Download Voltage Transfer Characteristic

Recall Last Lecture

DC Analysis and Load Line



Input load line is based on the equation derived
from BE loop.
Output load line is derived from CE loop.
To complete your load line parameters:



Obtained the values of IB from the BE loop
Get the values of x and y intercepts from the derived IC
versus VCE.
Draw the curve of IB and obtained the intercept points
IC and VCE (for npn) or VEC (for pnp) which is also
known as the Q points
Voltage Transfer
Characteristic
VO versus Vi
Voltage Transfer Characteristics - npn

A plot of the transfer characteristics (output voltage versus input
voltage) can also be used to visualize the operation of a circuit or
the state of a transistor.
Given VBEon = 0.7V,  = 120, VCEsat = 0.2V, Develop the voltage transfer
curve
Vo (V)
Cutoff
5
0.7
•In this circuit, Vo = VC = VCE
• Initially, the transistor is in cutoff mode because Vi is too
small to turn on the diodes. In cut off mode, there is no
current flow.
• Then as Vi starts to be bigger than VBEon the transistor
operates in forward-active mode.
Vi (V)
Active Mode

BE Loop
 100IB + VBE – Vi = 0
 IB = (Vi – 0.7) / 100

CE Loop
 ICRC + VO – 5 = 0
 IC = (5 – VO) / 4
 βIB = (5 – VO) / 4
β = 120
 IB = (5 – VO) / 480

Equate the 2 equations:
(Vi – 0.7) / 100 = (5 – VO) / 480
Vo = - 480 Vi + 836 A linear equation with
negative slope
100
Vo (V)
Cutoff
5
Active
Saturation
0.2
Vi (V)
0.7


1.7
x
5
To find point x,
the coordinate
is (x, 0.2)
However, as you increase Vi even further, it reaches a point where both
diodes start to become forward biased – transistor is now in saturation
mode.
In saturation mode, VO = VCEsat = 0.2V. So, what is the starting point, x, of
the input voltage, Vi when this occurs?
Need to substitute in the linear equation  Vi = 1.7 V
and VO stays constant at 0.2V until Vi = 5V
Voltage Transfer Characteristics - pnp
Vo (V)
saturation
Vo = 4.8
VEB
VEC β = 80
•Vo = VC and VE = VCC
•Hence, VEC = VCC – VO  VO = VCC - VEC
Vi (V)
• As Vi starts from 0V, both diodes are forward biased. Hence, the transistor
is in saturation. So, VEC = VECsat and Vo = VCC – VEC sat

As Vi increases, VB will become
more positive than VC, the junction
C-B will become reverse-biased.
The transistor goes to active mode.

The point (point x) where the
transistor start to become active is
based on the equation which is
derived from active mode operation

BE Loop
 200IB + 0.7 + Vi – 5 = 0
 IB = (4.3 – Vi ) / 200

CE Loop
 ICRC - VO = 0
 IC = VO / 8
 80 IB = VO / 8
 IB = VO / 640

VEB
VEC β = 80
Equate the 2 equations:

(4.3 - Vi) / 200 = VO / 640
Vo = - 640 Vi + 2752
200
A linear equation with
negative slope
Vo (V)
saturation
Vo = 4.8
VEB
VEC β = 80
Active
To find point x,
the coordinate
is (x, 4.8)
cutoff
2.8xV
4.3
5
Vi (V)
 By increasing Vi even more, the potential difference between VEB becomes less
than VEBON, causing junction E-B to become reversed biased as well. The diode
will be in cut off mode. VO = 0V

Using the equation derived:
Vo = - 640 Vi + 2752
200
when Vo = 0, then, Vi = 4.3 V