Sum5 Lec5 Transistors

Transistors

The following section analyzes the npn transistors

Transistor Basic Principles

- Three semiconductor, of which one p-type semiconductor is sandwiched between the two n-typed semiconductors

- Two back-to-back diodes, one forward biased (BE) while another one reverse biased (CB)

- When current passes through the transistor, the small amount of electrons injected from emitter will flow into the base junction

i_B+ i_C = i_E =^iC i_E

i_B =^iC where =

1₋

- Transistor is a voltage driven current source

- _VBE creates a large current, this will be attenuated when it arrives at collector, which is a site for buffer region - The collector region DOES NOT AFFECT the input part

(emitter region)

- The change in the current i^c is affected by VBE!!!!!

- A very small change in V^BE will cause changes in VCE! ( Notice that it is the voltage that causes the difference, NOT current!)

- Base current is generated directly from V^{BE, since vBE} is normally large, so the 1 in the diode equation is dropped out for approximation

i_B =^iC_≈^IES _e

v_BE V_T

Load Line Analysis

Using Kirchoff’s voltage law, by taking DC sources in account,

V_BB = i_BR_B+ v_BE V_CC = i_CR_C+ v_CE

- We can then find the Q point of the input voltage and the output voltage

- After that, AC source is taken into account, which will varies the intercepts of the graph

- Take note that the gain is out of phase, i.e., the voltage output is out of phase with the voltage input

- The transistor has this property of inverting the phase - However, input voltage of too close to the V^CC will cause

unwanted variation - For a transistor,

o _{If iC}=0, the transistor is at cut-off (Logic 0)

o _{If VCE}=0.2 V, the transistor is at saturation (Logic 0) QUESTION ANALYSIS

When dealing with finding iB range of output or Q points: - Use Kirchoff’s Voltage Law to construct DC equation

- Draw a straight line intersecting with the input curve, bring in AC variation to find the range of input voltage - Use another equation in the output to draw an intersection line with the output curve

- Observe the corresponding i^{B extremes} - Find the extreme output voltages - Find voltage gain

Small signal Analysis

(Assume that the AC signal is relatively small in the following discussion.)

Similarly, we can also find our AC resistant in transistor by taking only DC equivalent circuit and find the IBQ. After that the equation below can be used to determine r_π

r_π = ^VT I_BQ Where I_BQ= Ib at Q point

The equivalent AC model of BJT is:

Common Emitter Voltage Amplifier

- Similarly, due to the coupling capacitors, we can analyse this circuit by only seeing the DC point of view

- In this case, the voltage equation is

V_B = R_BI_B + V_BE + R_EI_E Where V_B = ^R2

R₁+R₂ ^{VCC and RB = R1|| R2}

- Substitute IE ⁼ + 1 IB , the following equation is obtained

� =� + � + � �^{� − � �} _� - Output DC voltage

V_CC = I_CR_C+ v_CE + I_ER_E PROBLEM ANALYSIS

When dealing with CE BJT with AC and DC inputs: - Analyse with DC equivalent circuit

- _{Obtain RB}, V_B and I_B

- _{Find r}π by using the equation above

- Construct AC equivalent circuit and replace the transistor with r_π at junction BE

- Find Voltage output

After that, we can construct the AC equivalent circuit and replacing the transistor as a single rπ

v_in = v_be = i_br_π and v_o =_−R^′_L i_b A_v = ^vo

v_in ⁼⁻ R^′_L

r_π - Note that the output and input voltage is OUT OF PHASE

- The Input Impedance of the transistor is desired to be large in order to have much greater voltage being divided

�� ⁼� ||�� ^and � = � - Power gain of the amplifier is defined as

A_i = ^io i_in ⁼

V_o Z_o

�_��

�_�� ^{= �}

�_��

� G = ^Po

P_in ⁼ i_ov_o

i_inv_in ^{= � �}

Analysis on CE BJT

- Normally we want the amplification to be large in amplifier. To do so, the only way to manipulate in this case is the resistant R_L and R_C, we want them to be HIGH.

- Normally, we want to have low output impedance, this means that the value of R^C should be AS SMALL AS POSSIBLE, so this contradicts the requirement.

- To obtain high input impedance, we should choose higher R^B - The voltage gain is higher than the current gain since RC ^Zin

- _RE in this circuit acts as a resistor which can make the i_C to be dependent to β! - _{If RE} ^RB

+1 and is very large (normal case), I_B = ^{VB− VBE}

R_B + _{+ 1 RE}^≈

� − � _� R_E

I_C= I_B =^{VB− VBE} R_E

- So with the help of R^E, the collector current is INDEPENDENT to the transistor used. - On the other hand, we can also reduce the variation of V^BE due to temperature effects

Common Collector Circuit

We can modify the circuit by taking output across Emitter terminal instead of Collector terminal as in CE BJT By doing so, we introduced R_E into the AC equivalent circuit, which gives rise to great changes in the voltage gain

- Notice that the output is taken across the E terminal, so the E is connected to the OUTPUT - C is then moved down at which it only provides current i^C

Input Analysis

By using Kirchoff’s voltage law, we see that

v_in = i_br_π + _{1 + ib}R_L^′ Z_it =^vin

i_b ^{= rπ+ (1 + )RL}

′

�_�⁼� ||�_�� Voltage Gain, A_v= ^vo

v_in ⁼

1 + R^′L

r_π + _{1 + R}^′_L ¹

- The input impedance now is much larger in the CE BJT because of the r^π is in series with the R’L

Output is now taken across E- terminal, not across C-terminal as indicated in the dotted line

R_E now takes part in AC equivalent circuit

Output Analysis

- Zero input voltage, remove output resistance and add a voltage v^x at the place of RL

Using the Ohm’s law equation,

v_x = Z_oi_{x→ Zo} =^vx i_x v_x =_{− ib}r_π + i_bR^′_s

i_b =₋ ^vx r_π+ R_S^′

--- (1) i_e+ i_x= _{1 + ib}+ i_x = ^vx

R_E

--- (2) Combining the two equations,

−^v_r^{x 1 +}

π^{+ R}S^′

+_�� = ^��

��

Z_o =^vx i_x ⁼

1 + r_π + R^′_S⁺

1

R_E ^{= Zot||RE} Z_ot =^{rπ+ RS}

′

1 + - _{Since RS} R₁, R₂, so we can conclude that R^′_{S ≈ RS} - _{Hence, Zo} = Z_ot||R_{E≈ Zot}, since R_E is generally very large

A_i = A_v ^Zi R_L

In common emitter BJT:

- The input impedance is larger than other BJT - The output impedance is smaller than other BJT