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I am trying to design a programmable electronic load for testing the PWM outputs of PLC controllers. The outputs operate between 5-36 V and 10-2000 Hz frequency. The output is supposed to tested for upto 300 mA current. Based on a current sink circuit from the Internet, we designed this particular circuit.

Electronic Load Schematic

  • V3 is the µC/RPi Output that we plan to use to control the current output at sense resistor R2.

  • VOutput1 is the PWM signal that is going to be generated by our PLCs.

Now I have a few questions regarding this circuitry.

  1. When i try to simulate this for lower values of current between 1 mA to 150 mA i always have certain overshoots for a few µSecs before it settles to a square wave signal. Why is it so? Here is the result for 20mA. enter image description here

  2. After reading few articles on the internet i figured out that it is the gate capacitance that could be causing this problem. In the initial stages of circuit designing i ended up using up mosfets with very small Ciss ( 8pf to 55pf on diff SMD Mosfets). On simulation they gave perfect results.However since i expect a power dissipation of upto 10 W at the Mosfet i have no option than to use a bigger Mosfet but then they have large Ciss which brings the Overshoots with them. Can anyone suggest a solution to this ? Could the OP-Amp controlling the gate voltage be blamed for this phenomenon?

  3. Instead of the LTC6752 i had previously used a normal OP Amp AD8031 as a comparator at U3. Now this particular change i did based on recommendation of a senior. He advised me to keep a normal OpAmp and use the NPN-PNP(BC547B-BC557C) transistor as switch.He said this is because the op amp which i used previously (LM239) was an open collector comparator. And for this particular purpose we need to use an Open Emitter Comparator for which i couldn't find a spice model and hence this solution. Now i am guessing LTC6752 is also an open collector comparator.Because when i replaced AD8031 with this model i didn't find any change in the output.Can anyone explain the reason for using this open emitter configuration here and is it recommended to use any op amp instead of LTC6752?

  4. I was looking at the simulation results and found out that the Feedbck signal goes shortly to 0V and then comes back to µOut value. This short duration when it goes to 0V and comes back to original value causes this Overshoot.The reason we used the Pnp-Npn-Comparator part in the circuit was to prevent AD823A to generate output when PWM signal is at 0V. Now i feel this is what is one of the cause to generate the Overshoots. Is there any way i could limit the feedbck to drop only to the value equivalent to µCout? This way the AD832A won't be generating overshoot before settling as a uniform square wave. I guess using a comparator may solve a problem but i am not sure how.

I am doing this project as a final year project at a company for my undergrad degree.I am not so well versed with electronics and hence the questions may seem naive, but i am hoping to get some pretty good responses that could take my work progress forward. Thank you very much in advance.

I will keep posting updates if i am able to come up with anything at work. :)

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  • \$\begingroup\$ Does increasing R2 for small currents help? \$\endgroup\$ – Janka Aug 30 '18 at 13:23
  • \$\begingroup\$ Hi there is a limit to increasing the value R2. Since µC can supply at max 3.3 V and we plan to control the circuit at that range going above 10 Ohms seems unlikely. But i have also observed that changing R2 does have an impact but i am not sure how to overcome the limitation of 3.3V \$\endgroup\$ – Sajeev Pillai Aug 30 '18 at 13:25
  • \$\begingroup\$ This isn't meant as a solution. You only need it for testing at very small currents, using 100Ω instead of 10Ω will scale the voltage measured. The problem I see is the amplifier feedback runs through a low-pass made up from R1 and C1. If the difference between the + and - inputs of U2 is very small (small current through R2), this will slow down the feedback, resulting in that overshoot. \$\endgroup\$ – Janka Aug 30 '18 at 13:27
  • \$\begingroup\$ Can't help you with your problem, just a remark to your circuit: If you have a supply (10 V) which is used to power everything, it might be worth to only place one voltage source and name the net "Vsupply" or something and then use the "name net" tool and name all the + nodes to "Vsupply". If you change the supply voltage, you only have to do it in one location and not 5 different ones. \$\endgroup\$ – Arsenal Aug 30 '18 at 13:31
  • \$\begingroup\$ @Arsenal thanks for the suggestion. Will keep in mind next time. \$\endgroup\$ – Sajeev Pillai Aug 30 '18 at 13:34
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This is a suboptimal circuit for the application. You shouldn't be trying to switch the load transistor on and off in time with the pulses — that's just silly.

Instead, you should be using a precision peak-detector circuit on the feedback from the sense resistor, so that the circuit only attempts to regulate the peak current. This eliminates all transient effects inside the load control circuit.

In other words:

  • Eliminate that entire upper circuit that runs from PlcOut to Feedbk
  • Replace R9 with the peak detector.
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  • \$\begingroup\$ Thanks for the edit and for this link. I will read up and try to make any changes if possible \$\endgroup\$ – Sajeev Pillai Aug 30 '18 at 14:06
  • \$\begingroup\$ Hi @Dave. As per your recommendation, i tried implementing a peak detector in place of R9. After reading the link and some more reading off the internet, i used a diode and a capacitor in place of R9. It seems i have used a very primitive detector circuit and the results are not satisfying. Do you have any suggestions ? Could you post an image or something as suggestion. I am having a bit of a hard time trying to constituting an op amp based peak detector in the feedback loop. Thank you very much :) \$\endgroup\$ – Sajeev Pillai Sep 3 '18 at 13:33
  • \$\begingroup\$ By adding an opamp and putting the diode in the feedback loop as shown in my original link above, you eliminate the drawbacks of the simple diode-capacitor setup. Be sure to use an opamp with rail-to-rail inputs and outputs if you're trying to operate from a single supply. \$\endgroup\$ – Dave Tweed Sep 4 '18 at 13:13

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