated.

How to figure power ratings: R17 is the audio output cathode current resistor. It will only be exposed to the tube current (AC+DC) which will be in the neighborhood of 50mA. The power that it will therefore "see" is I*I*R,or (50mA)*(50mA)*500 = 1.25 watts. Derate to 2 watts.

R18-R20 are in series and go from B+ to ground. So, they will all experience the same current as follows: 300V/(10K+3K+5K) = 16mA. For R18 (15K) the power it will see is therefore (0.016)*(0.016)*(10,000)=2.56W. Derate to 5 watts.

For R19 (3K) The power will be 0.83 watts. Derate to 2 watts.

For R20, the power will be 1.28 watts. Derate to 3 watts.

6/11/2014 3:28:45 PMfrank goddard

Wow!Thanks CV. That's just the info I need,

Frank

:R17 = 2 watts
:R18 = 5 watts
:R19 = 2 watts
:R20 = 3 watts
:
:The above values have already been derated.
:
:How to figure power ratings: R17 is the audio output cathode current resistor. It will only be exposed to the tube current (AC+DC) which will be in the neighborhood of 50mA. The power that it will therefore "see" is I*I*R,or (50mA)*(50mA)*500 = 1.25 watts. Derate to 2 watts.
:
:R18-R20 are in series and go from B+ to ground. So, they will all experience the same current as follows: 300V/(10K+3K+5K) = 16mA. For R18 (15K) the power it will see is therefore (0.016)*(0.016)*(10,000)=2.56W. Derate to 5 watts.
:
:For R19 (3K) The power will be 0.83 watts. Derate to 2 watts.
:
: For R20, the power will be 1.28 watts. Derate to 3 watts.
:

6/11/2014 5:27:22 PMCV

I incorrectly referred to R18 as being 15K. It is actually 10K. The math uses the (correct) 10K value so the computed power requirement for it is correct. Sorry for the confusion.