Chapter 3
Q. 3.12
Objective: Design the size of a power MOSFET to meet the specification of a particular switch application.
The load in the inverter circuit in Figure 3.45 is a coil of an electromagnet that requires a current of 0.5 A when turned on. The effective load resistance varies be tween 8 and 10 Ω depending on temperature and other variables. A 10 V power supply is available. The transistor parameters are k´_{n} = 80 µA/V^{2} and V_{T N} = 1 V.
Step-by-Step
Verified Solution
One solution is to bias the transistor in the saturation region so that the current is constant, independent of the load resistance.
The minimum V_{DS} value is 5 V. We need V_{DS} \gt V_{DS}(sat) = V_{GS} − V_{T N} . If we bias the transistor at V_{GS} = 5 V, then the transistor will always be biased in the saturation region. We can then write
I_{D} = \frac{k´_{n}}{2} \cdot \frac{W}{L} (V_{GS} − V_{T N})^{2}
or
0.5 = \frac{80 × 10^{−6}}{2} \left(\frac{W}{L} \right) \cdot (5 − 1)^{2}
which yields W/L = 781.
The maximum power dissipation in the transistor occurs when the load resistance is 8 Ω and V_{DS} = 6 V. Then
P(max) = V_{DS}(max) \cdot I_{D} = (6) \cdot (0.5) = 3 W
Comment: We see that we can switch a relatively large drain current with essentially no input current to the transistor. The size of the transistor required is fairly large, which implies a power transistor is necessary. If a transistor with a slightly different width-to-length ratio is available, the applied V_{GS} can be changed to meet the specification.