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I acquired an excellent vintage amplifier and started the journey of giving it a new life. I have managed to repair most of the damage and have come to a working device.

But there is one problem still to solve. In the attached image there are 9 resistors marked in red. With the help of a thermal camera, I discovered that the resistors are getting hot: 100-130°C. The temperature when the amplifier gets an input signal is the same as in standby without an input signal.

I measured the resistors in question and they are all in range. The main voltage rails are a bit higher, 62V is 65V, and 58V is 61V. The voltage drop across transistors is the same. Taking into consideration the transistor voltage drop, with the higher input voltage, the voltages on the resistor are in good condition.

So if anybody has an idea or a proposal for the next step I would really appreciate it.

Thermal images of the problematic spots:

enter image description here

enter image description here

enter image description here

Hotspots are marked with a red circle:

Hotspots are marked with a red circle

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    \$\begingroup\$ You have schematic with voltages marked on it. Did you measure the circuit which voltages match and which don't? \$\endgroup\$
    – Justme
    Aug 12, 2022 at 18:15
  • \$\begingroup\$ What’s the voltage across it and current through it? Is it oscillating? \$\endgroup\$
    – winny
    Aug 12, 2022 at 18:42
  • \$\begingroup\$ I measure the voltages and they are OK, It is just weird that they are using a lower power rating for those resistors. When you calculate what power rating they should have there is a big difference in what they used. \$\endgroup\$
    – Nano
    Aug 13, 2022 at 10:13

1 Answer 1

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You can calculate what the power dissipation should be from the voltages and resistances shown on the schematic.

$$P=\frac{E^{2}}{R}$$

For example R441 and R442 are shown as 12k, 1/2 W resistors. They are in series and the voltages at each end are shown as 57.2 V and -57.2 V.

So R is going to be \$12k\times2 = 24k\$
V is going to be \$57.2 V + 57.2 V = 114.4 V\$

Using the formula:

$$P= \frac{114.4^{2}}{24k} = 0.545 W$$

That power is divided equally between the two resistors so each one should dissipate 0.2725 W, a little over a quarter watt. They're rated at 1/2 W so they should be good, but might get a little warm.

You can calculate the expected dissipation of the others. R479 and R480 look like they should have 49 V across them (62 V - 13 V) so at 4.7k would be 0.51 W which is over their rating. I don't know why they would design it that way. You might want to replace them with higher wattage resistors.

Measure the voltages to make sure they are close to what they are shown as on the schematic, and check with a scope for oscillations, if you have one.

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  • \$\begingroup\$ So I measured DC voltage R441 which is +61.2V and R442 is 61.4V and I measured AC whit my scope ground to R441 and 422 with no oscillation. \$\endgroup\$
    – Nano
    Aug 12, 2022 at 21:32
  • \$\begingroup\$ The 58v rail is 62V and the 62V rail is 67V \$\endgroup\$
    – Nano
    Aug 12, 2022 at 22:01
  • \$\begingroup\$ If we take into consideration the higher rail voltage and take the voltage drop, the voltages are pretty much OK. So I should change all the resistors in the questions to a higher Power rating? \$\endgroup\$
    – Nano
    Aug 13, 2022 at 9:32
  • \$\begingroup\$ Your supply voltages are 7% to 8% high. The supplies are unregulated and the '62 V' supplies use voltage doublers, so they would vary with line voltage and can't be expected to be exact. If you 'recapped' it that could change the voltages as well. If everything else is working and it's just a matter of the resistors getting hot but not to the point where they're burning up, I would think changing them to higher wattage ones should be okay. I can't really say what the long term effects of the slightly higher voltages will be, it depends on how much of a safety factor was used in the design. \$\endgroup\$
    – GodJihyo
    Aug 13, 2022 at 17:32
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    \$\begingroup\$ I found something interesting The amplifier has an input voltage selector and has been set to 220V AC. I am from the EU so my main voltage is 230V AC so I set the selector switch to 240V AC. Now the 58V DC rail is 56.3 and the 62V DC rail is 60.1V DC. The temperature of the resistor has fallen down to 90-100°C. And the DC balancing and Idle current are more stable with less drift, the thing is the resistor is on the pre-drive board and drive board so thermal drift affects the stability of the amplifier. I think changing the resistors to a higher power rating can improve the stability much more \$\endgroup\$
    – Nano
    Aug 13, 2022 at 18:59

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