1. Prepared By: SHIYON P JOHNSON Department: MEDICAL ELECTRONICS
Subject Name: LIC
Subject Code:10EC46
Prepared By: SHIYON P JOHNSON
Department: MEDICAL ELECTRONICS
Date:03/03/2015
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2. VOLTAGE SOURCES
The voltage source circuit is simply a potential divider and voltage follower with transistor Q1 included at output.
A transistor is necessary for higher loads current levels as that supplied by op amp is only 25mA.
The maximum load current is now maximum emitter current for the transistor and op amp has to supply the much smaller transistor base current.
With the inverting input terminal of the opamp connected to the transistor emitter,the emitter terminal is maintained at the same voltage level as the non inverting terminal(at voltage V2).
Instead of using potential divider to provide reference voltage,a Zener didode can be used as it is independent of any variations in supply voltage.
Design involves:Selecting transistor to handle desired load current,then calculating potential divider resistors.If using zener diode,a diode desired breakdown voltage is first selected,then R1=VCC-VZ/IZ
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3. VOLTAGE SOURCES Low resistance voltage source
Precision Voltage Sources
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4. PRECISION VOLTAGE SOURCES
Problem: Zener diode can vary slightly because its current will change if the supply voltage does not remain constant.
This can be solved by using a output voltage source to maintain a constatnt current through the diode,thus providing a precise unchanging voltage ,and consequently,a precise output level.
If the required load current is too great for the op-amp to supply,a transistor can be included at the output.
Both the zener diode circuit(D1 and R1) and the potential divider(R2 and R3) are spplied from the op-amp output terminal.Since the op-amp inverting and noninverting terminal voltages must be equal when the op-amp is operating as a linear circuit, VR3=VZ and VR2=VR1.
FIG
Now if the output were to somehow increase by V0,the voltage change at the op-amp inverting input would be
VR3=(V0 x R3)/(R2+R3).
This would cause the output to move in a negative direction to correct the original increase in V0.
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5. CURRENT SOURCES iL iL Load Load vin R1 R1 i1 i1 _ _ OR + + + + vin _ _
General Equations:
iL = i1 = v1/R1
v1 = vin
The transconductance, gm = io/vin = 1/R1
Therefore, iL = i1 = vin/R1 = gmvin
The maximum load resistance is determined by:
RL(max) = vo(max)/iL
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6. CURRENT SOURCES iL Load Given: vin = 2 V, R1 = 2 k vo(max) = 10 V R1
Find: iL , gm and RL(max)
Solution:
iL = i1 = vin/R1 = 2 / 2000 = 1 mA
gm = io/vin = 1/R1 = 1 / 2000 = 0.5 mS
RL(max) = vo(max)/iL = 10 V / 1 mA
= 10 k 
Note:
If RL > RL(max) the op amp will saturate
The output current, iL is independent of the load resistance.
vin R1 i1 _ + + _
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7. CURRENT AMPLIFIER: Current to Voltage Converter RF iF _ + + iin vO -
General Equations:
iF = iin
vo = -iFRF
rm = vo/iin = RF
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8. Current to Voltage Converter
Current to Voltage Converter are used for low-power applications to produce an output voltage proportional to the input current.
Photodiodes and Phototransistors, which are used in the production of solar power are commonly modeled as current sources.
Current to Voltage Converters can be used to convert these current sources to more commonly used voltage sources.
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9. CURRENT AMPLIFIERS iin = iL vin = vo Load _ +
1. Possible ICIS Operational Amplifier Application +
iin = iL
Similar to the voltage follower shown below: _ +
vin = vo
vin _ + + vO -
Both these amplifiers have unity gain:
Av = Ai = 1
Voltage Follower
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10. INSTRUMENTATION AMPLIFIER
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11. Instrumentation amplifiers CONTD..
Instrumentation amplifiers are a combination of three op-amps that are typically
grouped into two stages.
The first two op-amps comprise the first stage and each is a noninverting amplifier. The second stage is a differential amplifier that may or may not have unity gain.
An instrumentation amplifier is beneficial for several reasons:
1. high input impedance, unlike the lower input impedance of a differential amplifier
by itself
2. high CMRR (see the pages ahead for a better understanding of CMRR); the source internal resistances of v1 and v2 do not affect the total resistance on each input arm
3. good for smaller, insignificant input signals
4. gain of the non-inverting amplifiers (first stage) can be varied by the rheostat (RC)
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12. Instrumentation amplifiers CONTD..
Now consider the KCL analysis for the first stage of the instrumentation amplifier.
It is helpful to redraw op-amps A and B and their corresponding circuitry for analysis.
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13. Instrumentation amplifiers CONTD..
The KCL for op-amp A is as follows:
Solve for VA,
The KCL for op-amp B is as follows:
Solve for VB,
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14. Instrumentation amplifiers CONTD..
The values vA and vB are the two inputs into the differential amplifier.
Refer back to the analysis of the differential amplifier and see that the output is one input minus the other,and this difference is multiplied by x, which is the gain.
In the case of the instrumentation amplifier, the output is (vA - vB) if the differential amplifier has unity gain, or x(vA – vB) if it has gain.
Either way, vA – vB equals:
which equals,
If R1 = R2, then
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15. Instrumentation amplifiers CONTD…
Notice that the first stage has a gain of
If the differential amplifier has a gain value (x), then the final output would be the product of the two gains multiplied by the difference between the two input voltages, or
Final output =
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16. PRECISION RECTIFIERS
The precision full-wave rectifier receives an ac signal and produces a fully rectified output.
The second op-amp (2) is a summation op-amp, and the first (1) is an inverting amplifier with two diodes (D1 and D2), which make the rectification possible.
The first op-amp inverts the signal but does not amplify it.
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17. PRECISION RECTIFIERS
The two conditions possible are a positive or negative alternation. On the positive alternation, the op-amp inverts the signal, producing a negative value that conducts through D2 and the feedback loop. The output (Vout1) is –Vin.
On the negative alternation , the op-amp again inverts the signal, and this time produces a positive value that does not conduct through D2, so the output is zero. Instead, the positive signal feeds D1 to avoid saturation.
Using an op-amp along with diodes for rectification is an asset because without the op-amp, the typical 0.7V drop across a diode has to be taken into account.
So , a small signal like 0.5V would not be rectified because the diode needs at least 0.7V to conduct.
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18. PRECISION RECTIFIERS
The KCL analysis for each op-amp in the precision full-wave rectifier is detailed below:
Op-amp 1 Positive alternation
Negative alternation: Vout1 = 0
Op-amp 2Positive alternation
Negative alternation
note: Vin is negative, so –(-Vin) = Vin
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