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Vintage Radio (domestic) Domestic vintage radio (wireless) receivers only. |
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21st Oct 2020, 3:50 pm | #61 |
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Re: Motorola 5A7A dead.
The 150ohm resistor gets warm. volts between it and the rec - 98.5
other side of it to r8=89 center of r8=48 end of r8=6v what does wrt mean? |
21st Oct 2020, 4:11 pm | #62 |
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Re: Motorola 5A7A dead.
wrt = with respect to.
What's the voltage across R8? One probe on each end.
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21st Oct 2020, 4:38 pm | #63 |
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Re: Motorola 5A7A dead.
I can explain 0 on pin 6. My meter has a zero for open, since the antenna was not connected its open, connected its 4meg
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21st Oct 2020, 6:14 pm | #64 |
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Re: Motorola 5A7A dead.
1 & 7 resistance of tubes
1r6 1=0 7=9 1u4 1=55 7=47 1s5 1=18 7=18 3s4 1=55 7=72 |
21st Oct 2020, 8:24 pm | #65 | |
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Re: Motorola 5A7A dead.
Quote:
With the aerial disconnected R1 is also in circuit, so the total should be 8.98M. This won't fix the low filament voltage problem though. Have you made the measurements requested in post #59?
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21st Oct 2020, 9:03 pm | #66 |
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Re: Motorola 5A7A dead.
I think I answered the tests in #59 with the exception the voltage across r-8 is 83vdc, thats the junction of r-4 and r-8 and the other side of the r8.
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21st Oct 2020, 10:14 pm | #67 |
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Re: Motorola 5A7A dead.
FYI I was tracking down the filament voltage issue and located r-20 but the 1u4 tube. The chassis on this radio is the HS-62a which has a 10 meg resistor off of the 1u4 tube pin 1 then onto a 50mmf cap (c-10) to the T-1. Now with the tube removed I measured the resistance across the c-10 and there was resistance. I am not sure if its the vom or circuit causing it but I am suspecting a bad cap. To the point I am thinking of replacing all resistors and mmf remaining caps in this set. They all look like junk.
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22nd Oct 2020, 7:39 am | #68 |
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Re: Motorola 5A7A dead.
R20 is connected to pins 1 and 5 of the 1U4.
I'd leave C10 well alone, low value capacitors seldom give trouble. Chances are you're measuring resistors in parallel with it. Only way to be sure is to lift one end. I'd replace R4, as it's drifted high in value, and see what that does for the filament voltages. If no joy replace the rectifier E2 with a silicon diode such as a 1N4007, marked end towards R4. However you'll then need to increase the value of R4 to compensate for the reduced forward volt drop of the diode when compared to the rectifier.
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22nd Oct 2020, 2:06 pm | #69 |
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Re: Motorola 5A7A dead.
How will I know what the new value of r4 should be?
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22nd Oct 2020, 2:08 pm | #70 |
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Re: Motorola 5A7A dead.
I'd start with something like 250R and reduce the value until you read 1.4V on the valve filaments. Battery valve filaments are easily blown.
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22nd Oct 2020, 2:55 pm | #71 |
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Re: Motorola 5A7A dead.
how many watts on that resistor?
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22nd Oct 2020, 3:02 pm | #72 |
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Re: Motorola 5A7A dead.
The current flowing through R4 will be a lot higher than the DC load current (a good rule of thumb is twice the load current) and that the resistors power dissipation rating should be chosen to suit, eg: if the load current is 0.06 Amps (60mA) then multiply that value by 2 eg: 0.12 Amps, and use that to calculate the power dissipation in the resistor (Isquared*R) eg: 0.12squared*R eg: if say R = 280 Ohms then Pdiss = 0.12squared*280 which would equal approx. 4 Watts dissipation so choose a power dissipation rating that's at least twice that value for a happier cooler life for the resistor and anything that's in close proximity to it.
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22nd Oct 2020, 4:00 pm | #73 |
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Re: Motorola 5A7A dead.
a 250 ohm ceramic resistor at 6 watts or any around those values are more then the radio cost. and to experiment around with resistors would be a financial disaster. I can but these radios for under $20. This one is headed for the trash.
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22nd Oct 2020, 6:52 pm | #74 |
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Re: Motorola 5A7A dead.
That would be a pity. It's yours to dispose of as you wish, though.
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23rd Oct 2020, 2:31 pm | #75 | |
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Re: Motorola 5A7A dead.
Quote:
I looked in on this thread before and had a funny feeling it was going to end this way - glad I didn't waste my time contributing to it. It was probably a 'man-made' fault caused by the original work. However, as said, it's completely the OP's choice to do whatever he wants to do and perhaps some of the readers of this thread have learnt (or learned for the US...like valves/tubes) something, so perhaps all is not lost |
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