A question about MoFo

Cpan

Member
2018-03-05 2:06 pm
Rome
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?

Thanks in advance,
Claudio
 

LBHajdu

Member
2002-02-18 7:53 pm
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

Member
2018-03-05 2:06 pm
Rome
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