Square root circuit and √((V1V2)²+(V3V4)²)

Square-root circuit

Taking a square root means multiplying the logarithm by ½: √Vi = exp(½ ln Vi).

Building blocks used (all voltages in volts, Vref = 1 V, K = scale factor such as VT or 1 V/decade):

• Log amp (L): output = −K ln(Vin), for Vin > 0
• Antilog amp (E): output = exp(Vin/K)
• Inverting amp of gain −m: output = −m × input

                                R/2
                          +---/\/\/---+
                          |           |
 Vi -> [LOG] --V1'--[R]---+--(-)\     |
                                  >---+--> V2'
                         GND--(+)/
 V2' -> [ANTILOG] -> Vo

Derivation:

1. Log amp: V1' = −K ln Vi
2. Inverting amplifier with gain −(R/2)/R = −½:

V2' = −½ × (−K ln Vi) = (K/2) ln Vi

3. Antilog amp:

Vo = exp(V2'/K) = exp(½ ln Vi) = Vi^(1/2)

Result: Vo = √Vi. With a general reference, Vo = √(Vref·Vi). For example, Vi = 9 V gives Vo = 3 V (Vref = 1 V). The input must be positive.

Circuit for V0 = √((V1·V2)² + (V3·V4)²)

Plan: form (V1V2)² and (V3V4)² with log–antilog stages, add them, then take the square root.

 V1->[LOG]-+
           +->[summer, gain -2]->[ANTILOG]->(V1V2)^2 -+
 V2->[LOG]-+                                         |
                                       [SUMMER] <----+
 V3->[LOG]-+                              ^    |
           +->[summer, gain -2]->[ANTILOG]+    | S
 V4->[LOG]-+              (V3V4)^2             v
          S -> [LOG] -> [gain -1/2] -> [ANTILOG] -> V0

Step by step:

1. Log amps give −K ln V1, −K ln V2, −K ln V3 and −K ln V4.
2. Inverting summer with gain 2 (input resistors R, feedback 2R):

Va = −2(−K ln V1 − K ln V2) = 2K ln(V1V2)

The antilog gives exp(Va/K) = (V1V2)². In the same way the second branch gives (V3V4)².

3. A non-inverting summer (or an inverting summer followed by an inverter) gives:

S = (V1V2)² + (V3V4)²

4. The square-root block (log, gain −½, antilog) gives:

V0 = √S = √( (V1V2)² + (V3V4)² )

All inputs must be positive (or first passed through precision rectifiers, since the result depends only on squares). The circuit uses 8 log or antilog amps (5 log, 3 antilog) plus the summing and scaling op-amps.