AD8500 output impedance

Document created by analog-archivist Employee on Feb 23, 2016Last modified by analog-archivist Employee on Feb 23, 2016
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We apply step signals on the AD8500 inputs (+ or -). Can this lead to frequency
differences of the integrator output signal? Therefore we would like to know
the output impedance of the device. Please find the schematics below (any
comments on the circuitry would be appreciated). We compare Vout1 and Vout2.
Circuitry around U1 and U3 are the same but in case of U3 the input signal is
directly connected and in case of U1 it is connected via the impedance
converter U2.
- The integrator cutoff frequency is 160Hz
- We apply 2V step signal on the input

Function generator settings:
Waveform:        Square wave
Frequency:        100MHz
High-Level:        2V
Low-Level:         0V
Duty-Cycle:       50%
Output:             High-Z

Measurement description:
Measuring the open-loop step response: we set the target voltage of
Voffset=1.48V for the integrator, then connect the outputs of the function
generator to Vin and GND. Supply voltage of AD8500 is 2.05V. At the square wave
edges the integrator tries to control the output to reach the target voltage.

 

Yes, the output impedance of all three AD8500 amplifiers will affect the output
impedance of the two integrator signal paths.

Other observations:
C6 and C7 on the outputs of the integrators appear to be incorrect. It appears
that these large caps are going to keep the outputs from moving.
The AD8500 has high output impedance (15-20 kOhms). It is meant to drive
megohms, not the 10k load at R1.

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