Trying to replicate an accelerator pedal using a differential amplifier

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SUMMARY

This discussion focuses on replicating the voltage signals from an accelerator pedal for a 6.0 L Diesel engine using a differential amplifier. The user initially faced issues with voltage discrepancies when implementing a single op-amp configuration. After modifying the schematic to include voltage follower circuits and troubleshooting, they achieved a linear output in simulation but encountered real-world inconsistencies. Suggestions included using rail-to-rail op-amps, adjusting resistor values, and ensuring equal Thevenin resistance for accuracy.

PREREQUISITES
  • Understanding of differential amplifiers and their configurations
  • Familiarity with op-amps, specifically LM358 and rail-to-rail types
  • Basic knowledge of voltage follower circuits and their applications
  • Experience with circuit simulation software for testing designs
NEXT STEPS
  • Research rail-to-rail op-amps suitable for low voltage outputs
  • Learn about Thevenin equivalent resistance in op-amp circuits
  • Investigate the impact of decoupling capacitors on op-amp performance
  • Explore resistor value adjustments to optimize differential amplifier outputs
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Electronics engineers, automotive engineers, and hobbyists working on sensor signal replication and differential amplifier applications.

jacadams
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I'm trying to replicate the voltages coming from an accelerator pedal, to the ECM, on 6.0 L Diesel engine. I acquired the three separate voltages using an oscilloscope. I have access to two analog output channels on the dyno. Out of the three voltages two them follow each other by a difference of 0.6 volts. Going from 1.58 to 4.26 and 1.00 to 3.64, equating to 0 to 100% throttle. I have uploaded the current schematic I'm trying to utilize to achieve this goal. V1 will always be 12v and V2 will be an analog output, going from 1.58 to 4.26. My original schematic was only utilizing one op amp as a differential amplifier, but the V-(pin 2 of U2A) was rising with V+. I was not getting a constant 0.6v difference between the two signals. So I then added two voltage follower circuits, as shown, to buffer the signals going into U2A. If I run the current schematic in my simulation software, everything works out great. I get a liner voltage out at Probe1 rising from 1.00v to 3.64, which is what I'm looking for. When I build the schematic in real life and test it I am getting a more constant voltage difference, but I'm not able to dial it in precisely. After some trouble shooting I found that I'm not getting the same voltage in as I'm getting out on probe 3 and probe 4. Instead of going from .568 v to .568 v it goes .568 to .738v. I've tried surfing for half a day on a solution and I'm not getting anywhere. Any help would be appreciated. Thanks!
 

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what type of op amps are you using? some have outputs that can't get lower than about .7 volts. Look for an op amp that is rail-to-rail, sometimes abbreviated R-R.

Alternatively, you could change the resistors to make the output of U1A and U1B double the value you have now, and then halve the difference by making R5 and R3 to 3k instead of 1.5k.

If your real life part is indeed a LM358 then, according to the datasheet, it should be able to go to as low as .005V and then i don't know what the problem is.

For best accuracy (on the scale of ~1mV, not enough to cause .17V of error), the input pins of the op amps should have the same thevenin equivalent resistance, in this case this would be accomplished by adding a 500Ohm resistor between pin 1 and 2 of U1A.

Don't forget to put decoupling caps on the supply pins of all of the ICs.
 
U2A subtracts the 568 mV from U1A instead of adding it.
Could you go to a +/- 6 volt power supply instead of a +12 volt power supply?
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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