New project mosfet amp with depletion mode front end.

I converted one of my old 150WRMS irfp240/9240 amps to have a depletion mode mosfet front end in the LTP.

I started with a 2SK170 front end but my rails were +/-45VDC which stopped that. So i had to search around and found some Microchip depletion mode mosfets at RS Components which work up to 300VDC.
Luckily, I checked pin out and Micrchip mosfets are opposite way around to 2SK170's so have to rotate them 180 degree's on my pcb which was made for 2SK170's.
Got my pcb's made at JLCPCB. Built up the first one and powered it up.
It was terribly distorted. So I set output DC offset to zero.
I then turned up bias to give about 10mA and the previously distorted sound transformed into a great sounding little amp.
The voices are very clear and I can hear what the words are in songs I couldnt previously.
So been a good project and very happy with it.
100-0285.jpg
 
Thanks Nigel.

A wee bit of a pity the graphic-capture resolution rendered all values and markings essentially unreadable to my septuagenerian eyes. Maybe younger peepers will be OK. Dunno… this ageing thing is getting … old.

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Better!

After I got over the eccentric layout (likely 'default' by your digital schematic entry app) I think I understand most of the thing.

One question tho': R3, R4, Q4, C3, D1 … I can't quite figure if that is a phase mirror or just a variable ΔV drop from the Q3 PNP 'upside down' VAS-phase-re-inverter, since tapping R9 is the first inversion/amp stage in a constant-current tailed LTP differential pair configuration. A configuration with solid feedback thru the 'other leg'.

Insight would be helpful. As in, where did you adjust to 10 mA?

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Out of curiosity, what is voltage across R1? … actually that's not a good question. It should ideally be 0.63 to 0.72 volts. The better question would be what is the voltage difference between pin 2 and pin 3 (G and S) on the DN–2530 left-side FET?

Datasheet (for what its worth) implies something around –2.5 V … but seems to have a lot of variability in specs. The 220 Ω R1 at 0.7 V essentially guarantees a constant current of about 1.6 mA per D-MOSFET.

And those low currents are NOT defined on any of the industry-sourced device PDFs. This doesn't mean one shouldn't try to run them at 1.5 mA! Rather … just have to measure the voltage, and broadcast it so others can plan for it.

Thanks, Nigel
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The current through both depletion fets is about 1.5mA.
With around 45 volts across the fet gives 0.07 watts.
I didnt want to cook the fet but at the same time have a bit of current going through it.
I am sure it could be a bit more as max power for fet is around 0.75 watts.
That gives up to 15mA but like I said before I didnt want to run the fets hot.
 
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Math slip … 45 V × 0.0015 A = 0.063 W or 63 milliwatts. Plenty of margin! I'd bet (tho' I wouldn't advise doing it) that the TO–92 MOSFET case barely gets warm.

As you say, you could run 'em hotter.

Changing out R1 to 68 Ω (a nice standard value) only raises the per FET current to 5 mA. 5 mA at 45 V is only 225 mW.

Now the MOSFETs would get warm. But not hot. A quarter watt is itsy-bitsy stuff.
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Did you work up the whole design, or from bits-and-pieces of other work? I am quite pleased with the creativity employed making the BJT-double-back constant-current sources. They're sweet. Probably rock-solid too, from a current stability perspective.

The nice thing about changing R1 to 68 Ω is that R9, the drain-load I→V resistor would run with a ΔV of about 5 V, substantially increasing the dynamic range of the input signal. Maybe it wasn't a goal, but I'd do the R1 swap, just to get that increased range.

Nice 'talking' with you Nigel.
Thanks for playing along.

⋅-=≡ GoatGuy ✓ ≡=-⋅

EDIT ... I see you fixed up the dissipation math decimal slip. Bravo
 
While swapping out the fets I changed R1 to 100R so it pulls a bit more current and it sounds good.
Thanks for the update.

Your Q1 quiescent current is essentially
IS = ½ VBE / R1
where
VBE is the nominal forward voltage of the NPN base-emitter junction
R1 is per your diagram, in Ω​
So, following that:
IS = ½ ( 0.7 V ÷ 220 Ω ) = 0.0016 A, 0.072 W, 1.6 V R9 drop
IS = ½ ( 0.7 V ÷ 100 Ω ) = 0.0035 A, 0.158 W, 3.5 V R9 drop
IS = ½ ( 0.7 V ÷ 68 Ω ) = 0.0051 A, 0.232 W, 5.2 V R9 drop
IS = ½ ( 0.7 V ÷ 47 Ω ) = 0.0074 A, 0.335 W, 7.5 V R9 drop​
So, there's a conveniently precomputed range of current-and-power values that might be of interest.

Best of Luck!
Feed the Kaboom gremlins!

⋅-=≡ GoatGuy ✓ ≡=-⋅