I am a digital design engineer and am struggling to come up with an analog circuit that could produce an (RC) curve like below. It does not have to be an exact RC curve. Something that comes close would do perfectly fine.

enter image description here



  • \$\begingroup\$ No problem. Did you want to put some XY values on that? And. What happens after this graph ? \$\endgroup\$ – Tony Stewart EE75 Nov 26 '19 at 5:11
  • \$\begingroup\$ One BJT and no opamp and no bipolar supply trivially produces this. Of course, the questions may be what the units are on each axis. ;) \$\endgroup\$ – jonk Nov 26 '19 at 5:37

This is a dangerous kind of want, as there has to be some sort of cutoff, you can't increase without bound (at least for long).

As far as replicating that behaviour, a real pole on the right side of the s-plane will have an step response of that nature. As for getting a pole there, it shouldn't be too hard, I don't think I've ever had a circuit naturally stable before compensating :P




L = tf(1, [1 -10]) % real pole on right half plane
xlim([0 .5])   

As for another method (I'd recommend this one I think), the voltage to current relationship of a diode is exponential, meaning if you increase the forward voltage applied, the current increases proportional to a*e^(V/b), where a and b are relatively constant (both a and b will vary with temperature, something a high current can cause to rise, so be careful with power dissipation!). Look up diode current equation for the full picture.

enter image description here

Convert this current to voltage (transimpedance amp) and you will have an exponentially growing voltage in response to a linear increase in input.



If you want to cut out some of that long initial tail, just start sweeping the diode voltage at around 200mV.

  • \$\begingroup\$ The combination of a diode and transimpedance amplifier forms the well-known circuit of an "anti-log converter". Figuratively speaking, the role of the op-amp here (and in all the op-amp inverting circuits) is to compensate the voltage drop VR = I.R across the resistor with voltage VOUT = -VR to obtain zero voltage (virtual ground). Even more figuratively speaking is to think of the op-amp output as of a "negative resistor -R that neutralizes the "positive" resistance R. The result is zero resistance "seen" by the diode (short connection). \$\endgroup\$ – Circuit fantasist Nov 26 '19 at 6:37

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