I am working on a problem that can be made really simple if the output of the transistor were to be a fixed power of the input. For ex., if the input is 8 V and power is 0.333, then the output voltage is 2 V.

An opamp circuit would be fine as well. Essentially, the circuit should be minimal (1 transistor or 1 opamp) and the output should be exponentiation.

  • \$\begingroup\$ You can get an exp(x) function by using an op amp and one transistor or one diode (I believe the transistor option is superior, but I can't remember why), along with a few passives. \$\endgroup\$ – Hearth Feb 2 '19 at 16:37
  • \$\begingroup\$ Re-reading the question though, it sounds like you want a polynomial of the input, not an exponential. That's also possible, though complicated. Certainly not a one-transistor or one-opamp job; look up analog multipliers. \$\endgroup\$ – Hearth Feb 2 '19 at 16:39
  • \$\begingroup\$ @Hearth I don't remember the physics explanation, but the lore I've heard is that a diode-connected transistor acts more like an ideal diode than a diode does, being good over several more decades of current. \$\endgroup\$ – TimWescott Feb 2 '19 at 16:46
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    \$\begingroup\$ So are you looking for a one-transistor exponentiation circuit so that you can use it to build a more complicated power-of-whatever circuit? Or are you conflating exponentiation with \$x^3\$? \$\endgroup\$ – TimWescott Feb 2 '19 at 16:49
  • \$\begingroup\$ Also: How do you want it to handle negative inputs? \$\endgroup\$ – Hearth Feb 2 '19 at 16:58

Look up the square root diode function circuit, but modify for higher current maybe using a 3x Darlington emitter follower. This drives an unity inverter Op Amp 1st stage with emitter follower feedback for current gain to drive the 2nd stage inverting square root circuit for non-INV sqrt. sqrt function where \$Vout=k\sqrt{Vin}=k\sqrt{Pd*Rin} \$, where R input resistor accepts the high current to achieve the constant power or quadratic voltage of base-emitters before saturation, where it becomes more linear. Here the intermediate input R receives the current and the op amp Vin+ ref =0 but needs two inv. stages to make a non inverting function from a high impedance source.

If you have a low impedance source input that can drive this current, then use the 1st stage on the output which is needed to provide linear voltage gain to make k * k2=1 since k is limited due to the voltage range of 3 Vbe’s in a triple Darlington from 0 to 1.5Vish volts or maybe use 2 Darlingtons for a 2V range or a linear voltage gain circuit.

 enter image description here Here is one example where labelled R values are equal for k=1 but others must be chosen not to allow output currents to saturate the Op Amp or the Vce=0 which limits the decades of range unless you inject clever offsets that cancel/avoid this.

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    \$\begingroup\$ Got a link? I've seen square root circuits on the same page as squarer circuits; should be some guidance there. \$\endgroup\$ – TimWescott Feb 2 '19 at 17:08
  • \$\begingroup\$ Anyone can search it, the answer is knowing what keywords to search for \$\endgroup\$ – Tony Stewart EE75 Feb 2 '19 at 17:10
  • \$\begingroup\$ What lazy users non-comprende =-1 \$\endgroup\$ – Tony Stewart EE75 Feb 2 '19 at 18:36
  • \$\begingroup\$ For limited accuracy applications, you can simply use a diode, which has an approximate exponential relationship of voltage vs. current. This will only be good over 2 or 3 orders of magnitude, but is is certainly doable if that range is adequate. The transimpedance connection shown is typically good over about 6 orders of magnitude. \$\endgroup\$ – WhatRoughBeast Feb 2 '19 at 18:37
  • \$\begingroup\$ Yes but not good for 1/3W \$\endgroup\$ – Tony Stewart EE75 Feb 2 '19 at 21:03

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