1 Review of Pre-requisites-2

Analysis of any transistor amplier circuit proceeds with the following steps:

DC analysis: Open all capacitors, short all inductors and calculate DC param-eters of interest such as IC , ID and VGS VT . The ac voltage source must beshort circuited and ac current sources must be open circuited. Assertain thatthe circuit is operating in forward active mode for BJT and saturation regionfor MOSFET. This should enable you to nd the parameters of small signalanalysis such as gm, r and ro.

AC analysis has the following steps:1. Step 1: Short all DC voltage sources and open all DC current sources.

2. Step 2: Short all capacitors and open all inductors.

3. Step 3: Replace the transistor with its own small signal equivalent.

In doing all the steps see if you can bring any simplication.

1.1 Solved Problems

1. Determine the small-signal voltage gain of a common-source circuit containinga source resistor shown in Fig.(1). The transistor parameters are VTN = 0.8V,Kn = 1mAV

2 and = 0. Assume that the capacitors are innite.

+5V

vi

165k 7k

50035k

vo

-5V

Figure 1: Problem 1

1

The DC equivalent circuit is shown in Fig.(2). This gure is arrived based onpotential divider rule which states that,

VG =VDDR2 + VSSR1

R1 +R2(1)

=5 35 5 165

165 + 35= 3.25 (2)

+5V

7k

500

-5V

-3.25V

Figure 2: Problem 1:DC equivalent circuit

Applying KVL around the gate source loop, we get:

VG = VGS + IDRS + VSS (3)

= VGS +KnRS(VGS VT )2 + VSS (4)3.25 = VGS + 0.5(VGS 0.8)2 5 (5)

Where we have assumed that the transistor is operating in saturation region.This has to be conrmed at some stage. Now, the above equation transformsto a quadratic equation:

0.5V 2GS + 0.2VGS 1.43 = 0 (6)Eq.(6) has two solutions namely VGS1 = 1.503V and VGS2 = 1.903V. Wediscard the second solution as VGS2 < VT , implying that the device is in ocondition which is false. The drain current is:

2

ID = Kn(VGS VT )2 = 0.494mA 0.5mA (7)Applying KVL over the drain source loop, we obtain the VDS = 6.25V. Now, wecan see that VDS VGS VT , asserting that the device is in saturation region.The transconductance of the MOSFET is gm = 2Kn(VGS VT ) = 1.4mS andro = as = 0. Having obtained these parameters, we obtain the small signalmodel using the steps outlined above. After applying the step 1, the circuitappears as shown in Fig.(3).

vi

165k 7k

50035k

vo

Figure 3: Problem 1: Step 1

We can see that 165k and 35k can be combined in parallel. After applyingthe second step and third step we obtain the circuit in Fig.(4).

The output voltage is,

vo = gmvgsRD = 1.4 7vgs (8)We also have,

vi = vgs + vs (9)

= vgs + gmRSvgs (10)

= vgs(1 + gmRS) (11)

vgs = vi1 + gmRS

(12)

The small signal voltage gain is,

3

vi7k

500

vo

28.875k gmvgsvgs

+

-

G

S

D

Figure 4: Problem 1: After step 3

AV =vovi

= gmRD1 + gmRS

= 1.4 71 + 1.4 0.5 = 5.76 (13)

2. For the circuit in Fig.(5), let = 100, VBE(on)0.7V and VA = 100V. Determinethe small-signal voltage gain. Determine the input resistance seen by the signalsource and the output resistance looking back into the output terminal. Assumethat the capacitors are innite.

V+=10V

V-

=-10V

500

RB=100k

10k

20k

RC

REvi

vo

Figure 5: Problem 2

Treating the capacitors as open circuit, the DC equivalent will contain thereistors RB, RC and RE . Applying KVL around the base-emitter loop, we get:

4

0 = IBRB + VBE(on) + IERR + VEE (14)

0 = IBRB + (1 + )IBRE + VEE + VBE(on) (15)

IB = VEE + VBE(on)RB + (1 + )RE

(16)

IB = 10 + 0.7100k + (1 + 100)20k

= 4.387A (17)

Now, gm =IbVT h

= 16.873mS, r =gm

= 5.927k and ro =VAIC

= 228k.Proceeding sequentially through all the steps we arrive at the small signal modelshown in Fig.(6). For the incremental gain we follow through the calculationsgiven under:

500

RB=100kvi

vo

r gmv ro RC=10k

B C

E

v

+

-

Figure 6: Problem 2: Small signal circuit

vo = gm(RC ro)v (18)v =

RB r500 +RB r vi (19)

Av = vovi

= gm(RC ro) RB r500 +RB r (20)

= 148V V. (21)

The input impedance is Ri = 500 + 100k r = 6.09k and the output resis-tance is Ro = Rc ro = 9.58k. Note in calculating the output impedance wemust nd the Thevinin resistance looking from the output terminals.

3. Determine the small-signal voltage gain of a circuit (shown in Fig.(7)) biasedwith a constant-current source and incorporating s source bypass capacitor. Thetransistor parameters are: VT = 0.8V, Kn = 1mAV

2 and = 0. Assume thatthe capacitors are innite.

5

+5V

-5V

0.5mARG=200k

RD=7k

vi

vo

Figure 7: Problem 3

Let us assume that the transistor is biased in saturation region and hence bydenition:

ID = Kn(VGS VT )2 (22)

VGS =

IDKn

+ VT = 1.507V (23)

We still need to assert that the transistor is operating in saturation region, tothat end we draw the DC equivalent circuit shown in Fig.(8).

Note that in the DC equivalent circuit the ac signal source must be grounded,if it were an ac current source it should have been open circuited. The terminalvoltages can be obtained by applying KVL around their respective loops:

VD = VDD IDRD = 5 7 0.5 = 1.5 (24)VGS = VG VS = VS = 1.507V (25)

VS = 1.507V (26)VDS = VD VS = 3.007V. (27)

We notice that the condition for saturation VDS VGS VT is satised. Thesmall signal equivalent circuit can be deduced from the sequence of steps andis shown in Fig.(9). The transconductance is gm = 2

KnID = 1.414mS.

We notice that vgs = vi and vo = gmvgsRD and hence the gain is,

6

+5V

-5V

0.5mA

RD=7k

D

S

G

Figure 8: Problem 3: DC equivalent circuit

RGvi RD

G D

S gmvgsvgs+

-

vo

Figure 9: Problem 3:AC equivalent circuit

Av =vovi

= gmRD = 1.414 7 = 9.9 (28)

7

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