The MOSFETS M1 and M2 are connected in parallel to carry a total curre...
Introduction:
In this question, we are given two MOSFETs (M1 and M2) connected in parallel to carry a total current of 20 A. We are also given the drain-to-source voltage (Vds) for each MOSFET, as well as the current sharing series resistance of 0.50 Ω. We need to find the drain currents of M1 and M2.
Analysis:
When MOSFETs are connected in parallel, their gate voltages are typically driven by a common voltage source. In this case, since the drain currents of M1 and M2 are not equal, the voltage across the series resistances will be different for each MOSFET. This voltage difference is given by Vds1 - Vds2.
Calculation:
Given:
Total current (Itotal) = 20 A
Drain-to-source voltage of M1 (Vds1) = 2.5 V
Drain-to-source voltage of M2 (Vds2) = 3 V
Current sharing series resistance (R) = 0.50 Ω
First, we need to find the voltage difference across the series resistances:
Vds_diff = Vds1 - Vds2 = 2.5 V - 3 V = -0.5 V
Since the MOSFETs are in parallel, the voltage across each MOSFET is the same as the total voltage drop across the series resistances:
Vds = Itotal * R = 20 A * 0.50 Ω = 10 V
Now, we can find the drain current of each MOSFET using Ohm's Law:
I1 = Vds1 / R = 2.5 V / 0.50 Ω = 5 A
I2 = Vds2 / R = 3 V / 0.50 Ω = 6 A
Conclusion:
The drain current of M1 is 5 A and the drain current of M2 is 6 A. The MOSFETs are connected in parallel, sharing the total current of 20 A. The voltage across the series resistances is 10 V, which is the same for both MOSFETs. The voltage difference between the drain-to-source voltages of M1 and M2 is -0.5 V.
The MOSFETS M1 and M2 are connected in parallel to carry a total curre...
First, we need to understand that the current through each MOSFET is determined by the voltage across its drain-to-source and the series resistance.
The voltage across the series resistance of M1 (V_R1) is the total voltage (V_DD) minus the drain-to-source voltage of M1 (V_DS1). Similarly, the voltage across the series resistance of M2 (V_R2) is V_DD minus the drain-to-source voltage of M2 (V_DS2).
However, we don't know the total voltage V_DD. But we do know that the total current (I_DD) is the sum of the currents through M1 and M2 (I_D1 and I_D2), and that these currents are determined by Ohm's law (I = V/R).
So, we can write two equations:
I_D1 = (V_DD - V_DS1) / R
I_D2 = (V_DD - V_DS2) / R
And we know that I_DD = I_D1 + I_D2.
Substituting the expressions for I_D1 and I_D2 from the first two equations into the third equation gives:
I_DD = (V_DD - V_DS1) / R + (V_DD - V_DS2) / R
Rearranging this equation gives:
V_DD = (I_DD * R + V_DS1 + V_DS2) / 2
Substituting the given values gives:
V_DD = (20A * 0.5Ω + 2.5V + 3V) / 2 = 7.75V
Now we can substitute V_DD back into the first two equations to find I_D1 and I_D2:
I_D1 = (7.75V - 2.5V) / 0.5Ω = 10.5A
I_D2 = (7.75V - 3V) / 0.5Ω = 9.5A
So, the drain currents of M1 and M2 are 10.5A and 9.5A, respectively.
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