Download two load line of vdb amplifier

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Solar micro-inverter wikipedia , lookup

Electrical ballast wikipedia , lookup

Public address system wikipedia , lookup

Power factor wikipedia , lookup

Electric power system wikipedia , lookup

Electrical substation wikipedia , lookup

Power inverter wikipedia , lookup

Stray voltage wikipedia , lookup

Electrification wikipedia , lookup

Three-phase electric power wikipedia , lookup

Variable-frequency drive wikipedia , lookup

History of electric power transmission wikipedia , lookup

Rectifier wikipedia , lookup

Power engineering wikipedia , lookup

Islanding wikipedia , lookup

Voltage optimisation wikipedia , lookup

Voltage regulator wikipedia , lookup

Resistive opto-isolator wikipedia , lookup

Mains electricity wikipedia , lookup

Current source wikipedia , lookup

Semiconductor device wikipedia , lookup

Pulse-width modulation wikipedia , lookup

Tube sound wikipedia , lookup

Two-port network wikipedia , lookup

Power electronics wikipedia , lookup

Audio power wikipedia , lookup

Alternating current wikipedia , lookup

Distribution management system wikipedia , lookup

Power MOSFET wikipedia , lookup

Buck converter wikipedia , lookup

Switched-mode power supply wikipedia , lookup

Amplifier wikipedia , lookup

Opto-isolator wikipedia , lookup

History of the transistor wikipedia , lookup

Transistor wikipedia , lookup

Current mirror wikipedia , lookup

Transcript
MAHATMA GANDHI INSTITUTE FOR
TECHNICAL RESEARCH AND
EDUCATION
Guided by:Anita Mam
Sapna Mam
TWO LOAD LINE OF VDB AMPLIFIER
Fig (a) shows Voltage Divider Biased (VDB) amplifier.
In VDB two resistors R1&R2 is used instead of VBB. In dc circuit there is no
capacitor.
For finding ,VBB = R2
*Vcc
R1+R2
Fig (b) shows ac equivalent circuit. As emitter in ac gnd so no effect.
a
b
DC LOAD LINES & AC LOAD LINES:The Saturation &Cut off points are different in ac &dc load line. The collector
&Emitter resistance are less than dc resistance. The ac &dc load lines intersect
at Q point.
DC LOAD LINE:
Q point vary by R2.
R2>>Transistor in Saturation &current given as
Ic(sat) = Vcc
RC+RE
R2<< Transistor in Cutoff & vtg given as
VCE(cut off) = Vcc.
Fig (c) shows DC Load Line.
c
AC LOAD LINE:
The Collector current is
IC= ICQ + VCEQ – VCE
rc
Transistor in saturation, VCE=0rc
ic(sat)=ICQ + VCEQ
rc
Transistor in cut off , Ic=0
vce(cut off) = VCEQ + ICQrc
d
The ac load line has higher slope than dc load line so max. peak to peak o/p always
less then Vcc.
Fig(d) shows AC Load Line.
CLASS A OPERATIONAL AMPLIFIER
Class A operation of an amplifier means that the transistor operates in the active
region at all times.
Fig(a) shows Class A Amplifier. Output is not clipped. Collector current flows
for 360° throughout the cycle. Some parameters useful in Class A Amplifiers
are Power Gain Output Power, etc.
With a class A amplifier, the designer usually tries to locate the Q point
somewhere near the middle of the load line so signal do not distorted.
In output we get the signal amplified . The efficiency is 25%.
DEFINATIONS:1. OUTPUT POWER
Fig (a)The output power is given by,
Pout = Vrms2
RL
Usually, we measure the output voltage in peak-to-peak volts with an
oscilloscope. In this case, a more convenient equation to use for output
power is: Pout=Vout2
8RL
Fig(b) The maximum output power is given by,
P(out)=MPP2
8RL
2. TRANSISTOR POWER DISSIPATION:
When no signal drive amplifier then ,
PDQ = VCEQICQ.
When signal present the power dissipation of transistor decrease bze some power
converts in signal power.
3. CURRENT DRAIN:
The dc voltage source supply a dc current Idc to the amplifier. This dc current
has two components: the biasing current through the voltage divider and the
collector current through the transistor. The dc current is called the current
drain of the stage.
4. EFFICIENCY:The dc power supplied to an amplifier by the dc source is:
Pdc = VCCIdc.
The efficiency is defined as n=Pout
*100%
Pdc
The efficiency of any amplifier is between 0 and 100 percent.
The higher the efficiency, the better the amplifier is at converting dc
power to ac power.
CLASS B OPERATIONAL PUSH PULL AMPLIFIER
Class B operation is different. It means that collector current flows for
only half the cycle (180°), as shown in Fig. designer locates the Q point at
cutoff. Then, only the positive half of ac base voltage can produce collector
current. This reduces the wasted heat in power transistors.
Push-pull means that one transistor conducts for half a cycle while the other
is off, and vice versa.
On the positive half cycle of input voltage, the secondary winding of T1 has
voltage v1 and v2, as shown. Therefore, the upper transistor (Q1)conducts
and the lower one(Q2) cuts off. The collector current through Q1 flows
through the upper half of the output primary winding. This produces an
amplified and inverted voltage, which is transformer-coupled to the
loudspeaker.
On the next half cycle of input voltage, the polarities reverse. Now, the lower
transistor (Q2) turns on and the upper transistor(Q1) turns off. The lower
transistor amplifies the signal, and the alternate half cycle appears across the
loudspeaker.
Since each transistor amplifies one-half of the input cycle, the loudspeaker
receives a complete cycle of the amplified signal.
Advantages and Disadvantages:Advantages are as below
Each transistor is at cutoff when there is no input signal
Improved efficiency where there is an input signal.
The maximum efficiency of a class B push-pull amplifier is 78.5 %.
Disadvantage are as below
use of transformers. Audio transformers are bulky and
expensive. So nowadays not used more.