Quiz 1

Question 1

An NMOS device with a threshold voltage of 1.7V and VGS of 2.5V operates in saturation. The VDS is at least that many Volts.

Answer 1

VDS ≥ VGS - VTH = 2.5-1.7 = 0.8V

Question 2

This current of an NPN is ‘in-going’ and this one is ‘out-going’:

Answer 2

IB/IC is ‘in-going’ and IE is ‘out-going’

Question 3

This current of a PNP is ‘in-going’ and this one is ‘out-going’:

Answer 3

IE is ‘in-going’ and IC/IB is ‘out-going’

Question 4

A BJT is said to be ‘diode connected’ when these two terminals are tied together:

Answer 4

Base and collector

Question 5

Diode connection of a BJT prevents this mode of operation:

Answer 5

Saturation

Question 6

A non-ideal VCVS has AV=150 and rout=10kΩ. The device performance is equivalent to that of a VCCS with gout = 100uS and Gm of that many [mA/V]

Answer 6

Gm = AV / rout = 150 * 10*10^3 = 15mA/V

Question 7

The output of these single-transistor amplifiers is taken at the collector terminal of the BJT:

Answer 7

CE and CB

Question 8

The input of these single-transistor amplifiers is applied to the base terminal of the BJT:

Answer 8

CE and CC

Question 9

A CE amplifier uses an NPN with an IC of 250uA. For a supply voltage of 15V these values of RB1 and RB2 are ‘good’:

Answer 9

RB1 = VBB/(0.1IC) = 2/ (0.1250uA) = 80k

RB2/RB1 = VCC/VBB - 1 -> RB2 = (80k)((15/2)-1) = 520k

Question 10

A CB amplifier uses an NPN with an IC of 250uA. For a supply voltage of 15V these values of RE and RC values are ‘good’:

Answer 10

RE = (VBB-0.7/IC) = 2-0.7/250u = 5.16k
RC = [(VCC-VBB)/2]/IC = [(15-2)/2]/250u = 26k

Question 11

Well-designed CE amplifier uses a PNP and operates from a single 12V supply. The VB of the PNP has this approximate value:

Answer 11

VB = 12 - 2(volt drop) = 10V

• PNP is always 2V drop so switch for NPN and it starts at 12 - 2 *

Question 12

The transfer characteristic of a BJT relates these two quantities:

Answer 12

Collector Emitter Voltage (VCE) and Collector Current (IC)

Question 13

Spice uses this two-port description (y, z, h, g) to model the ‘small-signal’ behavior of transistors:

Answer 13

Y - parameters

Question 14

To describe the small-signal behavior of BJTs some datasheets list these two-port parameters:

Answer 14

rpi, ro, beta, gm

Question 15

This type of a gain cannot be achieved without transistors or vacuum tubes:

Answer 15

Power gain, Voltage OC gain

Question 16

This BJT quantity (β, VA, VT) is independent of technology and device physical dimensions.

Answer 16

Vt

Question 17

To produce an ‘AC equivalent circuit’ the following elements are replaced with “short”:

Answer 17

DC voltage sources and Capacitors

Question 18

Diode conducts ID of 200μA and has VD(on) of 0.67V at T=25oC. At T=50oC its VD(on) has approximately this value:

Answer 18

-2mV/°C -> VD(on)=0.67-0.002(25)=0.62V

Question 19

The input signal is applied to the gate terminal of a MOSFET device. The amplifier must be:

Answer 19

Common-Source (CS)

Question 20

The amplifier configuration is ‘Common-Emitter’. The transistor has gm of 30 mA/V and the load is 10kΩ. The (loaded) voltage gain of this amplifier is no larger than:

Answer 20

AV = gm * rout = 3m [A/V] * (ro||RL) = 0.03(RL) = 0.03(10000) = -300

Question 21

Common-Source amplifiers have 50mVpp input linear range. If the drain bias current is 150μA, the transconductance of the MOSFET is approximately:

Answer 21

input linear range = (VGS - VTH)/5 = 25mVp -> (VGS - VTH) = 0.125V

gm = (2ID)/(VGS - VTH) = (2150u) / 0.125V = 0.0024 A/V

Question 22

The BJT in a CE amplifier is biased at IC=300μA and the emitter node is by-passed using a large-value capacitor. If the open-circuit voltage gain is 100 mV/mV, RC has this value approximately:

Answer 22

Av(oc) = gm * rout = Ic/Vt * Rc
Rc = 26mV/0.3mA * 100 = 8.67k

Question 23

The BJT in a CE amplifier is biased at IC=250μA. If the input linear range is 50mVpp the degeneration resistance must be approximately that many Ohms:

Answer 23

Vin(max) = Vt(1.3gmRdeg - 1)
25mVp = 0.026(1.3(Ic/Vt)Rdeg -1)
Rdeg = (25m/26m + 1) / (1.3* (0.25m/26m)
= 157 ohm

Question 24

A BJT device has transconductance gm of 2mA/V and current gain β of 50. The input resistance rπ of the BJT is approximately:

Answer 24

rπ = 𝛽 / gm = 50 / 2mA/V = 25kΩ

Question 25

Common-Emitter amplifier carries 2mA and has an Early Voltage of 50V. The voltage drop over the RC “bias” resistor is 5V. The output resistance of this CE amplifier is approximately:

Answer 25

rout = ro || RC = (Va/Ic) || (5V/2mA) = (50/2mA) || (5V/2mA) = 2.27k CALC

Question 26

Common-Emitter amplifiers are emitter-degenerated using resistor RE. The voltage drop VRE is 170mV. The RE is NOT by-passed. The maximum AC signal that one could apply to the base of the BJT and still consider the circuit to be linear has magnitude on the order of:

Answer 26

Vin(max) = Vt(1.3*gm*Rdeg - 1) Vin(max) = Vt(1.3*(IC/Vt) * (170m/Ic) -1) *Ic cancels Vin(max) = Vt(1.3* 1/Vt * 170mV - 1) =0.195Vp CALC

Question 27

A transistor conducts 1mA of collector current. If the architecture has RE of 500Ω (not by-passed), the gm(eff) of the BJT-RE topology is approximately:

Answer 27

gm(eff) = gm/(1+gm*RE) = (Ic/Vt) / (1+ (Ic/Vt) * RE) = (1m/26m)/1+(1m/26m)*500 = 0.002 A/V

Question 28

The amplifier configuration is Common-Base. If the BJT is biased at 0.5mA, the input resistance of the CB is on the order of:

Answer 28

Rin (CB) = 1/gm = 1/(0.5mA / 26mV) = 52Ω

Question 29

By rounding component values to the closest E12 value, the designer is likely to introduce that much error (worst case):

Answer 29

E12 Values → worst case error = 10%

Question 30

The standard component values approximate this type of a number sequence:

Answer 30

(Geometric) E24 - 5% Tolerance