I have to switch through mosfet with 5V@80mA with Rdson = .4ohm.

Now

1. Power dissipation
2. Current Consumption
3. Input Power
4. Output Power
5. Efficiency

Thanks..!!!

 

Ummm... Have you tried to make any of these calculations yourself? Why do you care what the values are if you don't know what the words mean?

1. P = I^2 • R all inputs given
2. Given as 80mA
3. P = V•A all inputs given
4. Eqn. (3) minus Egn. (1)
5. Eqn. (4) / Eqn. (3) x 100%

 

Ummm... Have you tried to make any of these calculations yourself? Why do you care what the values are if you don't know what the words mean?

1. P = I^2 • R all inputs given
2. Given as 80mA
3. P = V•A all inputs given
4. Eqn. (3) minus Egn. (1)
5. Eqn. (4) / Eqn. (3) x 100%

I know the meaning of terminoligy still I had doubt- thats why I asked

1- I agree

2- 80mA is load current so u can say current consumption. If one need to deliever load current 80mA at 5V ..so is that same with input current...!!!

Current consumption means what is the input current Mosfet will draw.

3. Now input power
4. Out power.

Please interrupt if I am wrong...!!!!

 

2- 80mA is load current so u can say current consumption. If one need to deliever load current 80mA at 5V ..so is that same with input current...!!!

Current consumption means what is the input current Mosfet will draw.

Does charge accumulate within the MOSFET? No. So that means current in equals current out.

The slight exception to this is the gate of the MOSFET, which has a small capacitance. So a tiny charge actually is required to enter the MOSFET to turn it on, and this charge must be drawn off to turn the MOSFET back off. This becomes relevant at high (>50kHz?) frequency, as this charge/discharge current adds up in proportion to frequency.

This has nothing to do with the load current, which is typically much larger.

 

Does charge accumulate within the MOSFET? No. So that means current in equals current out.

The slight exception to this is the gate of the MOSFET, which has a small capacitance. So a tiny charge actually is required to enter the MOSFET to turn it on, and this charge must be drawn off to turn the MOSFET back off. This becomes relevant at high (>50kHz?) frequency, as this charge/discharge current adds up in proportion to frequency.

This has nothing to do with the load current, which is typically much larger.

Thank you so much...!!!