Design of various signal conditioning circuits are covered in this presentation.

1. Analog Signal Conditioning
By:
Dr. Sebastian George

2. Temperature is to be transmitted in the range 00C to 1000C. Sensor available is RTD
Pt-100. Design the necessary signal conditioning circuitry to convert the
temperature range into 0-5 V. Provide “zero” and “span” adjustments.
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Example 1-RTD Signal Conditioning

3. Solution
 RTD Pt-100 has a resistance of 100 Ω at 00C
 At 1000C, its resistance will be R100 = R0 [1+αt] = 100[1+0.0039 x 100] = 139 Ω
α is the temperature coefficient of resistance whose value is 0.0039/0C for
platinum.
 The RTD is connected in a Wheatstone Bridge circuit which is excited by +5V
DC.
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4. Solution Contd… 4

5. Solution Contd…
 The bridge unbalance voltage at 00C, ΔV = 0V
 At 1000C, ΔV = [(5 x 139) / (139+100)] – [(5 x 100) / (100+100)] = 0.4V
 Hence, the temperature variation of 00C – 1000C is converted into 0 – 0.4 V
 This is further converted into 0 – 5V by the instrumentation amplifier.
 Therefore, the gain of the amplifier will be (5 – 0) / (0.4 – 0) = 12.5
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6.  For the instrumentation amplifier, the gain is
1 + (2R1/R3) = 12.5
Let R1 = 10 K
Therefore, R3 = 1.739 K
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Solution Contd….

7. Circuit Diagram 7

8. Zero and Span Adjustments
 For calibration purpose, the circuit needs ‘Zero’ and ‘Span’ adjustments.
 The Zero adjustment is nothing but the offset adjustment which is required in the
bridge circuit to compensate for the tolerances in the resistance values of the
remaining three arms of the bridge. This is accomplished by the 100 Ω variable
resistance connected in the third arm of the bridge circuit.
 The Span adjustment is nothing but the gain adjustment of the instrumentation
amplifier which is accomplished by a 2.2 K pot connected in the amplifier circuit.
 For calibration of the signal conditioning circuit, the zero and span adjustments are
to be done multiple times keeping the RTD in melting ice (00C) and boiling water
(1000C).
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9. Example 2- I/V Converter
Design a signal conditioning circuit which coverts 4 – 20mA into 0 – 5V.
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10. Solution 10

11. Solution
 Vout = mVin + Vo where m is the gain and Vo is the offset.
 Now substituting two different values of Vin and Vout , we get :
0 = m(0.4) + V0 …(1)
5 = m(2) +Vo …(2)
 From equations (1) and (2), we get:
m = 3.125 and Vo = -1.25
 Therefore, Vout = 3.125 Vin – 1.25
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12. Circuit Diagram 12

13. Gain and Offset Adjustments
 For obtaining a gain of 3.125, a fixed resistance of 10K and a variable
resistance of 47K are used as seen in the circuit diagram. The 47K pot will
act as the ‘Span’ adjustment.
 An offset of -1.25 V is derived from a +5V power supply using a voltage
divider. The 10K pot will act as the zero adjustment.
 For calibration of the I/V converter, both zero and span adjustments may
have to be done multiple times.
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14. Example 3- V/I Converter
 Design a signal conditioning circuit which converts 0-5 V into 4-20mA.
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15. Solution 15
 Vout = mVin + Vo where m is the gain and Vo is the offset.
 Now substituting two different values of Vin and Vout , we get :
0.4 = m(0) + V0 …(1)
2 = m(5) +Vo …(2)
 From equations (1) and (2), we get:
m = 0.32 and Vo = 0.4
 Therefore, Vout = 0.32 Vin + 0.4

16. Circuit Diagram
 The gain of 0.32 can be adjusted
with the pot 47K.
 Offset of 0.4V is derived from +5V
power supply using a voltage divider
consisting of a fixed 47K resistance
and a variable 4.7K pot.
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17. Howland Circuit
 A Howland circuit is used for getting 4-
20 mA from 0.4 – 2V.
 As all the four resistances are of 100 Ω
each, an output current of 4 -20 mA is
obtained corresponding to a voltage
input of 0.4-2V.
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18. Circuit Diagram 18
Zero
adjustment
0-5V

19. Zero and Span Adjustments
 As shown in the circuit diagram, a 47K pot is used for ‘Span’ adjustment and a 4.7K
pot is used for ‘Zero’ adjustment.
 For calibration of the V-I converter, both the Zero and Span have to be adjusted
multiple times.
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Thank You!