Transistor

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Transcript Transistor

Transistor
BJT Transistors:
NPN
Transistor
PNP
Transistor
Sandwiching a
P-type layer
between two ntype layers.
Sandwiching a
N-type layer
between two ptype layers.
How a “NPN” Transistor works?
The base-emitter diode
(forward) acts as a
switch. when v1>0.7 it lets
the electrons flow toward
collector. so we can
control our output current
(Ic) with the input
current (Ib) by using
transistors.
C
backward
B
E
Forward
Transistors have three terminals:
Collector
Base
Emitter
Active:
Always on
Ic=BIb
Transistors work in 3 regions
Saturation
:Ic=Isaturation
On as a switch
Off
:Ic=0
Off as a switch
Transistor as a Switch
• Transistors can be used as switches.1
Transistor
Switch
• Transistors can either
conduct or not conduct current.2
• ie, transistors can either be on or off.2
Transistor Switching Example15
X
12V
Variable
Voltage
Supply
• When VBE is less than 0.7V the transistor is off
and the lamp does not light.
• When VBE is greater than 0.7V the transistor is on
and the lamp lights.
Transistor Circuit : Light-Controlled Circuit
• This transistor circuit contains
a Light-Dependent Resistor.
• Because of the LDR, this circuit
is dependent on light.
• The purpose of this circuit is to
turn on the LED when the light
reaches a certain intensity.
Input = Voltage Divider
Process = Transistor
Output = LED
1)
2)
3)
4)
5)
6)
LED = Off.
Cover LDR.
RLDR .
VLDR .
Transistor switches on.
LED = On.
Transistor as an amplifier:
Transistors are often used as amplifiers to increase input
signal in radios, televisions and some other applications
.The circuit may be designed to increase the current or
voltage level.
The power gain is the product of current gain and voltage
gain (P=V*I).
Amplifier example:
As you see, the transistor is
biased to be always on. The
input signal is amplified by
this circuit. The frequency
of output is the same as its
input, but the polarity of the
signal is inverted.
The measure of
amplification is the gain of
transistor.
Example:
Input Amplitude =0.2v
Output amplitude=10v
Gain=10/0.2=50
Field Effect Transistors
JFET
MOSFET
CMOS
How a JFET transistor
works?
When the gate is negative ,it
repels the electron in the Nchannel. So there is no way for
electrons to flow from source to
drain.
When the negative voltage is
removed from Gate ,the
electrons can flow freely from
source to drain .so the
transistor is on.
How a
MOSFET
Transistor works?
In MosFET, the Gate is insulated from p-channel or
n-channel. This prevents gate current from flowing,
reducing power usage.
When the Gate is positive voltage ,it allows electrons to
flow from drain to source .In this case transistor is on.
How a CMOS transistor works?
N-channel & P-channel MOSFETs can be
combined in pairs with a common gate .
When Gate (input) is high
,electrons can flow in N-channel
easily . So output becomes low.
(opposite of input)
When Gate (input) is low ,holes
can flow in P-channel easily. So
output becomes high.
(opposite of input)
Opamp
Schematic diagram of lm741
Ideal Opamp
Operational Amplifier (OP AMP)
Basic and most common circuit
building device. Ideally,
1. No current can enter terminals
V+ or V-. Called infinite input
impedance.
A
2. Vout=A(V+ - V-) with A →∞
Vo = (A V + -A V )
= A (V + - V )
3. In a circuit V+ is forced equal
to V-. This is the virtual
ground property
4. An opamp needs two voltages
to power it Vcc and -Vee. These
are called the rails.
-
-
INPUT IMPEDANCE
Input
Circuit
Output
Impedance between
input terminals =
input impedance
WHY?
For an instrument the ZIN should
be very high (ideally infinity) so
it does not divert any current
from the input to itself even if
the input has very high
resistance.
e.g. an opamp taking input from a
microelectrode.
OUTPUT IMPEDANCE
Impedance between output terminals
=
output impedance
WHY?
Input
Circuit
Output
For an instrument the ZOUT should
be very low (ideally zero) so it can
supply output even to very low
resistive loads and not expend
most of it on itself.
e.g. a power opamp driving a motor
OPAMP: COMPARATOR
Vout=A(Vin – Vref)
If Vin>Vref, Vout = +∞ but practically
hits +ve power supply = Vcc
A (gain)
very high
If Vin<Vref, Vout = -∞ but practically
hits –ve power supply = -Vee
Application: detection of QRS complex in ECG
VREF
VIN
Vcc
-Vee
OPAMP: ANALYSIS
The key to op amp analysis is simple
1. No current can enter op amp input terminals.
=> Because of infinite input impedance
2. The +ve and –ve (non-inverting and inverting)
inputs are forced to be at the same potential.
=> Because of infinite open loop gain
3. These property is called “virtual ground”
4. Use the ideal op amp property in all your
analyses
OPAMP: VOLTAGE FOLLOWER
V+ = VIN.
By virtual ground, V- = V+
Thus Vout = V- = V+ = VIN !!!!
So what’s the point ? The point is, due to the
infinite input impedance of an op amp, no
current at all can be drawn from the circuit
before VIN. Thus this part is effectively
isolated. Very useful for interfacing to high
impedance sensors such as microelectrode,
microphone…
OPAMP: INVERTING AMPLIFIER
1. V- = V+
2. As V+ = 0, V- = 0
4. I1 = (VIN - V-)/R1 = VIN/R1
3. As no current can
enter V- and from
Kirchoff’s Ist law,
I1=I2.
5. I2 = (0 - VOUT)/R2 = -VOUT/R2 => VOUT = -I2R2
6. From 3 and 6, VOUT = -I2R2 = -I1R2 = -VINR2/R1
7. Therefore VOUT = (-R2/R1)VIN
OPAMP: NON – INVERTING
AMPLIFIER
1. V- = V+
2. As V+ = VIN, V- = VIN
4. I1 = VIN/R1
3. As no current can
enter V- and from
Kirchoff’s Ist law,
I1=I2.
5. I2 = (VOUT - VIN)/R2 => VOUT = VIN + I2R2
6. VOUT = I1R1 + I2R2 = (R1+R2)I1 = (R1+R2)VIN/R1
7. Therefore VOUT = (1 + R2/R1)VIN
SUMMING AMPLIFIER
If
Recall inverting
amplifier and
If = I1 + I2 + … + In
VOUT = -Rf (V1/R1 + V2/R2 + … + Vn/Rn)
Summing amplifier is a good example of analog circuits serving as analog
computing amplifiers (analog computers)!
Note: analog circuits can add, subtract, multiply/divide (using
logarithmic components, differentiate and integrate – in real time and
continuously.
DRIVING OPAMPS
•For certain applications (e.g. driving a motor or
a speaker), the amplifier needs to supply high
current. Opamps can’t handle this so we modify
them thus
Irrespective of the
opamp circuit, the small
current it sources can
switch ON the BJT
giving orders of
magnitude higher
current in the load.