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Pilot Light Replacement (Pioneer SA-7500)

Volmon

Volmon

New Member
Hi, All. I recently purchased my a Pioneer SA-7500 (my second) and the only thing that seems to be wrong with it - at all - is a dead pilot light. Asking as a novice, is this a difficult repair? Also, where is the best place to buy this particular replacement part? Thanks in advance! Cheers - Lucio (AU)
 
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The lamp is an 8v 50mA lamp, run off of a very much more than 8v AC circuit and uses a highly specific resistance value and wattage resistor to allow it to "sip from the firehose".

A different lamp current will upset that balance and could end up burning out the replacement lamp (or LED).

Thus we (baaahhh me, I will ) will have to carefully reverse engineer the resistance value for the lamp you do use.

MOST replacement lamps are more like 35 or 30 mA.

620 ohms 2 watts 0.050 Amps .... so : ohms law says:

620 x 0.050 = 31 volts dropped by resistor,, 31v x 0.050 = 1.55 watts of power

thus 0.035A or 35mA

31v / 0.035 = 885 ohms 31 * 0.035 = 1.085 watts

a LED would be 10 or 20mA (or less if it is too bright) and 3 volts

31 + 8v - 3v = 36v / 0.010 = 3600 ohms 36v x 0.010 = 0.36 watts

Thus you see, it's not a simple question, nor a simple answer
 
markthefixer said:
The lamp is an 8v 50mA lamp, run off of a very much more than 8v AC circuit and uses a highly specific resistance value and wattage resistor to allow it to "sip from the firehose".

A different lamp current will upset that balance and could end up burning out the replacement lamp (or LED).

Thus we (baaahhh me, I will ) will have to carefully reverse engineer the resistance value for the lamp you do use.

MOST replacement lamps are more like 35 or 30 mA.

620 ohms 2 watts 0.050 Amps .... so : ohms law says:

620 x 0.050 = 31 volts dropped by resistor,, 31v x 0.050 = 1.55 watts of power

thus 0.035A or 35mA

31v / 0.035 = 885 ohms 31 * 0.035 = 1.085 watts

a LED would be 10 or 20mA (or less if it is too bright) and 3 volts

31 + 8v - 3v = 36v / 0.010 = 3600 ohms 36v x 0.010 = 0.36 watts

Thus you see, it's not a simple question, nor a simple answer
Click to expand...
 
Is it okay to run a bulb with less mA draw than the original without changing the resistor to match the original load?
 
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