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I am restoring my Sansui AU 6600. On the f 2097 driver and power supply board there are four(two for each channel) 3W cement resistors rated at 0.47 ohm

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I took all four out of the circuit and tested them and the meter reading is between 0.68 and 0.71 as you can see from a photo.

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My question is, as I don't the tolerance looks like they are about 40% over the 0.47 listed in the part list.

Is that OK? Does that mean that after over 40 years the resistors have changed the resistance? Do I need to replace them?

Here is what they look like:

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Hi thank you for your reply. I've red your reply, I must be honest but my knowledge is not sufficient to fully understand it. However just for curiosity attached a photo showing what happened to the resistor R47. Plus a photo of the schematic. This amp was in very bad shape (still is I am recapping it at the moment)as I got it from a friend who had it repaired several times over the years....anyway so the channel (if i remember well the left where the resistor R47 is ) is gone and the two power transistors (2sc898 and 2sa758) are blown out. Actually I didn't find those exact ones but instead there were motorola (genuine or counterfit??) Mj15025 and MJ15022 . Sorry I am new to this site I tried to give you 5 points but was allowed 2 only. Cheers

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  • \$\begingroup\$ So in 40 years, these resistors have increased in value by 40%. I'd just reuse them, and if you can improve the heat flows out of the areas, to get another 40 years lifespan. Oh, these affect crossover distortion. What do do? \$\endgroup\$ – analogsystemsrf May 11 '19 at 15:38
  • \$\begingroup\$ I have carbon composition resistors that have increased 30% over 40+ years. But, this is probably a wire-wound, less likely to age. I suspect that JRE is correct, you need to measure more carefully. \$\endgroup\$ – Mattman944 May 11 '19 at 17:15
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The datasheet for that resistor says that it has a 10% tolerance. That's the k after the value (0.47 ohm k.)

So, it's allowed to be between 0.423 and 0.517 ohms.

That you measured 0.71 ohms doesn't mean that the resistor is bad, though.

Short your multimeter leads, and take note of the resistance. Subtract that from the values you measure for your resistors.

The leads of the multimeter have resistance, and there's a little resistance at the contact points. Normally, you don't care. What's 0.3 Ohms when you're measuring 100k ohms? The values you are working with, though, are so small that you need to account for things you can normally ignore.

Some multimeters have a zero function for this kind of thing. Short the leads, press the zero button, and the following readings have the lead resistance subtracted.

Your UT61E has this function. Short the leads, press "rel" and it will subtract the lead resistance for you.

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  • \$\begingroup\$ Hi JRE Thank you very much for your reply , very helpful I did as you said and the reading is 45 so I think is ok to reuse it . I attached a photo of the reading. Cheers \$\endgroup\$ – Anthony May 11 '19 at 17:36
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If you zoom out on your schematic a little I expect that, because you say you have four resistors, we will see another pair of output transistors, each with an emitter resistor.

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Figure 1. When doubling up output transistors it is common to add some emitter resistance. Source: TubeCAD.

To increase the output power of an amplifier the output transistors can be doubled up. The problem is that even with a good heatsinking arrangement the temperature dependent Vbe will vary between the two parallel transistors and the one with the lower Vbe will start to hog the current. This will raise the temperature, lowering Vbe further and probably lead to failure of that transistor.

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Figure 2. IE versus Vbe. Source BJT characteristic curves.

The emitter resistors are to balance the currents. Yours are nominally 0.5 Ω so will have a voltage drop of 0.5 V/A. This will be much more significant than variations in Vbe.

I wouldn't expect much trouble with these but since you have them out I suggest that you make matched pairs - as best you can - and install a pair on the NPNs and the other pair on the PNPs. They should then function as intended.

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  • \$\begingroup\$ Hi transistor thank you for your reply I added a comment in the description in the question as my comment was too long, you can also see the photos I attached . many thanks \$\endgroup\$ – Anthony May 11 '19 at 18:10

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