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I've recently posted this question regarding input impedance when working with DC signal on HCPL-7840 Isolation Amplifier. The answers were very clarifying and helpful, but I've come to realize I need more specific help.

Just a disclaimer: I decided to use HCPL-7840 because it was the best option available on my chosen supplier that seems to fufil my needs.

My goal is to design a simple isolated amplifier for an NTC


  • NTC characteristics considered:

    Resistance [kΩ] Temperature [ºC]
    10 30
    15 20
    20 10
    25 0
  • HCPL-7840 characteristics considered:

    • The information on the "Recommended Operating Conditions" table of datasheet
    • Equivalent Input Impedance RIN — 500 — kΩ
    • Figures 12 and 14 of datasheet

  • Based on these informations, I designed the following simple circuit:
    • It's composed of a simple voltage divider, where R1 is fixed precise resistor, and R2 is the NTC.

enter image description here

  • As soon as I assembled the board and started making measurements, I noticed that the voltages on TP1 were not exactly what I expected to see

    • I was expecting a simple voltage divider between R1, R2
    • I figured the difference should be because of internal resistence of HCPL (Rin), between pins 2 and 3
    • This Rin is suposelly connected in parallel with R2 (between TP1 and GND iso)
  • I than started to conduct some experiments to empirically try and figure out the value of Rin. This way I could calculate adjusted values for R1 and R2 and keep the voltage on TP1 within the input limits of HCPL linear range

    • The experiment consists of trying different combinations of R1 and R2 and measuring the voltage at TP1.

    • The results of this experiments drive me crazy because different combinations of R1, R2 make me infer different values for Rin each time.

    • The following table summarizes the experiment results:

      Index Vdd [V] R1 [Ω] R2 [Ω] V measured @ TP1 [V]
      1 5.40 2150 100 0.213
      2 5.45 2150 68 0.158
      3 5.56 2150 40 0.101
      4 5.55 246000 11600 0.186
      5 5.70 246000 5800 0.119
      6 5.77 717500 11600 0.082
      7 5.79 717500 23500 0.113
      8 5.84 717500 51300 0.142
    • Vdd and V measured @ TP1 were measured with osciloscope.

    • R1 and R2 were measured with Ohmmeter before the experiment.

    • Outcomes:

      • Given Vdd, R1 and V measured @ TP1, I was able to calculate what would be the "real R2" that the circuit is submited to.
      • This "real R2" would be the parallel between R2 (from the table) and Rin
      • Because I have R2, I'm able to calculate Rin
    • After this procedure I ended up with a lot of different values of Rin for each experiment case, and I don't know how to proceed!!!

Any help is appreciated.

Thanks!

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  • \$\begingroup\$ transferring an analog signal over an opto is challenging. You'l get much better linearity using the opto to transfer a frequency or pulse width. You could probably use a NE555 in both of those cases. Assuming you go with a pulse width modulated signal, analog filtering on the output of the opto will create an analog signal. Of course, you could use a single chip micro and just transmit uart data across the opto. Or go Bluetooth or WiFi and negate the need for an opto. \$\endgroup\$
    – Kartman
    Apr 29 at 10:44

2 Answers 2

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Figure 14 of the datasheet suggests that the input current into the HCPL-7840 should be fairly linear (resistor like) for input voltages in the range of -200mV to +200mV. But acconding to note 12 on page 18 of the datasheet the input stage is actually a "switched capacitor", not a real resistor, so there is going to be current spikes. I would suggest putting a 0.1uF ceramic capacitor across R2.

I would also suggest making changes that make your circuit less sensitive to input current on the HCPL-7840 input pins.

  • One option is to buffer the + input of the HCPL-7840 with another rail-to-rail op-amp configured as a voltage follower.
  • Alternatively, you could lower the values of R1 and R2 enough that the input current into the HCPL-7840 doesn't affect the measurement as much.
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from datasheet ...

Description
The Broadcom® HCPL-7840 isolation amplifier family is designed for current sensing in electronic motor drives. In a typical implementation, motor currents flow through an external resistor and the resulting analog voltage drop is sensed by the HCPL-7840. A differential output voltage is created on the other side of the HCPL-7840 optical isolation barrier. This differential output voltage is proportional to the motor current and can be converted to a single-ended signal by using an op-amp as shown in the recommended application circuit.

You need using it as a "differential" device. Not single input(ok)/output(not ok).
So you need using something as figure 4 in the datasheet.

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