MOSFET Power Amplifier Coursework help

[Ryth_cs]

Hey, I am an A2 Electronics student and need to prove that my MOSFET Power Amplifier is functional, image is attached.

I was told to use the P=V^2/R formula to prove that the power at the input is greater than that of the output.

I currently have no clue on how to use this formula to prove this, any help would be greatly appreciated.

I can give additional information of needed.

 

Cheers,

Ryan

[Crunchy Dragon]

12 minutes ago, Smiffy_ said:

the P=V^2/R formula

Power = Voltage squared, divided by Resistance?

 

I haven't taken an electronics course in a few years and my knowledge is pretty limited.

[Ryth_cs]

Just now, Crunchy Dragon said:

Power = Voltage squared, divided by Resistance?

 

I haven't taken an electronics course in a few years and my knowledge is pretty limited.

I'm aware of the formula just not on how to apply it to solve my problem.

[Crunchy Dragon]

Just now, Smiffy_ said:

I'm aware of the formula just not on how to apply it to solve my problem.

I'm asking if that was the correct interpretation of the formula since I wasn't sure. Sorry for not pointing it out.

[Ryth_cs]

2 minutes ago, Crunchy Dragon said:

I'm asking if that was the correct interpretation of the formula since I wasn't sure. Sorry for not pointing it out.

Oh yes it is, sorry for not realising.

[Unimportant]

1 hour ago, Smiffy_ said:

I was told to use the P=V^2/R formula to prove that the power at the input is greater than that of the output.

Surely you mean to prove that power at the output (relay) is greater then the input (MOSFET gate).

 

There's some information missing, such as the relay's coil resistance and the MOSFET Rds(on) but let's go worst case.

 

Let's ignore the fact that the MOSFET itself will have a huge input resistance at the gate and model the Gate-Source in the worst possible way, a short circuit. Even then we have at least the 2.1M gate resistance.

 

Assuming the gate is also being driven with 5V, and seeing the MOSFET Gate-Source as a short with 0 voltage drop, the full 5V has to drop over the 2.1M resistor, we get:

P = 5^2 / 2.1M = 0,000011905 Watt

of power dissipated in the gate resistor. As a side note, current will stop flowing into the MOSFET gate once it's been fully charged (barring some leakage current), but let's ignore that as well.

 

As for the output side, let us assume we have a really crappy MOSFET with a Rds(on) of 10 ohms. As for the relay's coil resistance, let's go by this datasheet that says 50 ohms.

The relay coil and MOSFET Rds(on) are in series and thus form 60 ohms, we get:

5^2 / 60 = 0,42 Watts of power dissipated in the output circuit total.

Of which (0,42 / 60) * 50 = 0,35 Watts in the relay and...
(0,42 / 60) * 10 = 0,07 Watts in the MOSFET.

And so, power in the output loop is much greater then in the input loop.