Last Year Revision Notes

Question 1

Electron flow direction in npn transistor biased in active mode

Answer 1

emitter to collector

this is the default mode, the names of these nodes decribe the electron flow of this default mode.

Question 2

BJT: How should you connect a external voltage source to the EB junction of a transistor to enable forward bias?

Answer 2

You want to connect the positive terminal (high voltage) to the p-side and the negative terminal to the n-side. This goes for both npn and pnp.

Question 3

What is the high potential side of a battery?

Answer 3

The positive (long side) of the battery.

Question 4

PN junction: What does biasing mean?

Answer 4

Application of external voltage across the terminals.

Question 5

For an npn transistor what direction do electrons flow?

Answer 5

The electrons flow from emitter to collector.

As such, the current flows from the collector to the emitter.

Question 6

For a npn transistor to be in active mode, what are the junction requirements?

Answer 6

The EB junction needs to be forward biased and CB junction needs to be reverse biased.

Question 7

For an npn transistor, for it to function, how should the battery terminals be connected relative to the emitter and the collector?

Answer 7

Negative terminal(short side) needs to be connected to the emitter and positive terminal(long side) needs to be connected to the collector.

This makes sense as electrons need to flow from emitter to the collector.

Question 8

How do you tell what the anode and cathode are of a LED(diode)?

Answer 8

The anode is the longer end of the LED, and thus the cathode is the shorter one.

Anode is the positive side and cathode the negative.

For the LED to function the anode (positive side) needs to be connceted to the side of electron flow.

This is because the anode is biased positively and attracts the electrons.

Question 9

What direction does the electric field go in a PN diode?

Answer 9

Goes from the Postive side to the Negative.

i.e. electric field arrows point from positive to negative

Question 10

What does it mean when a PN junction is in forward bias?

What happens?

Answer 10

In forward bias the positive side of the battery terminal is connected to the P-type region and the negative side is connected to the N-type region.

The positive terminal repels the holes in the P-type region and the negative terminal repels the electrons in the N-type region. As a result, the width of the depletion region decreases. Once the forward bias reaches a potential of 0.7 for silicon junction diodes, the potential barrier will be overcome and current will start to flow.

Question 11

What does it mean when a PN junction is reverse biased?

Answer 11

In reverse bias the negative side of the battery terminal is connected to the P-type region and the positive side is connected to the N-type region.

The negative terminal attracts the holes in the P-type region and the positive terminal attracts the electrons in the N-type region. As a result, the width of the depletion region increases.

The reverse bias will keep on increasing until it finally reaches the reverse breakdown voltage, commonly known as the ‘Zener’ breakdown or Avalanche Region.

Question 12

For an npn transistor to be in active mode, why does the EB junction have to in forward bias and the CB junction in reverse bias?

Answer 12

To encourage electrons through the p-region we need to overcome the barrier potential in the EB junction. By adding a positive voltage (forward bias) to the junction, the electrons at the emitter are repelled inwards, this reduces the depletion region width. This reduction in width gives the opportunity for electrons to pass straight through to the other N-type region without recombination in the P-type. That being said, the other N-type region will soon start to repel incoming electrons unless an external bias is applied. By applying a negative potential (reverse bias) to the CB junction we can attract electrons out of the second N-type region and thus allow for current flow throughout the transistor.

Question 13

What is the i_c - v_BE characteristic for a npn transistor?

Answer 13

i_c = I_se^vbe/Vt

Question 14

Common-emitter current gain equation?

What is the commonly quoted value for one of the variables?

Answer 14

I_C = β*I_B

β is commonly quoted as being 100.

Question 15

What is the common-base current gain equation?

Answer 15

I_C = α*I

Question 16

Why can’t you just flip a BJT transistor around and expect it to work?

Answer 16

The emitter of a BJT is more doped than the collector so flipping it will not work. This doping is to encourage current flow in one direction.

Question 17

For an npn BJT in active mode what regions are at a higher voltage?

Answer 17

In the EB junction the p-type region is at a higher voltage than the n-type. In the p region because it is connected to the positive terminal of the battery, holes will be repelled in. On the other hand, the negative terminal will repel the electrons into the emitter and push them towards the p-type region.

The CB junction is in reverse bias and this means the n-type is at a higher voltage than the p-type. As a result, the electrons are attracted out of the n-type region because of the positive terminal. This then allows for electrons from emitter to flow in.

Question 18

What does electron flow move with respect to low and high potential?

Answer 18

Electron flow moves from low potential (voltage) to high potential (voltage).

Question 19

Why is electron-hole recombination so minimal in an npn BJT?

Answer 19

Because the base is made really thin.

If the base is made thin enough, the incoming electrons from the emitter cannot be stopped and move straight through to the collector.

The recombination of electrons causes a small current I_B into the base i.e. holes moving into the base.

Question 20

How much bigger are I_E and I_C than I_B in a npn BJT?

Answer 20

Generally, it is considered that I_E is 101 times bigger than I_B and I_C is 100 times bigger than I_B .

The difference in value is due to a few of the electrons recombining with holes in the base.

Question 21

For a npn BJT, what happens if output impedance is really high, what does that mean for the current through the transistor?

Answer 21

For a really high impedance no matter how large the voltage across the CE, the current stays roughly the same.

Question 22

For an npn BJT in active mode what is the equation that relates all the currents (base, collector and emitter)?

Answer 22

I_E = I_B + I_C

This makes sense as some of the electrons from the emitter combine with holes so current I_C must be lower than I_E .

Question 23

Why use a coupling capacitor for the input and output of a transistor?

Answer 23

To remove a DC bias/offset.

Question 24

When does a MOSFET plateau?

Answer 24

When V_ds = V_ov

If it plateaus then the saturation region has been reached.

Question 25

For MOSFETs in the saturation region, how do the V_GS, V_T and V_OV relate?

Answer 25

V_GS = V_T + V_OV

Question 26

How do you calculate V_DSMin for a MOSFET transistor to operate in the saturation region?

Question 27

What is the basic design of a common drain MOSFET?

Answer 27

Question 28

Common Source Basic Design

Answer 28

Question 29

What happens in the cutoff region?

Answer 29

V_GS < V_T

as a result i_D = 0

Question 30

What is the equation for i_D in the triode region?

Answer 30

Question 31

What is k^’_n and what is it equal to ?

Answer 31

It is the process transconductance parameter

= µ*C_OX

Question 32

What is k_n and what is its equation?

Answer 32

It is the transconductance parameter of the nMOS-FET.

= µ*C_OX * (W/L)

Question 33

For a very small V_DS relative to V_GS - V_T what is the equation for channel resistance?

Answer 33

r_DS = V_DS/i_D = 1/(k^’_n * W/L * v_ov)

Question 34

What is the equation for i_D in the saturation region?

Answer 34

i_D = ½*µ* C_OX * (W/L) * (v_GS - v_T )²

v_GS - v_T = v_OV

Question 35

Small signal equivalent circuit for CS

Answer 35

Question 36

Small Signal Equivalent for CS with a source resistance.

Answer 36

Question 37

Small Signal Equivalent for CG

Answer 37

Question 38

Small Signal Equivalent for CD

Answer 38

Question 39

What are the R_in values for CS, CS with R_S, CG and CD?

Answer 39

For CS it is infinity

For CS with R_S it is infinity

For CG it is 1/g_m

For CD it is infinity

Question 40

What are the R_o values for CS, CS with RS, CG and CD?

Answer 40

For CS it is R_D||r_o

For CS with RS it is R_D

For CG it is R_D

For CD it is 1/g_m

Question 41

What is Av =Vo/Vi for CS, CS with R_S, CG and CD?

Answer 41

For CS it is −𝑔_𝑚(𝑅_𝐷||𝑟_𝑜)

For CS with R_S, − (𝑔_𝑚𝑅_𝐷)/ (1 + 𝑔_𝑚𝑅_s)

For CG it is 𝑔_𝑚𝑅_D

Question 42

What is G_v =V_o /V_sig for CS, CS with R_S, CG and CD?

Answer 42

For CS it is −𝑔_𝑚(𝑅_𝐷||𝑟_𝑜)

For CS with R_S it is − 𝑔_𝑚𝑅_𝐷 /(1 + 𝑔_𝑚𝑅_S)

For CG it is 𝑅_𝐷 / (𝑅_𝑠𝑖𝑔 + 1/𝑔_m)

For CD it is 𝑅_𝐿 / ( 1 /𝑔_𝑚 + 𝑅_L)

Question 43

What does fF mean…..

Answer 43

femtofarad…

which means 10^-15 …

Question 44

What do you need to remember about calculating kn or k’n ?

Answer 44

That u_n is given in the units cm² this means you need to multiply by 10^-4 not 10^-2 to obtain m² …

You need to remember this every time you’re dealing with units that are squared or of a higher power.

Question 45

How do you find r_o of a MOSFET?

Answer 45

Find the slope of the v_DS - i_D graph and take the inverse of the value to get r_o .

Question 46

How do you calculate λ for a MOSFET?

Answer 46

1/V_A

So 1 over the early voltage value.

Question 47

MOSFET find the voltage at node

questions a to e.

Answer 47

Question 48

MOSFET find voltage at node.

questions f to h.

Answer 48

Question 49

What relates r_o with V_A ?

Answer 49

r_o ≈ V_A / I_D

Question 50

What is the Unit of g_m for

Answer 50

A/V

Question 51

What is the source absorption theorem?

Answer 51

The source absorption theorem states that the current source I_X appearing between two nodes whose voltage difference is V_X can be replaced by an impedance Z_X = V_X / I_X = 1 / g_m

Question 52

What do you need to remember about capacitors in parallel?

Answer 52

They add

Question 53

What gains do 20db, 40db and 60db relate to?

Answer 53

10, 100 and 1000 respectively.

Question 54

What do you need to remember about bode phase plot zeros and poles?

Answer 54

The frequency values are one order of magnitude smaller.

So a s/100 would break at 10 on the bode phase plot

s/10⁵ would break at 10⁴ and so on.

Question 55

What do you need to remember about zeros and poles at the origin for a bode phase plot?

Answer 55

They give rise to +90 and -90 degree phase shifts respectively.

This means if you find as that isn’t being divided you can apply this shift.

Question 56

For a npn transistor, if you want an active mode, what should be the state of the EB junction and CB junctions?

Answer 56

The EBJ needs to be forward biased.

The CBJ needs to be reverse biased.

Question 57

What sort of transistor is this?

Answer 57

npn transistor

arrow going out.