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#### Cpan

Hi All
I'm starting to take the parts to build the MoFo amplifier.

I'd like to try the BIG MoFo, but I'm a little worried about the "heat management" of that configuration, so I'd like to fall back to the normal MoFo just in case it's too over-heating. The (maybe silly) question is: if I use the 193V inductor, can I bias the amplifer to work at 19V/1,7A or I have to change the inductor to 193T in order to realize the "small" version?

Claudio

#### MEGA_amp

Paid Member
this belongs in the Pass Forum. Yes, you can use the 193V inductor with 19V/1,7A, and it will work perfectly fine.

#### Cpan

this belongs in the Pass Forum. Yes, you can use the 193V inductor with 19V/1,7A, and it will work perfectly fine.

Ops, excuse me for the wrong section. Hope an admin could move the thread in the right forum.

Anyway, thanks for you're reply. So, if it's the case, I will try the 193V!

#### LBHajdu

Hi Cpan,
The 193V has a DC resistance of 1.00 ohm vs 0.7ohm for the 193T so not too much of a difference, but let us calculate how much voltage we need to bais up to the 1.7A:
V = I * R
1.19v = 1.7A * 0.7ohm
1.7v = 1.7A * 1.0ohm

So you lose 0.51V volts (1.7v - 1.19v) of voltage swing with the 193V vs 193T. The gate threshold voltage for the irfp240 is between 2-4V volts so it's going to need a little over 4v + 1.7v = 5.7 on the gate to bais up. My simulator (LTspice) says 6.5v to bais up. On the benefits side, the change in bais is less with the 1.0ohm load.

I think what is clear is that when picking a resistor and running at a low rail voltage one must be careful not to pick an inductor with too high a resistance. If the inductor had a DCR of 3.5 ohms the simulator says you need 10.5volts to bais the gate. The source needs to be at 5.9V ( = 3.5ohm * 1.7A).

I hope that helps you understand.

#### Cpan

Hi Cpan,
The 193V has a DC resistance of 1.00 ohm vs 0.7ohm for the 193T so not too much of a difference, but let us calculate how much voltage we need to bais up to the 1.7A:
V = I * R
1.19v = 1.7A * 0.7ohm
1.7v = 1.7A * 1.0ohm

So you lose 0.51V volts (1.7v - 1.19v) of voltage swing with the 193V vs 193T. The gate threshold voltage for the irfp240 is between 2-4V volts so it's going to need a little over 4v + 1.7v = 5.7 on the gate to bais up. My simulator (LTspice) says 6.5v to bais up. On the benefits side, the change in bais is less with the 1.0ohm load.

I think what is clear is that when picking a resistor and running at a low rail voltage one must be careful not to pick an inductor with too high a resistance. If the inductor had a DCR of 3.5 ohms the simulator says you need 10.5volts to bais the gate. The source needs to be at 5.9V ( = 3.5ohm * 1.7A).

I hope that helps you understand.

Many many thanks for your consideration. They're really interesting! I will take it into account during my works.

Claudio

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