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AD835ARZ Datasheet(PDF) 12 Page  Analog Devices 

AD835ARZ Datasheet(HTML) 12 Page  Analog Devices 
12 / 16 page AD835 Rev. D  Page 12 of 16 SQUARING AND FREQUENCY DOUBLING Amplitude domain squaring of an input signal, E, is achieved simply by connecting the X and Y inputs in parallel to produce an output of E2/U. The input can have either polarity, but the output in this case is always positive. The output polarity can be reversed by interchanging either the X or Y inputs. When the input is a sine wave E sin ωt, a signal squarer behaves as a frequency doubler because () ( ωt ) U E U ωt E 2 cos 1 2 sin 2 2 − = (6) While useful, Equation 6 shows a dc term at the output, which varies strongly with the amplitude of the input, E. Figure 24 shows a frequency doubler that overcomes this limitation and provides a relatively constant output over a moderately wide frequency range, determined by the time constant R1C1. The voltage applied to the X and Y inputs is exactly in quadrature at a frequency f = ½πC1R1, and their amplitudes are equal. At higher frequencies, the X input becomes smaller while the Y input increases in amplitude; the opposite happens at lower frequencies. The result is a double frequency output centered on ground whose amplitude of 1 V for a 1 V input varies by only 0.5% over a frequency range of ±10%. Because there is no squared dc component at the output, sudden changes in the input amplitude do not cause a bounce in the dc level. AD835 1 2 3 4 8 VOLTAGE OUTPUT R2 97.6 Ω R1 R3 301 Ω X2 VP W Y1 X1 Y2 +5V C1 –5V VN Z 7 6 5 VG Figure 24. Broadband ZeroBounce Frequency Doubler This circuit is based on the identity θ = θ θ 2 sin 2 1 sin cos (7) At ωO = 1/C1R1, the X input leads the input signal by 45° (and is attenuated by √2, while the Y input lags the input signal by 45° and is also attenuated by √2. Because the X and Y inputs are 90° out of phase, the response of the circuit is () () ( t U E t E t E U W ω = ° + ω ° − ω = 2 sin 2 45 sin 2 45 sin 2 1 2 ) (8) which has no dc component, R2 and R3 are included to restore the output to 1 V for an input amplitude of 1 V (the same gain adjustment as previously mentioned). Because the voltage across the capacitor (C1) decreases with frequency, while that across the resistor (R1) increases, the amplitude of the output varies only slightly with frequency. In fact, it is only 0.5% below its full value (at its center frequency ωO = 1/C1R1) at 90% and 110% of this frequency. 
