K8010 - Question regarding power dissipation of R73


I just finished building two K8010 kits.
After doing some first measurements everything seems to work pretty well.

But I encountered an issue with R73 (15 ohms / 5 Watt), which is located right after the rectifier
in the HV section of the power supply.

In the schematic there are two voltages stated, which are 395V and 390V. I can measure these two
voltages in both amps. So everything seems to be fine from the documentation and design point of view.

395V - 390V = 5V voltage drop over this resistor
So there is a current of: I = 5V / 15 ohms = 0,33 A
This calculated current of 0,33A is also okay, if we keep in mind, that each of the four KT88 tube is
powered with approx. 75mA, which gives us a total of 0,3A for the KT88 tubes plus the required
current for the ECC81 and ECC83.

This voltage drop and current results in a power dissipation of
P = 5V * 0,33A = 1,66 Watt

The resistor is a 5 Watt type. So this seems to be okay either.

I soldered this resistor with a distance of approx. 1cm to the PCB to allow a better
air flow.

During my measurements I noticed, that this resistor gets really hot!

I used a IR-Thermometer to measure the surface temperature of this resistor and
got 213° celsius!

I think this temperature is “a little bit” too high from long term durability point of view?
Or is it just okay to have a working temperature of more than 200° celsius on this resistor?

Due to the fact, that the 5V drop over this resistor is also stated in the schematic,
I’m sure that I’ve made no error during my build.

Best regards

Jochen Hammer

It is a ceramic resistor, should not be a problem.