ELECTRONIC INSTRUMENTS
Electronic voltmeter
FIGURE 4.1
Transistor Voltmeter;Emitter Follower
Question
PBL : ELECTRONIC INSTRUMENTATION
FIGURE 4.1
• VE=Vin-VBE=10 V-0.7 V=9.3V
• Im=VE/(RS+Rm)≅9.3 V/9.3k Ω=1mA
• With transistor
• IB ≅Im/hfe=1mA/100=10 µA
• Ri ≅Vin/IB ≅10V/10 µA=1MΩ
• Without transistor= Ri=Rs+Rm=9.3 kΩ
FIGURE 4.2
Assignment (14/Sept)
• Derive VE1-VE2=Vm. Draw the equivalent
circuit.
Solution
When Vin=1 V
VE1=Vin -VBE =1 V -0.7 V =0.3 V
VE2
=Vp - VBE =0 V – 0.7 V =-0.7 V
Vm=VE1-VE2 = 0.3-(-0.7)=1V
When Vin=0.5 V
VE1=Vin -VBE =0.5 V -0.7 V =-0.2 V
VE2
=Vp - VBE =0 V – 0.7 V =-0.7 V
Vm=VE1-VE2 = -0.2-(-0.7)=0.5V
When Vin=7 V
VE1=Vin -VBE =7 V -0.7 V =6.3 V
VE2
=Vp - VBE =0 V – 0.7 V =-0.7 V
Vm=VE1-VE2 = 6.3-(-0.7)=7V
FIGURE 4.4
R1
EG=E x (Rc+Rd)/(Ra+Rb+Rc+Rd)
=7.5 V x (60k+40k) / (800k +100k +60k +40k)
= 0.75 V
Vs = EG-VGS=0.75 V –(-5 V) =5.75 V
VE1=Vs – VBE=5.75 V – 0.7 V=5.05 V
VE2=VP – VBE=5 V – 0.7 V =4.3 V
V = VE1 – VE2 =5.05 V – 4.3 V
=0.75 V = EG
Im = V / RS + Rm =0.75 V/1kohms
=0.75 mA
75% of full scale
Op-Amp Voltage-Follower Voltmeter
Input attenuator
FIGURE 4.7
Vout = Im x Rm =100µA x 10kΩ = 1V
R3 = E/ I4 =20mV/ 0.2 mA =100Ω
R4 = (Vo – E) / I4 = (1 V -20 mV) / 0.2 mA
=4.9 kΩ
FIGURE 4.8
The small voltage drop across
shunt resistor Rs is amplified
before being applied to
deflection instrument.
This approach is just as applicable to
the measurement of low level alternating
Currents as it is to direct-current measurement
FIGURE 4.14
FIGURE 4.9
FIGURE 4.10

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