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High CMRR Instrumentation Amplifier (Schematic and Layout) design for biomedical applications

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Instrumentation amplifiers are intended to be used whenever acquisition of a useful signal is difficult. IA’s must have extremely high input impedances because source impedances may be high and/or unbalanced. bias and offset currents are low and relatively stable so that the source impedance need not be constant. Balanced differential inputs are provided so that the signal source may be referenced to any reasonable level independent of the IA output load reference. Common mode rejection, a measure of input balance, is very high so that noise pickup and ground drops, characteristic of remote sensor applications, are minimized.Care is taken to provide high, well characterized stability of critical parameters under varying conditions, such as changing temperatures and supply voltages. Finally, all components that are critical to the performance of the IA are internal to the device. The precision of an IA is provided at the expense of flexibility. By committing to the one specific task of
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Measurement of the high resistance by using loss of charge method.

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Aim:- Measurement of the high resistance by using loss of charge method.
Apparatus:- Multimeter –1no
Ammeter-(0-200ma)-1no
Voltmeter –(0-30v)-1no
Capacitor-10uf-1no
Resister-100K-1no
Power supply-(0-30v)-1no
.
Circuit Diagram:-
Observation Table:-
S.
NO.
Time (sec) V(without
R)
V(with R) Log10(V/v)
without R
Log10(V/v)
with R
Theory:-
In this method the resistance which is measured is connected in parallel with
the capacitor C and the electronic voltmeter V. The capacitor is the charged up to some
suitable voltage by means of the battery having the voltage V and is then allowed to
discharge through the resistance.
The terminal voltage is observed over the considerable period of the time
during discharge.
Let ,
V=initial voltage on the charged capacitor .
.v=instantaneous discharging voltage.
I=the discharging capacitor current through the unknown resistor at time “t”.
Q=the charge still remaining in the capacitor.
I=dq/dt=-cdv/dt since[I=V/R]
V/R+C dv/dt=0 (1)
1/RC dt+1/V.dV=0
integrating both sides
t/RC+logev+K=0 (2)
K is const. of integration
At initial condition
When T=0 , v=V from equ. (2)
K=-logeV
now equ. (2) becomes
t/RC+logev-logeV=0
therefore. Loge (v/V)=-t/RC
v/V=e-t/RC
v=V*e-t/RC
Taking log on both sides
logev=logeV+logee-t/RC
R=t/C*loge(V/v)
R=0.4343*t/C*log10(V/v)
Procedure:-
1) Connections is make as per the circuit diagram.
2) Close the switch s and keep s2 open the capacitor charge by own leakage method.
3) New open reading and voltmeter .as its own resistance.
4) Note down the reading of the voltmeter Vs equal interval of the time
Result:- High resistance of the resistance is calculated by using loss of charge method.
Viva Questions:- 1)Why this method is called as loss of charge method?
2)What errors occur while performing this practical?

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