In the given circuit, the drain current (ID) through the MOSFET can be determined using Ohm's Law:
ID = (VG - VD) / RD
Where:
ID is the drain current
VG is the gate voltage
VD is the drain voltage
RD is the drain resistor
Given VG = 0V, R1 = 95Ω and RDS(on) = 4Ω, we can calculate ID as follows:
ID = (0 - VD) / (95 + 4)
ID = -VD / 99
We know that RDS(on) is the drain-source resistance when the MOSFET is fully ON. Therefore, when the MOSFET is fully ON, the drain current will be at its maximum value.
The maximum drain current can be calculated using the formula:
ID(max) = (VG - VD) / RDS(on)
Since VG = 0V, we have:
ID(max) = -VD / RDS(on)
ID(max) = -VD / 4
Since we want to find VD, we set ID = ID(max) and solve for VD:
-VD / 99 = -VD / 4
Multiply both sides by 99 and 4 to get rid of the fractions:
4 * -VD = 99 * -VD
-4VD = -99VD
Subtract 99VD from both sides and simplify:
-4VD + 99VD = 0
95VD = 0
Divide both sides by 95:
VD = 0V
Therefore, the voltage at the drain (VD) is 0V.
In 2n7000 MOSFET switching operation. Given VG=0V, R1=95Ω and RDS(on) is 4Ω. Determine the voltage at the drain, VD.
Express your answer in mA and in 2 decimal places.
Vg is gate
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