Question 10.8: Objective: Design a MOSFET current source circuit to meet a ...
Objective: Design a MOSFET current source circuit to meet a set of specifications.
Specifications: The circuit to be designed has the configuration shown in Figure 10.17. The bias voltages are V^{+} = 2.5 V and V^{−} = 0. Transistors are available with parameters k´_{n} = 100 µA/V² , V_{T N} = 0.4 V, and λ = 0. Design the circuit such that I_{REF} = 100 µA, I_{O} = 60 µA, and V_{DS2}(sat) = 0.4 V.
We have V_{DS2}(sat) = 0.4 = V_{GS2} − 0.4, so that V_{GS2} = V_{GS1} = 0.8 V.
Then for transistor M_{2},
\left(\frac{W}{L} \right)_{2} = \frac{I_{0}}{\left(\frac{k´_{n}}{2} \right) (V_{GS2} − V_{T N} )^{2}} = \frac{60}{\left(\frac{100}{2} \right) (0.8 − 0.4 )^{2}} = 7.5
For transistor M_{1},
The value of V_{GS3} is found as
V_{GS3} = (V^{+} − V^{−}) − V_{GS1} = 2.5 − 0.8 = 1.7 V
Then for transistor M_{3}, we find
Trade-offs: As with other designs, slight variations in transistor parameters (k´_{n} , W/L, and V_{T N}) will change the current values slightly. See Test Your Understanding exercise TYU 10.5.
Comment: In this design, the output transistor remains biased in the saturation region for
V_{DS} > V_{DS}(sat) = V_{GS} − V_{T N} = 0.8 − 0.4 = 0.4 V
Design Pointer: As with most design problems, there is not a unique solution. The general design criterion was that M_{2} was biased in the saturation region over a wide range of V_{DS2} values. Letting V_{GS2} = 0.8 V was somewhat arbitrary. If V_{GS2} were smaller, the width-to-length ratios of M_{1} and M_{2} would need to be larger. Larger values of V_{GS2} would result in smaller width-to-length ratios.
The value of V_{GS3} is the difference between the bias voltage and V_{GS1}. If V_{GS3} becomes too large, the ratio (W/L)_{3} will become unreasonably small (much less than 1). Two or more transistors in series can be used in place of M_{3} to divide the voltage in order to provide reasonable W/L ratios (see end-of-chapter problems).