hw 3 solution

Solution of Homework problems in Chapter 11.

Chapter 11, Solution 3.

Known quantities:The threshold voltage, VT = 2 V, of an enhancement-type NMOS that has its source grounded and a 3 V DC source connected to the gate.Find:The operating state if:a) .b) .c)

Analysis:

a)

The transistor is in the triode region.

b)

The transistor is either in the triode or in the saturation region.

c)

The transistor is in the saturation region.


Known quantities:The threshold voltage, VT = 1.5 V, of the NMOS transistor shown in Figure P11.9. .

Find:The voltage levels of the pulse signal at the drain output, if vG is a pulse with 0 V to 5 V.

Analysis:

1) Since VT = 1.5 V, with vG = 0 V, vGS < VT, the transistor is cut off. Therefore, vD = 5 V.

2) When vG = 5 V, the transistor turns on since VG > VT.

Assume that the transistor is in the saturation region:

iD = k (vGS - VT)2 = 0.4 (5 - 1.5)2 = 4.9 mA. Therefore, vD = 5 - 4.91 = 0.1 V.

Since VD = 0.1 V < VGS – VT = 5 1.5 = 3.5, it is contradictory to the assumption.

Thus, the transistor is working in the triode region.

iD[mA] = k[2(VGS – VT)VDS – VDS2]=0.4[2(5-1.5) VDS – VDS

2]=2.8 VDS – 0.4VDS2 : Eq (1)

iD = 5-VDS : Eq.(2)

From Eq. (1) & (2), 0.4 , so VDS = 7.92 V or 1.58V .

Since VDS < VGS – VT = 3.5 V for triode region, VDS is 1.58 V when VG = 5V.


Known quantities:Circuit shown in Figure 11.10.

Find:Find the current .

Analysis:In the circuit of Figure P11.10,

, therefore the transistor is in the ohmic region. We can compute the drain current to be:


Known quantities:In the circuit shown in Figure P11.11, the MOSFET operates in the active region.

Find:a) b) The largest allowable value of for the

MOSFET to remain in the saturation region.

Analysis:

iD

VGS

VGD

VSD+

(a) Since VD = 3V & ID = 5mA, RD = VD/ID= 3V/5mV= 600

(b) Since the transistor is in the saturation region and , we have

, so VGS = 0 V or -2V.

For p-channel MOSFET in the saturation region, VGS VT and VGD = VGS – VDS VT.

Thus, . Since the source is at 10 V, the gate voltage must be 8 V (VGS = VG – VS, 2V=VG – VS).Saturation region operation would be maintained when VD exceeds VG by |VT| (or VGD = VGS – VDS VT),

Therefore, .

Alternatively, VG = VGD +iDRD, VGD = VG – iDRD where VGD VT. VG – iDRD VT RD (VG – VT)/ iD

can be found to be

simply from (a), RD,max = 18 k


Known quantities:The i-v characteristic of Figure P11.13(a), and the circuit in Figure P11.13(b):

Find:The current iDQ the voltage vDSQ, and the region of operation of the MOSFET.

Analysis:The operating point can be determined using the load line method.

By superimposing the load line on Figure P11.13(a), and by noticing that , we obtain

The MOSFET is in the saturation region.

Q point


Known quantities:The circuit in Figure P11.13(b):

Find:The current iDQ the voltage vDSQ, and the region of operation of the MOSFET.

Analysis:Assuming that the MOSFET is in the saturation region, the quiescent drain current is

The drain-to-source voltage is

Since hypothesis was correct


Known quantities:The circuit in Figure 11.9 in the text:

Find:The current iDQ, the voltage vDSQ, the resistance RS, and the operating region of the MOSFET.

Analysis:Using Thevenin equivalent,

We can write the equations

Assuming saturation conditions, the current iD can be written as

Notice that the problem has more unknown than equations.

Suppose that the transistor is in the saturation region. Then, VGSQ VT & VGDQ = VGSQ-VDSQ VT

From eq.(1), 18 – RSiDQ 4 RSiDQ 14,From eq.(2), RDiDQ – 18 4 10103iDQ 22, so iDQ 2.210-3 AThus, RS ~ 6.36 k, iDQ 2.210-3 A. Then, since VDD = (RS+RD)iDQ+VDSQ,VDSQ = VDD – (RS+RD)iDQ 36 – (6.36+10) 2.2 = ~ 32.4 V. So, VDSQ > 0 for the saturation region.RS ~ 6.36 k, iDQ 2.210-3 A, and VDSQ > 0, meaning that there are many solution of RS, iDQ, & VDSQ with which the transistor is in the saturation region.

One solution: we can impose the vDSQ to ensure saturation conditions as

Since VGSQ > 4 V, VGSQ = 5 V.Remark: The other solution of the algebraic equation is not acceptable because < VT.

The resistance RS is given by

and the drain current

Suppose that it is in the triode region.Then, VGSQ VT & VGDQ = VGSQ – VDSQ VT

From eq.(1), 18 – RSiDQ 4 RSiDQ 14,From eq.(2), RDiDQ – 18 4 10103iDQ 22, so iDQ 2.210-3 AThus, RS ~ 6.36 k & VDSQ 0 but VDSQ VGSQ – VT One solution is as follows:RS = 5 k, iDQ = 2.3 mA, VGSQ = 18 – 52.3= 6.5V.Thus, VDSQ VGSQ – VT = 6.5 – 4 = 2.5V.KVL in SD circuit : 36V = 102.2+VDSQ + 52.2VDSQ = 1.5 V.


Known quantities:The circuit in Figure 11.9:

Find:The current iDQ, the voltage vDSQ, and the voltage vGSQ.

Analysis:Using Thevenin,

We can write the equations

Assuming saturation conditions, the current iD can be written as

The other solution is not acceptable because less then VT.

It follows


Known quantities:The power MOSFET circuit shown in Figure P11.17.

Find:a) If , find the range of for which

the VCCS will operate.b) If , determine the range of for

which the VCCS will operate.

Analysis:The MOSFET should be working in the saturation region, so

a) , so

b)

Solve the above two equations, we can have

. Thus, or

Since , the second one is reasonable, so


Known quantities:The Class A amplifier shown in Figure P11.19

Find:c) Determine the output current for the given

biased audio tone input, . Let

and .d) Determine the output voltage.e) Determine the voltage gain of the signal.f) Determine the DC power consumption of the resistor and the MOSFET.

Analysis:a) The MOSFET should be working at the saturation regionSo

b)

c)

d) We can ignore the cosine part signal when calculating the DC power consumption.


Known quantities:The source-follower amplifier shown in Figure P11.20.

, and

Find:a) Determine the load current .b) Determine the output voltage.c) Determine the voltage gain of the

signal.d) Determine the DC power consumption of the resistor and the MOSFET.

Analysis:The MOSFET should be working at the saturation region, So the

a)

c)

d) We can ignore the cosine part signal when calculating the DC power consumption.


Known quantities:A “push-pull amplifier” can be constructed from matched n-and-p-channel MOSFETs, shown in Figure P11.24.

Find:Determine and .

Analysis:If , the output of the operational amplifier is positive infinity, so both MOSFETs will work in the triode region and the will increase, until it reaches ;

If , the output of the operational amplifier is negative infinity, so both two MOSFET will work in cutoff region, and the will increase, until it reaches ; So is the only

equilibrium in the system, so . Correspondingly,


Known quantities:The two-stage amplifier shown in the circuit of Figure P11.26.

Find:a) Determine the and for b) Determine the and for c) Determine the and for

Analysis:Assume the drain current in the MOSFET in the left hand side is and in the right hand side

. Obviously, both MOSFET are working in saturation region.

a)

Solve it and we can have Obviously, the first solution is reasonable, so and

b)

Solve it and we can have

Obviously, the first solution is reasonable, so and a) Basically, the signal in c) part is a comparably small cosine signal superposed by the signal

the in a) part. If we ignore the harmonics larger than 1st order, we can approximately have the following solution

hw 3 solution

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