# small signal common emitter amplifier | Small Signal CE Amplifier model analysis

By   June 30, 2020

Small Signal CE Amplifier , small signal common emitter amplifier , Small Signal CE Amplifier model analysis equivalent circuit of pdf .

CE amplifiers are very popular to amplify the small signal AC. After a transistor has been biased with a Q point near the middle of a DC load line, AC source can be coupled to the base. this produces fluctuations in the base current and hence in the collector current of the same shape and frequency. the output will be enlarged sine wave of same frequency.

The amplifier is called linear if it does not change the wave shape of the signal. As long as the input signal is small, the tranistor will use only a small part of the load line and the operation will be linear.

On the other hand, if the input signal is too large. the fluctuations along the load line will drive the transistor into either saturation or cut-off. this clips the peaks of the input and the amplifier is no longer linear.

The CE amplifier configuration is shown in fig. 24.

The coupling capacitor (CC) passes an AC signal from one point to another. At the same time it does not allow the DC to pass through it. hence, it is also called blocking capaitor.

For example, in fig. 25 the AC voltage at point A is transmitted to point B. for this series reactance XC should be very small compared in series resistance RS. the circuit to the left of A may be a source and a series resistor or may be the thevenin equivalent of a complex circuit. similarly, RL may be the load resistance or equivalent resistance of a complex network. the current in the loop is given by

I = Vin / (RS + RL)2 + (XC)2

= Vin / R2 + X2

As a frequency increases XC = (1/2 fc) decreases and current increases until it reaches to its maximum value Vin/R. therefore, the capacitor couples the signal properly from A to B then, XC << R. the size of the coupling capacitor depends upon the lowest frequency to be coupled. normally, for lowest frequency XC 0.1R is taken as design rule.

The coupling capaciitor acts like a switch, which is open to DC and shorted for AC.

The bypass capacitor CB is similar to coupling capacitor except that it couples an ungrounded point to a grounded point. then CB capacitor looks like a short to an AC signal and therefore, emitterb is said AC grounded. A bypass capacitor does not disturb the DC voltage at emitter because it looks open to DC current. As a design the rule XCb 0.1 RE at lowest frequency.

Analysis of CE Ampliffier

In a transistor ampliffier, the DC source sets up quiescent current and voltages. the AC source then, produces fluctuations in these current and voltages. the simplest way to analyze this circuit is to split the analysis in two parts DC analysis and AC analysis. one can use superposition theorem for analysis.

AC and DC Equivalent Circuits

For DC equivalent circuit, reduce all AC voltage sources to zero and open all AC current sources and open all capacitors. with this reduced circuit shown in fig. 26 (a) DC current and voltages can be calculated.

For AC, equivalent circuit reduces DC voltage source to zero and open current source and short all capacitors. this circuit is used to calculate AC  currents and voltage as shown in fig. 26(b).

The total current in any branch is the sum of DC and AC currents through that branch. the total valtage across any branch is the sum of the DC voltage and AC voltage across that branch.

Phase Inversion

Because of the fluctuation is base current; collector current and collector voltage also swings above and below the quiescent voltage. the AC output voltage is inverted with respect to the AC  input voltage, means it is 1800 out of phase with input.

During the positive half cycle base current increases, causing the collector current to increase. this produces a large voltage drop across the collector resistor, therefore the voltage output decreases and negative half cycle of output voltage is obtained. conversely, on the negetive half cycle of input voltage less collector current flows and the voltage drop across the collector resistor decreases and hence collector voltage increases, we get the positive half cycle of output voltage as shown in fig.27.

1. Common-Collector Amplifier

If a high impedance source is connected to low impedance amplifier then, most of the signal is dropped across the internal impedance of the source. to avoid this problem common-collector amplifier is used in netween source and CE amplifier. it increases the input impedance of the CE amplifier without significant change in input voltage.

Fig. 28 shows a common-collector (CC) amplifier. since, there is no resistance in collector circuit therefore, collector circuit therefore, collector is AC grounded. if is also calld grounded collector amplifier. when input source drives the base, output appears across emitter resistor. A CC amplifier is like a heavily swamped CE amplifier with a collector resistor shorted and output taken across emitter resistor.

Vout = Vin – VBE

Therefore, this circuit is also called emitter follower, because VBE is very small. As Vin increases, Vout increases.

IfVin is 2 V,                                      Vout = 1.3 V

IfVin is 3 V,                                      Vout = 2.3 V

Since, Vout follows exactly the Vin therefore, there is no phase inversion between input and output. the output circuit voltage equation is given by

VCE = VCC – IERE

Since,                IE >> IC

IC = VCC – VCE / RE

This is the equation of DC load line. the DC load line is shown in fig. 28.

Voltage Gain

Fig. 29(a) shows an emitter follower driven by a small AC voltage. the input is applied at the base of transistor and output is taken across the emitter resistor. fig. 29(b) shows the AC equivalent circuit of the amplifier. the emitter is replaced by AC resistance R’E.

The AC output voltage is given by

and                Vout = RE IE

Vin = IE (RE + R’E)

Therefore,            A = RE / RE + R’E

Since                  R’E >> RE

A >> 1

Therefore, it is a unity gain amplifier. the practical emitter follower circuit is shown in fig. 29 (c).

The AC source (VS) with a series resistance (RS) drives the transistor base. because of the biasing resistor and input impedance of the base, some of the AC signal is lost across the source resistor.

The AC equivalent circuit is shown in fig. 29(d).

The input impedance at the base is given by

Zin (base) = Vin/ IB

= IE (R’E + RE) / IB

= BIB (R’E + RE) / IB

= B (R’E + RE)

Since, R’E is very small in comparison with RE.

Zin(base) ~ BRE

The total input impedance of an emitter follower includes biasing resistors in parallel with input impedance of the base.

Zin = R1||R2||B(R’E + RE)

Since BRE is very large as compared to R1 and R2.

Thus,                       Zin~ R1||R2

Therefore, input impedance is very high.

Applying thevenin’s theorem to the base circuit of fig. 29(d). it becomes a source Vin and a series resistance (R1||R2||RS) as shown in fig. 29(e).

Vin = (R1||R2||RS) + IE (R’E + RE)

Or      IE = Vin / RE + R’E + R1||R2||RS / B

The impedance of the amplifier seen from the output terminal is given by

Vout = AVin = RE    VS

RE +R’E + R1||R2||RS / B

~ Vin (if RE is very large)

Example 8.  Find the Q point of the emitter follower circuit of given figure with R1 = 10 K and RE = 20 K. Assume the transistor has a B of 100 and input capacitor C is very-very large.

sol. We first find the thevenin’s equivalent of the base bias circuitry.

RB = R1||R2 = 6.67 K

VBB = R1 VCC / R1 + R2 = 12 (104) / 30×103 = 4 V

From the bias equation, we have,

IC = ICQ = VBB – VBE = 4 – 0.7  = 4.95 mA

RB / B + RE      6670/100 +600

Example 9. Find the output voltage swing of the circuit of given above figure.

Sol. The Q point location has already been calculated in example 1. We found that the quiescet collector current is 4.95 mA.

The output voltage swing = 2.IC Peak (RE||Rload)

= 2 (4.95 x 10-3) (300) = 2.97 V

This is less than the maximum possible output swing. continuing the analysis,

VCE, Q = VCC – VCQRE = 9.03 V

V’CC = 10.5 / 300 = 35.1 mA

The load lines for this problem are shown in figure.

h-parameters

Small Singnal Low Frequency Transistor Models

All the transistor amplifiers are two port networks having two voltages and two currents. The positive directions of voltages and currents are shown in fig. 30 (a).

Out of four quantities two are independent and two are dependent. if the input current i1 and output voltage v2 are taken independent then other two quantities i2 and v1 can be expressed in terms of i1 and v2

v1 = f1 (i1, v2)

i2 = f2 (i1, v2)

The equations can be written as

v1 = h11 i1 + h12 v2

i2 = h21 i1 + h22 v2

Where, h11, h12, h21 and h22 are called h-parameters.

h11 = v1/ i1   /v2 = 0

= hr

= fraction of output voltage at input with input open circuited or reverse voltage gain with input open circited to AC (dimensions)

h21 = i2/i1  /v2 = 0

= hf

= negative of current gain with output short circuited to AC.

The current entering the load is neegative of I2. This is also known as forward short circuit current gain.

h22 = i2/ i1  / i2 = 0

= ho

= output admittance with input open circuited to AC

If these parameters are specified for a particular configuration, then suffixes e,b or c are also included, e.g., hfre, hib and h-parameters of common-emitter and common-collector amplifiers. using two equations, the generalized model of the amplifier can be drewn as shown in fig. 30(b).

The Hybrid Model for a Transistor Amplifier

Let us consider CE configuration as shown in fig 31(a) the variables, ib, ic, vc and vb represent total instantaneous currents and voltages ib and vc can be taken as independent variables and vb, ic as dependest variables.

Using taylor’s series expression and neglecting higher order tems, we obtain

vb = fi/ ib / vc              ib + fi/vc / ib    vc

ic = f2/ib / vc               ib + f2/vc / ib   vc

The partial derivatives are taken keeping the collector voltage or base current constant. the vb, vc, ib, ic represent the small signal (incremental) base and collector current and voltage and can be represented as

vb, ib, vc, ic.

vb = hie ib + hrevc

ic = hfeib + hoe vb

Where,       hie = f1/ib /vc       = vb/ib /vc

hre = f1/vc /ib    = vb/vc /ib

hfe = f2 /ib /vc        = ic/ib /vc

hoe = f2/v2 /ib        = vb/vc /ib

The model for CE configuration is shown in fig. 31(b).

Determination of h-Parameters

To determine the four h-parameters of transistor amplifier, input and output characteristics are used. input characteristic depicts the relationship between input voltage and input current with output voltage as parameter. the output characteristic depicts the relationship between output voltage and output current with input current as parameter. fig.32(a). shows the output characteristics of CE amplifier.

hfe = ic/ib /vc    = ic2 – ic1 / ib2 – ib1

The current increments are taken around the quiescent point Q which corresponds to ib = IB and to the collector voltage VCE = VC

hoe = ic / vc /ib

The value of hoe at the quiescent operating point is given by the slope of the output characteristic at the operating point (i.e., slope of tangent AB).

hie = vb/ib = vb/ib /vc

hie is the slope of the appropriate input on fig. 21(b) at the operating point (slope of tangent EFat Q).

hre = VB/VC    = VB/VC  /IB       = VB2 – VB1 / VC2 – VC1

A vertical line on the input characteristic represents constant base current. the parameter hre can be obtained from the ratio (VB2 – VB1) and (VC2 – VC1) for at Q.

Typical CE h-parameters of transistor 2N1573 are given below.

hie   = 1000

hre = 2.5 *10-4

hfe  = 50

hoe = 25 mA/V

H-Parameters Analysis of a Transistor Amplifier Using h-Parameters

To form a transistor amplifier, if is only necessary to connect an external load and signal source as indicated in fig. 33(a) and to bias the transistor properly.

Consider the two-port network of CE amplifier. Rs is the source resistance and ZL is the load impedance h-parameters are assumed to be constant over the operating range, the AC equivalent circuit is shown in fig. 33(b) (phasor notations are used assuming sinusoidal voltage input). the quantities of interest are the current gain, input impedance, voltage gain and output impedance.

Current Gain

For the transistor amplifier stage A, is defined as the ratio of output to input currents.

Af  = IL /IB = -IC / IB

(IL + IC = 0,   IL = – IC)

IC = hfeIB + hoeVC

VC = IL ZL = – IC ZL

IC = hfe IB + hoe (-IC ZL)

Or                  IC / IB =  hfe/ 1+ hoe Zl

Al = – hfe / 1 + hoe ZL

Input Impedance

The impedance looking into the amplifier input terminals (1, 1′) is the input impedance Zi

ZI = VB/ IB

VB = hieIB + hreVE

VB / IB = hie + hre VC/IB

= hie – hreICZL / IB

ZI = hie + hreAIZL

hie – hrehfeZL / 1 + hoeZl

Zi = hie – hrehfe / yl + hoe   (since, YL = 1 /ZL)

Voltage Gain

The ratio of output voltage to input voltage gives the gain of the transistors.

A= VC/VB = ICZL / VB

AV = IB AIZL / VB = AIZL /ZI

It is defined as ,

YO = IC/VC /VS = 0

IC = hfeIB + hoeVC

IC /VC = hfe IB / VC + hoe

when,       VS = 0, RS IB + HIE IB + HREVC = 0

IB/VC = – hre / RS + hIE

YO = hoe – hrehfe / RS + HIE

Voltage amplification taking into account source impedance (RS) IS given by

AVS = VC/VS = VC /VB + VB/ VS                 (VB = VS / R2 + ZI = ZI)

= AV   ZI / ZI + RS

= AIZL / ZI + RS

Av is the voltage gain for an ideal voltage source (RV = 0) Consider input source to be a current soure IS in parallel with a resistance RS as shown in Fig . 34 (a).

In this case, overall current gain AIS is defined as

AIS = IL / IS

= – IC / IS

= – IC / IB * IB / IS                   (IB = IS* RS / RS + ZI)

= AI * RS / RS + ZI

RSOO, AIS –AI

To analyze multistage amplifier, the h-parameters of the transistor used are obtained from manufacture data sheet. the manufacture data sheet usually provides h-parameters in CE configuration. these parameters may be converted into CC and CB values. for example, fig. 34(b) hre in terme of CE parameter can be obtained as follows:

For CE transistor configuration,

VBE + hieIB  + hreVCE

IC = hfeIB + hoe VCE

The circuit can be redrawn like CC transistor configuration as shown in fig 35.

VBE = hie IB + hRE VEC

IC = hFE IB + hOE VEC

hrc  = VBE / VEC / IB = 0

= VBE + VEC/ VEC  / IE = 0

= (VBE / VEC + 1 )  / IB = 0

Since , IB = 0,    VBE = HRE, VC = – hREVEC

hrc = 1 + (hreVEC / VEC)

= 1 – hre

Similarly,  hfe = IE / IB / VEC = 0   = – (IB + IC) / IB  / VEC = 0

= – ( 1 + hfe)

given below.

hie = 2k, hfe = 50, hre = 6*10-4, hoc = 25 mA/V

hic = 2k, hfe = -51, hre = 1, hre = 25 mA/V

The small signal model of the transistor amplifier is shown in figure below.

Sol. In the circuit, the collector resistance of first stage is shunted by the input impedance of last stage. therefore, the analysis is started with last stage. it is convenient to first compute current gain, input impedance and voltage gain. then, output impedance is calculated starting from first stage and moving towards end.

Ai2 = -hfe / 1 + hoc ZL = 51 / 1 + 25*10-6 * 5 * 103

= 45.3

Ri2 = hic + hreAi2ZL

= 2 * 103 + 1 * 45.3 *5 *103

= 228.5 K               (high input impedance)

AV2 = VO / V2 = AIZL / ZI2

= 45.3 * 5 / 228.5 = 0.99 = 1

RL1 = RC1||RI2

= 5 * 228.5 / 5 + 228.5 = 4.9K

AI1 = – hfe / 1 + hoeRL =  -50 / 1+25*10-6*4.9*103 = 44.5

RI1 = hie + hreAI1RL1

= 2-6*10-4*44.5*4.9 = 1.87K

Voltage gain of first state is

AV1 = AI1RL1 / RI1 = -44.5*4.9/1.87 =-116.6

Yo1 = hoe – hfehre / hie + RS

= 25*10-6 – 50*6*10-4/2*103+1*103

= 15*10-6 mho

Ro1 = 1/ yo1 =66.7k

R’O1 = RO1||RC1

= 66.7||5=4.65K

The effective source resistance R’S2 For the second stage is R01||RC1. Thus, RS2 = R’01 = 4.65 K

Frequency Response

Yo2 = hoe – hfehrc / hic + Rs2

=25*10-8 – (-51) (1) /2*103 + 4.65*103

= 7.70*10-3 A/V

Ro2 = 1/y02 = 130

Ro2 = Ro2||RC2

= 0.13||5K =127

Overall current gain of the amplifier is Ai and is given by

Ai = – IE2 / Ib1

= – Ie2 / Ib2 * Ib2 / Ie1 * IC1 / IB1

= – AI2 . IB2/IC1 * AI1

The equivalent circuit of the amplifier is showe in fig. 3. from the circuit, it clear that current ic1 is divided into two parts.

Therefore, IB2 / IC1 = -RC1 / RC1 + RI2

AI = AI2 . AI1 . RC1 / RC1 + RI2

= 45. 3* (-44.5) * 5 / 228.5 + 5 = – 43.2

AV = VO / V1 = VO / V2 * V2 / V1

= AV2 . AV1

= 0.99*(-11.6) = 115

Overall voltage gain of the amplifier is given by

AVS = – VO / VS = AV . RI1 / RI1 + RS

= – 115* 1.87 / 1.87 +1 = – 75.3