What's this and why does it get so hot? Vintage Marantz related

jemrey81
jemrey81 Posts: 161
edited May 2012 in Electronics
Hey guys, can you tell me what this is and if it's supposed to heat up? It starts getting hot as soon as the unit is turned on. Sorry, I should have circled the thing in the picture... its the gray thing (easier to see in the second photo) that has 3W150 on it (upside down).

IMAG0765.jpg

IMAG0767.jpg


thanks :redface:
Post edited by jemrey81 on
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  • BlueFox
    BlueFox Posts: 15,251
    edited May 2012
    Looks like a heavy duty 150 ohm resistor that can handle 3 watts. It is supposed to get hot. Need a schematic to see what it is actually doing.
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  • bmbguy
    bmbguy Posts: 416
    edited May 2012
    It's a 150 ohm, 3 watt resistor. They chose 3W there for some reason -- obviously they knew it would have to dissipate some heat, and that's why it gets warm.

    Judging from the pics, looks like it's in the power supply.
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  • [Deleted User]
    [Deleted User] Posts: 7,658
    edited May 2012
    Hello,
    That is a 3 Watt, 150 Ohm resistor. Do you happen to have a DC Volt meter? If so, with the unit turned off put one of the meter's clip leads on each end of the resistor, that way you can determine the voltage drop across it (once the unit is turned back on). Once you have that number, square it and divide by the resistor value.
    Regards, Ken
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  • jemrey81
    jemrey81 Posts: 161
    edited May 2012
    Ok cool.... I was just going through it and cleaning out all the years of dust and was checking everything out while it was on. I guess I just need to put some bulbs in and she'll be ready to go.

    Thanks guys :)
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  • jemrey81
    jemrey81 Posts: 161
    edited May 2012
    Kenneth Swauger wrote: »
    Hello,
    That is a 3 Watt, 150 Ohm resistor. Do you happen to have a DC Volt meter? If so, with the unit turned off put one of the meter's clip leads on each end of the resistor, that way you can determine the voltage drop across it (once the unit is turned back on). Once you have that number, square it and divide by the resistor value.
    Regards, Ken

    Thanks for that info... I'll check that out!
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  • bmbguy
    bmbguy Posts: 416
    edited May 2012
    Kenneth Swauger wrote: »
    Once you've made that measurement multiply that times the resistance value (150 Ohms) and see if it exceeds 3 Watts.
    Regards, Ken

    Well, not exactly... Power = (Voltage squared) / Resistance.

    Ack, my mistake -- are both correct?
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  • mhardy6647
    mhardy6647 Posts: 33,808
    edited May 2012
    P = I^2*R
    Ken's right.
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  • [Deleted User]
    [Deleted User] Posts: 7,658
    edited May 2012
    I saw my mistake as soon as I posted it, what I gave was the current flow, not power. That number would then have to be multiplied times the voltage drop for power.
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