4
\$\begingroup\$

I'm working on the final stage of my guitar amp. This stage needs to take the line-level signal (~1vpp-~2vpp) from the FX_RCV jack and swing it hard enough to drive the EL34 output tube. This stage of the amplifier is solid state to save on $.

I'm using an LND150 (500v depletion mode MOSFET) to boost the voltage and an IRF820 (500v enhancement mode MOSFET) to supply enough current to drive the output tube (when it goes into saturation).

When I fired the FX loop recovery section up, I had one resistor nearly explode (R29), and another smolder (R28) (Always wear safety glasses around high voltages!).

My wiring appears to be correct :( Is something wrong with my design around the LND150?

EDIT

I figured it out! My wiring is so not correct! The gate on the LND150 is the middle pin, and the gate on the IRF820 is the left pin. This means ALL of my MOSFETs are wired incorrectly. I made the assumption that did not vary from the transistors I've used in the past vs these MOSFETs and the lesson is to RTFM!

claideamh-v1.4.1.png

20160924_134701.jpg

\$\endgroup\$
11
  • 1
    \$\begingroup\$ What are the voltage and power ratings of your resistors? \$\endgroup\$
    – marcelm
    Sep 24, 2016 at 20:01
  • 2
    \$\begingroup\$ Wild speculation: you put in a 100Ω resistor as R28 by accident. When the FET turned on, this dumped all of the energy in C22 (and maybe C21) through R28-Q3-R29. Q3 immediately died, failing shorted. The remaining energy gets divided over R28 (100Ω) and R29 (470Ω). Most of the energy is dissipated in the larger R29, which explodes. \$\endgroup\$
    – marcelm
    Sep 24, 2016 at 20:14
  • 1
    \$\begingroup\$ Help us out here with some DC voltage measurements, Q3 grid and anode voltages, and Q4 grid and cathode for a start. \$\endgroup\$
    – user16324
    Sep 24, 2016 at 20:24
  • 1
    \$\begingroup\$ We do need some measurements including the state of the components. I'm for instance looking at what happens if Q4 shorts out, as well as Q3. R29 looks like it got a lot of current through it, and with 366 volts to play with, well... \$\endgroup\$
    – Ian Bland
    Sep 24, 2016 at 20:37
  • 1
    \$\begingroup\$ You can check the conductance of the Mosfets with the power off, with a multimeter :) I think this is why people dealing with valves often power the circuit via a Variac, heh. \$\endgroup\$
    – Ian Bland
    Sep 24, 2016 at 20:46

1 Answer 1

Reset to default
3
\$\begingroup\$

Have you calculated the power dissipation for R28 and R29?

The schematic is a bit fuzzy, but I take R28 to be 100K and R29 to be 470K. Those are large values, but you have the supply voltage given as 366V, which makes the total power rather high despite the high resistance.

I find 0.2 Watt for R29, and less than 0.1 Watt for R28. These are worst case (Q3 completely conducting,) but I think you need resistors rated for higher power. Like, 1 Watt.

Now that the values have been corrected, things change drastically.

R28 will dissipate 1.3Watt worst case. R29 will dissipate 0.06W worst case.

If R28 fails to a lower resistance, then the power through R29 goes up, which would destroy it as well.

\$\endgroup\$
11
  • 2
    \$\begingroup\$ Those resistors in the picture are already pretty fat, they look like 1/2W to me. Even if they're not, 0.5W through a 1/4W resistor will not cause instant fire and/or explosions. So I suspect something else was wrong. \$\endgroup\$
    – marcelm
    Sep 24, 2016 at 19:50
  • 1
    \$\begingroup\$ @marcelm: Judging by appearance, those are metal film resistors. The smaller one would be 1/8 Watt and the large one 1/4 Watt. \$\endgroup\$
    – JRE
    Sep 24, 2016 at 19:55
  • 1
    \$\begingroup\$ The maximum dissipation I can find through that part of the circuit is 3.7mA (due to the 100k resistor) which works out as 1.72W for R28 and .01W for R29. That's if the Mosfet is fully on. But maybe I'm missing something. \$\endgroup\$
    – Ian Bland
    Sep 24, 2016 at 20:00
  • 1
    \$\begingroup\$ @JonathanS.Fisher if R28 were to fail short (or nearly) that would pump a lot of current through the FET and R29. (edit) Is the FET failed short? Then you've just got 366V across the two resistors... \$\endgroup\$
    – Ian Bland
    Sep 24, 2016 at 20:07
  • 2
    \$\begingroup\$ R29 is 470Ω, not 470kΩ \$\endgroup\$
    – Jonathan S. Fisher
    Sep 24, 2016 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.