Help with Substituting Transistors

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Hi all, I have trouble obtaining the original transistors of the attached amplifier. So I want to use equivalent transistors. please tell me which transistors are better for using in the amplifier.
(The original amplifier uses +/-95V rails but I'm going to use it with +/-50V)

For:
MJE15034) TIP41 or 2SC669 ?
MJE15035) TIP42 or 2SB649 ?
IRF510) IRF540 or IRF 630 or IRF640 or IRF730 or IRF740 or IRF840 ?


Best regards

Ermac
 

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the irf510 is used as a Vbe multiplier.
In this location a BJT can be used.
A To92 can be glued or clamped to the heatsink.
or inserted into a close fitting hole drilled in the heatsink
or
a To126 device can be used bolted to the heatsink or bolted on top of one of the output devices.

The Vbe multiplier transistor must monitor the temperature of the output devices as closely as possible.

Change the 10k VR to 2k and add a resistor in series with it. Try 160r

If your VR is accidentally set to 0r0 then the maximum voltage from the Vbe multiplier will be ~{2200/160 + 1} * Vbe ~=8.8V
The minimum voltage will be ~{2200/(160+2000) + 1} * Vbe ~=1.2V
this will ensure the FETs turn off at minimum Vbe voltage and turn on at maximum Vbe voltage.

But, I repeat my warning mirrored by others.
You are in for a mighty fright that will be expensive and potentially very dangerous when this amplifier blows up.
Start with something much simpler and much cheaper and of much lower voltage.
 
AndrewT said:
the irf510 is used as a Vbe multiplier.
In this location a BJT can be used.
A To92 can be glued or clamped to the heatsink.
or inserted into a close fitting hole drilled in the heatsink
or
a To126 device can be used bolted to the heatsink or bolted on top of one of the output devices.

The Vbe multiplier transistor must monitor the temperature of the output devices as closely as possible.

Change the 10k VR to 2k and add a resistor in series with it. Try 160r

If your VR is accidentally set to 0r0 then the maximum voltage from the Vbe multiplier will be ~{2200/160 + 1} * Vbe ~=8.8V
The minimum voltage will be ~{2200/(160+2000) + 1} * Vbe ~=1.2V
this will ensure the FETs turn off at minimum Vbe voltage and turn on at maximum Vbe voltage.

But, I repeat my warning mirrored by others.
You are in for a mighty fright that will be expensive and potentially very dangerous when this amplifier blows up.
Start with something much simpler and much cheaper and of much lower voltage.


Would you please tell us why other MOSFETs (eg IRF740) cannot be used?

Kind regards, Ermac
 
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