Transcript DC load line - Portal UniMAP

POWER AMPLIFIER
(Additional Lecture Notes)
EKT 104
DC and AC Equivalent Circuits
+VCC
+VCC
IC
RC
RC
R1
R1
RL
vce
vin
vin
R2
R2
RE
R1//R2
IE
RE
rC = RC//RL
Bias Circuit
DC equivalent
circuit
AC equivalent
circuit
rC
Load Lines
• Every amplifier has two loads: a
DC load and an AC load.
• DC load line– all possible DC
combinations of IC and VCE.
• AC load line – all possible AC
combinations of iC and vCE.
• AC load line involves Rc||RL in
its determination, hence it is
steeper than the DC line
*Q point is shared by both DC and
AC load lines. However, it is
determined on the DC load line.
DC Load Line
IC(sat) = VCC/(RC+RE)
The straight line is known as the DC load
line
DC Load Line
IC
(mA)
The collector current IC and the collectoremitter voltage VCE must always lie on the
load line, depends ONLY on the VCC, RC and
RE
(i.e. The dc load line is a graph that
represents all the possible combinations
of IC and VCE for a given amplifier. For
every possible value of IC, and amplifier
will have a corresponding value of VCE.)
VCE(off) = VCC
VCE
What is IC(sat) and VCE(off) ?
Purpose of the DC biasing circuit
• To turn the device “ON”
• To place it in operation in the region of its
characteristic where the device operates
most linearly, i.e. to set up the initial dc
values of IB, IC, and VCE
DC Biasing Circuits
The ac operation of an amplifier
depends on the initial dc
values of IB, IC, and VCE.
+VCC
By varying IB around an initial dc
value, IC and VCE are made to
vary around their initial dc
values.
DC biasing is a static operation
since it deals with setting a
fixed (steady) level of
current (through the device)
with a desired fixed voltage
drop across the device.
v in
RC
RB
v out
ib
vce
ic
Q-Point (Static Operation Point)
• When a transistor does not have an ac input, it
will have specific dc values of IC and VCE.
• These values correspond to a specific point on the
dc load line. This point is called the Q-point.
• A quiescent amplifier is one that has no ac signal
applied and therefore has constant dc values of IC
and VCE.
Q-Point (Static Operation Point)
• The intersection of the dc bias
value of IB with the dc load line
determines the Q-point.
• It is desirable to have the Qpoint centered on the load line.
Why?
• When a circuit is designed to
have a centered Q-point, the
amplifier is said to be midpoint
biased.
• Midpoint biasing allows
optimum ac operation of the
amplifier.
8
DC Biasing + AC signal
• When an ac signal is applied to the base of
the transistor, IC and VCE will both vary
around their Q-point values.
• When the Q-point is centered, IC and VCE
can both make the maximum possible
transitions above and below their initial dc
values.
• When the Q-point is above the center on
the load line, the input signal may cause
the transistor to saturate. When this
happens, a part of the output signal will be
clipped off.
• When the Q-point is below midpoint on
the load line, the input signal may cause
the transistor to cutoff. This can also cause
a portion of the output signal to be clipped.
DC Biasing + AC signal
10
AC Load Line
What does the ac load line tell you?
• The ac load line tells the maximum
possible output voltage swing for a
given amplifier---- the maximum
possible peak-to-peak output voltage
(Vpp ) from a given amplifier.
• The maximum undistorted Vpp is
referred to as the compliance of the
amplifier.
• The smaller of the two swings limits
the maximum undistorted collector
current for a given amplifier.
AC Load Line
IC(sat) = VCC/(RC+RE)
DC Load Line
IC
(mA)
VCE(off) = VCC
VCE
• The ac load line of a given
amplifier will not follow the
plot of the dc load line.
• This is due to the dc load of
an amplifier is different from
the ac load.
IC(sat) = ICQ + (VCEQ/rC)
ac load line
ac load line
IC
IC
Q - point
dc load line
VCE(off) = VCEQ + ICQrC
VCE
VCE
* (AC Saturation Current Ic(sat) , AC Cutoff Voltage VCE(off) )
AC Load Line: The Upper Swing
• The current can
swing from the Q
point value to ic(sat).
- In this example, from
1.1mA to 3.52 mA.
• vCE can change from
the Q point value to
zero.
- In this example, from
4.94 V to 0 V.
AC Load Line: The Lower Swing
• The current can swing from the Q
point value to zero.
- In this case, from 1.1mA to 0 mA.
• vCE can change from the Q point
value to vCE(off).
- In this example, from 4.94 V to 7.18 V.
• The voltage swing is determined by
ICQrC.
- In this case, the maximum value of
ICQrC is (1.1 mA)(2.04k Ohm)=2.24 V.
* This means that as the collector
current swings between 1.1 mA and
zero, the value of vCE will vary from
4.94 V to 7.18 V
Calculating Compliance
• The smallest of the two (upper
swing and lower swing) determines
the maximum possible peak
voltage that can pass undistorted
throughout our amplifier.
• In the example, the maximum upper
swing peak voltage is 4.94 Vpk and
the minimum lower swing peak
voltage is 2.24 Vpk .
• Two times of the maximum possible
peak voltage will give the maximum
peak-to-peak transition value of the
output voltage.
Calculating Compliance
•
•
•
•
•
The maximum peak-to-peak swing is given
by:
PP=2VCEQ OR PP=2ICQrC
In the example, max peak-to-peak value is:
2(4.94 Vpk)=9.88 Vp-p OR
2(2.24 Vpk)= 4.48 Vp-p
Since 4.48 Vp-p is the smaller of the two---it
is the compliance of the amplifier.
The Q point is below the mid-point of AC
load line. Cutoff clipping happens when the
output voltage exceeds the compliance of
the amplifier as shown in Waveform (A).
Waveform (B) is limited to the compliance of
the circuit (4.48 Vp-p)---it is not clipped and
is undistorted.
Cutoff Clipping
• The Q point is
below the
midpoint and the
output voltage is
clipped off at the
value of Vce(off).
Cutoff Clipping
Saturation Clipping
• The Q point is
above the
midpoint and the
output voltage is
clipped off when
the amplifier hits
saturation. At this
point, the value of
Vce Is virtually 0.
Vce(off)
Saturation Clipping
Efficiency
• The ideal power amplifier would deliver 100% of the power it
draws from the power supply to the load.
• We know that this is not true and that components in the
amplifier will all dissipate some power that is being drawn
from the supply.
Where
PL
h = *100
PS
PL = average ac power to the load
PS = average power supplied by the source (VCC)
AMPLIFIER DC POWER
• The DC source supplies
direct current to the
voltage divider and to the
collector circuit.
• The total supply current is
the divider current plus
the quiescent collector
current. ICC=ICQ+I1
• The total dc power that
the amplifier draws from
the power supply is found
as: Ps=VCCICC
AC Load Power
• The ac load power is the power that is transferred to
the load.
• The ac load power can be calculated as follows
Exercise 1
• For the transistor in the commonemitter circuit in Figure Ex. 1, the
parameters are: β=80, PD,max =10W,
VCE(sus) =30V, and IC,max =1.2A.
• (a) Design the values of RL and RB for
VCC = 30 V. What is maximum power
dissipated in the transistor?
• (b) Using the value of RL in part (a),
find IC,max and VCC if PD,max = 5 W.
• (c) Calculate the maximum
undistorted ac power that can be
delivered to RL in parts (a) and (b) for
the assumption that iC ≥0 and 0≤vCE
≤VCC.
Figure Ex.1
Exercise 1: Solution (a)
Exercise 1: Solution (b)
Exercise 1: Solution (c)