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# HRckts

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High Input Resistance circuits:
The ideal voltage amplifier should have infinite input impedance and zero output impedance.
The CC and CE with Re basic amplifiers have these properties.
The Input impedance of these amplifiers is
Ri = hie+(1+hfe)Re using the simplified model (assuming that hfeRe &lt;&lt; 0.1)
As Re &reg; &yen; this equation suggests that Ri &reg; &yen; .
However, as Re &reg; &yen; the assumption hfeRe &lt;&lt; 0.1 is no longer valid. And the more accurate
equation is
Ri = hie + (1+hfe)Re / (1+hoeRe) which &reg; hfe/hoe
This is a theoretical limitation on Ri.
There are other practical limitations also.
1. As Re increases the bias current causes a larger voltage drop across it. For middle of operating
range
VCE = VRe = VCC/2. We thus require larger impractical power supply voltage
2. In integrated circuits Re occupies chip area. Larger the value, greater is the chip area occupied,
leaving less for other components.
3. Bias resistance appear in parallel with the Ri and with typical values of a few 100K the parallel
combination RI is now decided by the bias resistance, and is hence lower.
Solutions:
The problems (1) and (2) are due to the fact that we are thinking of Re as Ohmic physical
resistance.
If Re is an equivalent resistance, created by a relatively smaller physical resistance in another CC
circuit, this problem can be solved. (The DARLINGTON PAIR circuit). The bias resistance
problem (3) may be solved by the BOOTSTRAPPING technique.
DARLINGTON PAIR:
The Darlington pair is a cascade of two common collector circuits
as shown.
With input resistance of Q2 acting as Re for Q1.Q1 thus sees a
large equivalent Re, but the physical resistance producing this
effect is R2 which is much smaller. Emitter of Q1 is at VR2 +VBE2
and is reasonably low voltage, not requiring a high voltage power
supply.
E.g.
If IC1 = 0.1 mA, IC2 = 1 mA, R2 = 5 K , hie = 1K and hfe = 100
then Re for first stage is
1K + (1+100)5K = 506K, and VE1 is just 5K&acute;1mA + 0.7V=5.7V.
Had we used a physical resistance of 500K as R e for the first stage, and omitted the second stage,
theVE1 would have been 500K&acute;0.1mA = 50V, requiring a power supply of 100 V if V ce =Vre =
Vcc/2.
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For the second stage, AI2 = 1 + hfe and
Ri2 @ (1+hfe)R2
For the first stage load is Ri2 and hoeRi2 is likely to be &gt; 0.1
So
AI1 = (1+hfe) / (1+hoeRi2) = (1+hfe) / (1+hoe(1+hfe)R2) @ (1+hfe) /(1+hoehfeR2)
Overall current gain is
AI1AI2 =(1+hfe)2 /(1+hoehfeR2)
And
RI = hie + AI2Ri2 =(1+hfe)2 R2/(1+hoehfeR2)
The RI for the given parameters is (1+100)2&acute; 5K / (1+ 25&acute;10-6 &acute;100&acute;5K) @ 50M
The Darlington pair has a drawback that emitter current of first stage is amplified by the current
gain of the second stage, and so first stage drift gets amplified, making the circuit drift- prone.
In general the second stage has a higher collector current (hfe2 &acute; Ie1) and so the h parameters of the
two transistors cannot be considered as the same.(h parameters depend on bias point)
Cancelling the effect of the bias resistance by Bootstrapping:
Effect of bias resistance can be minimised by bootstrapping the CC circuit as shown.
Here R1 and R2 are the bias resistances. giving base bias through R3. C1 short-circuits the output to
point Y at the junction of the bias resistance.
Thus signal voltage at Y = Vout
At point X the signal voltage is Vin.
Let the signal current in R3 be Ieff
Then
Ieff = (Vin - Vout ) / R3
So effective resistance of the combination seen by the source
is
Reff = Vin /Ieff = Vin R3/(Vin - Vout ) = R3 / (1 – Av)
where Av = voltage gain Vout/Vin
for the CC and Darlington circuits, Av @ 1 thus Reff = R3/ e where e = 1 – Av and is a very small
number.
This means effective resistance of the bias combination is increased greatly.
e.g. If Av is 0.95 and R3 is 50K , R1 = 100K and R2 = 100K then without bootstrapping, bias
combination would give 50K + (100K parallel with 100K) = 50 + 50 = 100 K resistance.
With Bootstrapping, the resistance is 50K / (1 – 0.95) = 50K &acute; 20 = 1000K = 1M.
Current Sources:
The requirement for high effective (equivalent) resistance can also be met by the CURRENT
SOURCE. The ideal current source has an infinite impedance, and so it may be used in place of Re ,
ensuring correct bias current and providing high resistance simultaneously. Current sources may
also be used as loads in high gain circuits (Active Loads)
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3
Two Transistor Current Mirror:
Also called current repeater used frequently for IC transistor biasing
Q1 and Q2 are identical transistors, Q1 is controlling and Q2 is controlled transistor.
As Vbe of both are same, IC1 = IC2, and as hFE’s are equal IB1 = IB2 = IB
Now IR = Ic1 + 2Ib = Ic1 + 2(Ic1 / hFE)
Thus IC2/IR = IC1/IR = hFE / (2+ hFE)
Also IR = (VCC – VBE )/R
Hence
IC2 = =[ hFE / (2+ hFE)][ (VCC – VBE )/R]
I
For typical values of hfe the difference between IR and IC2 is
negligible
so the circuit is a current mirror, reflecting value of
Ic1
Ic2
IR in the controlled side.IC2. Adding more transistors (or using
multi collector fabricated transistor currents in several
Ib
Ib
branches can be controlled by the Q1 transistor., with all
transistors delivering same currents. Hence the circuit is a
CURRENT REPEATER circuit. For N controlled transistors
the value of ICn ==[ hFE / (N + 1+ hFE)][ (VCC – VBE )/R].
If hfe is low, then the IR = IC2 approximation is not valid, and the circuit will not be an accurate
current mirror / repeater.
In that case the Three transistor current mirror should be used..
Ic2
R
Ic3
Ic4
R
Ic2
Ic3
Ic4
Q1
Q1
Three Transistor current mirror
The three transistor current mirror is used to obtain current mirror action when the hfe of the
transistors is low.
For the circuit shown
As Vbe of both Q1 and Q2 are same, IC1 = IC2 = IC, and as hFE’s are
equal IB1 = IB2 = IB
IR = IB3 + IC2 = (IE3 / 1+hfe) + hfeIB
But IE3 = 2IB
Hence IR = IB([2/(1+hfe)]+hfe)
And IC = hfeIB
Hence
IC/IR = hfe(1+ hfe)/( hfe2 + hfe + 2)
Thus even with hfe of 10 the ratio is close to 1.
This is a better current mirror for low hfe transistors
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4
IR is given by (VCC - 2VBE )/ R
WILSON SOURCE:
This is also a three-transistor circuit,
And VBE of the Q2 and Q3 being equal, I B2 = IB3= IB and IC2 = IC3=
IR
IC
Now,
IR
= IB1+IC3 = IB1+ hfeIB = IE1/(1+hfe) + hfeIB
= (2IB+ IC)/ (1+hfe) + hfeIB
=(2IB+ hfeIB)/ (1+hfe) + hfeIB
Q3
=IB[(2+ hfe)/ (1+hfe) + hfe]
IC1
= hfe IB1 = hfe IE1/(1+hfe) = hfe (2IB+ IC)/ (1+hfe)
= hfe (2IB+ hfeIB)/ (1+hfe)
= IB[hfe(2+ hfe)/ (1+hfe)]
Then
IC1/IR = hfe(2+ hfe)/( hfe2 + 2hfe + 2) and IR is given by (VCC - 2VBE )/ R
Again this is useful when hfe of the transistors are low.
IC
R
Q1
Q2
CURRENT SOURCES FOR LOW CURRENT APPLICATIONS
For getting low currents like 50 uA the current sources above require too large a value of R.
(several 100 K)
Example: In a two transistor current mirror it is required to get a output current 50 uA. Find R if hfe
= 100. And VCC = 15V.
Here 0.05 = 100/(100+2) &acute;(15-0.7)/R
Hence R = 280 k.
The value is too high
Here the WIDLAR source is used.
VBE2 =VBE1 + (IB1 + IC1) Re
VBE2 - VBE1 = (IB1 + IC1) Re
Now IC1 = aFIESeVbe1/VT
and
IC2 = aFIESeVBE2/VT , also,
I c1
= e (VBE1 -VBE 2 ) / VT
I c2
Hence VBE1-VBE2 = VTln (IC2/IC1)
&aelig;I &ouml;
ln&ccedil;&ccedil; c 2 &divide;&divide;
&aelig;
1 &ouml;&divide; &egrave; I c1 &oslash;
I c1 &ccedil;1 +
&ccedil; h fe &divide;
&egrave;
&oslash;
The reference current IR1 = IC2 + IB2 + IB1 = IC2(1+(1/hfe)) + (Ic1/hfe)
The last term an be neglected as IC1 &lt;&lt; IC2
Thus Re =
VT
And IC2 = (VCC- VBE2)/R for hfe &gt;&gt;1
From this IC2 is determined, and Using desired value of IC1 , Re can be found
In the given problem taking R = 14K IC2 =
100 (15 - 0.7 )
=1.01 mA
(100 + 1) 14
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HRckts.doc
And Re =
5
1.01
0.025
ln
= 1.49K
1 &ouml; 0.05
&aelig;
0.05&ccedil;1 +
&divide;
&egrave; 100 &oslash;
Thus much smaller resistor values are required for the same current.
VOLTAGE SHIFTERS
The differential voltage at the input of a differential amplifier may be AC or DC. and the
corresponding output too may be AC or DC. Therefore Coupling capacitors and transformer
coupling cannot be used to interconnect stages.
It is also required that for 0 input output should be 0.
In the normal differential amplifier circuit the output is taken from the collector and this has a DC
component equal to the bias VC. with 0 input this is the quiescent voltage at the collector. To make
the final output voltage 0, capacitors and transformers cannot be used, so the standing VC has to be
shifted down to 0V. For this VOLTAGE SHIFTER circuits are used.
DIODE SHIFTER:
If the VC is an integral multiple of 0.7 V, say 3.5 V, then 5 diodes may be connected in series with
the collector to give a drop of 3.5 V
This method is used only when the drop required is K &acute; 0.7 with K an integer.
VBE MULTIPLIER
In the circuit shown, The voltage VBE = 0.7 V = VR2. Hence current in R2 =
0.7/R2 If we neglect base current, then voltage across R 1 and R2 = (R1+
A
R2)&acute;(0.7/R2). Thus voltage drop from A to B = (R1+ R2)&acute;(0.7/R2).
This can now be any multiple of VBE,, as decided by R1 and R2 ,hence the circuit
R1
is a VBE MULTIPLIER
R2
Common Collector with Split resistor
The CC circuit shown in figure below has its emitter
resistance split into two.i.e. R1 and R2 Here output is
taken from the junction of the resistors (B)
B
If the bias current is say 1 mA and if 5 V drop is required from base to
output (B) then 0.7 + VR1 = 5 hence VR1 = 4.3V and R1 = 4.3 K.
Thus R1 can be adjusted to give required shift ., but the gain reduces to
value &lt; 1, as the standard CC output will be at A and we are taking
actual output at B with a voltage division ratio R2/R1+R2. The
standard CC gain is @ 1 so now we have gain &lt; 1.
This can be avoided by substituting a current source in place of R2.
Then voltage division ratio becomes &yen; /(R1+&yen;) =1
A
R1
B
R2
A
Vcc
R1
B
Giving a circuit as shown if the figure., the voltage shift is between
Base of the upper transistor and output point B
Current source is designed to supply the required bias current, and R1 is chosen to obtain the
required drop.
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6
LOW OUTPUT RESISTANCE STAGE:
The output resistance of the differential amplifier should be ideally 0
Nearest to 0 would be the small output resistance of the CC amplifier, hence CC may be used as the
output stage. As DC output is to be zero for zero signal, the emitter resistance cannot be grounded
and so must be connected to -VEE so that the emitter can be at 0 V. More typically complementary
emitter-follower output stage is used
Note the complementary transistors (NPN-PNP) The current source supplies the bias to the base and
the two diodes maintain about 1.4 volts between the bases since the upper transistor requires +0.7V
and the lower -0.7 V for conduction
If signal VI goes +ve NPN Q1conducts, Q2 is cut-off and load gets current through Q1
If VI goes negative PNP Q2 conducts, Q1 is cut-off and the current flows through load into Q1.
Thus conduction occurs for each half cycle in alternate transistors. The diodes eliminate crossover
distortion (no output in range -0.6 to 0.6 as neither transistors would be conducting.) In place of the
diodes, VBE multiplier may also be used.
Two small equal current limiting resistors can be inserted at each emitter, with a small increase in
output impedance.
Vcc
Vcc
Vcc
VBE
multiplier
Current
limiter
V
V
Q
Vi
Q
Vee
Vee
Vee
Totem-Pole output stage
Similar to complementary output stage but uses transistors of same type.
Q1 amplifies the input and passes on the signal to the TOTEM POLE
arrangement of Q2 and Q3. If input is +ve current in Q2 is decreased and in Q3
Q2
increases thus current is taken from the load reducing vo. If input is –ve, reverse
Q1
behaviour causes current to be delivered to load, increasing vo for large +ve vin
Q2 is cut-off and Q3 saturates, so vo = VEE-VCE sat and for large -ve input Q2
saturates, Q3is cut-off and vo = VCC- VCesat.
Q3
Output may be stabilised by negative feedback.
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