ESE 372 / Spring 2013 / Lecture 14 Last time: biased by .

Bias

Q * But make sure that BJT is in FA, i.e. VCE > 0.2-0.3V

Equivalent circuit based on Hybrid-π model, caps are short circuits for signals

1 ESE 372 / Spring 2013 / Lecture 14 Common Emitter Amplifier biased by current source.

Voltage gains

MAX Finite output resistance puts upper limit on amplifier AV0  gm rO

2 ESE 372 / Spring 2013 / Lecture 14 Common Emitter Amplifier biased by current source.

Input/output impedances

Not high enough - problem

Net voltage gain

Depends on β – not too good 3 ESE 372 / Spring 2013 / Lecture 14 Common Emitter Amplifier biased by current source.

Short circuit current gain

Common emitter current gain

4 ESE 372 / Spring 2013 / Lecture 14 Example

npn - BJT with β 100 and VA 100V.

Bias current IE 1mA, VCC  VEE 10V

R B 100k, R C  8k, R S  5k, R L  5k

1. Check if BJT is in FA regime I 1mA I  E  10μ B β 1 101

VB  10μ 100k  1V  VE  1V  0.7V  1.7V  10V.

VC 10V 1m8k  2V  VCE  2V  1.7V  3.7V  0.3V 2. Calculate small signal parameters IQ 1mA mA 3. Find voltage gain g  C   40 m R in R L Vth 25mV V G  A  V R  R V0 R  R β 100 in S L out rπ    2.5kΩ V gm 40mA V A  g R || r  296 V0 m C O V VA 100V rO  Q  100kΩ R in  R B || r  2.4k, R out  R C || rO  7.4k IC 1mA  V  V G V  0.32 296 0.4  38  V  V

30mV p-p signal corresponds to about 10mV p-p variation of VBE. This produces about 1.2V p-p across load. 5 ESE 372 / Spring 2013 / Lecture 14 Example – output voltage swing

8 V

3.5 V

-1.5 To remain in FA

VCE > 0.2 V

Maximum amplitude of the undistorted sine wave at the output is limited by maximum negative voltage swing, i.e. by 3.5 V.

For 1mA bias current and 10 V what could we do to increase the amplitude of undistorted sine wave?

What is the maximum value of this amplitude?

6 ESE 372 / Spring 2013 / Lecture 14

CE amp biased by current source and with RE.

Case 1

R E  0  R in  R B || r  r  AV0  gm  R C || rO

Case 2 (neglect rO)

R E  0

R in  R B || r  R E1  r  R C AV0   r  R E  1 

Negative feedback RE improves at the expense of gain

7 ESE 372 / Spring 2013 / Lecture 14 amplifier.

No need for CE !

Equivalent circuit for AC analysis

8 ESE 372 / Spring 2013 / Lecture 14 Common Base amplifier.

Redraw equivalent circuit in more convenient

form and neglect rO for beginning.

9 ESE 372 / Spring 2013 / Lecture 14 Common Base amplifier.

Voltage gain

Noninverting amplifier with open circuit voltage gain value similar to that of CE amp

Short circuit current gain

i.e. no current gain ! 10 ESE 372 / Spring 2013 / Lecture 14 Common Base amplifier.

Input impedance

Very small – PROBLEM for voltage amplifier

Output impedance

11 ESE 372 / Spring 2013 / Lecture 14 Common Base amplifier.

Net voltage gain

Could we guess this without analysis? Observe current buffer action of CB amp

Equivalent circuit of amp with current gain equal to one. 12 ESE 372 / Spring 2013 / Lecture 14 CB amplifier with BJT having finite output small signal resistance.

Voltage gain

13 ESE 372 / Spring 2013 / Lecture 14 CB amplifier with BJT having finite output small signal resistance.

Input impedance

Impedance When transformation

max

14 ESE 372 / Spring 2013 / Lecture 14 amplifier Again.

No need for CE ! Bias current IE will determine gm, rπ and rO

Also R and C can be eliminated B C1 Redraw equivalent circuit in more convenient form

15 ESE 372 / Spring 2013 / Lecture 14 Common Collector amplifier

+

When

i.e. no voltage gain ! 16