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THE EQUATIONI have a circuit which has an exponential function as shown in the figure. I tried to put an anti-log opamp with the diode and BJT transistor, but did not get what I wanted. enter image description here Schematic

I think the whole problem is in the "exp" part.

THE INTITIAL VALUES ARE a=2 , b=4 & c=6

Desire outputs enter image description here enter image description here

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  • \$\begingroup\$ Is this your circuit? Are you sure it stands for exponential? It doesn't make sense given the context... \$\endgroup\$ – Eugene Sh. Nov 18 '19 at 20:49
  • \$\begingroup\$ the equation from which it is related is 'e' a natural log. \$\endgroup\$ – zweo Nov 18 '19 at 20:50
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    \$\begingroup\$ I mean the rest of the circuit is discrete elements, while "exp" is a totally different level of abstraction. \$\endgroup\$ – Eugene Sh. Nov 18 '19 at 20:51
  • \$\begingroup\$ yes it is..any idea \$\endgroup\$ – zweo Nov 18 '19 at 20:52
  • \$\begingroup\$ by the way this is not the whole circuit i have attached a part of it because of the fact that the other parts of the circuit are loud and clear except the ''exp'' thing... \$\endgroup\$ – zweo Nov 18 '19 at 20:57
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Your state transition equations have no input, so the output functions depend only on the initial values of the state variables and the parameters a, b and c. Of course, some of parameters a,b and c can be considered as inputs.

Opamps have several practical limitations. Their frequency range is limited, because transistors are slow, the spectrum of the input signals should stay low enough for acceptable gain. There are more slowness phenomenas. The changing rate (volts/second) of the outout voltage is limited to "slew rate" which is shown in the datasheet. Integrators charge and discharge capacitors, so output current limits are also essential limitations. Offset voltage make integrators drift away to the max or min output voltage sooner or later, offset voltage is like a constants DC input for integrators.

You can find easily antilog amp schematics which do not expect any certain transistor type. This link https://voer.edu.vn/m/operational-amplifiers/56a79505 contains one with quite plausible DC analysis. That 2 opamp+2 transistor circuit probably is the same you have already seen. It assumes a pair of identical transistor. Vt =kT/q = the thermal voltage unavoidably brings in the temperature. It must be stable for consistent operation. But the reverse saturation current of the PN junction is eliminated from the formula.

The shown antilog amp works properly only in limited input voltage range because - the exponential exp(V/Vt) in Shockley's diode current law is assumed to be much larger than one (=V is much more than 26 mV, for ex. V>125 mV) - non-ideal leaks must be neglible when compared to diode current given by Shockley's law - resistive voltage drop in diode materials is neglible when compared to voltage obeying Schockley's law

You must store the numerically calculated state variable time functions and check if integrators can really output them in terms of voltage, opamp frequency range, opamp slew rate and max opamp output current.

You should check is the drift of the integrators really neglible during the run period

You should make circuit simulation can your antilog amps really handle the whole swing range of Y and Z.

Then something beyond my math capabilities: This is a nonlinear system. A chaotic behaviour is possible.

Me and surely some others are curious what practical functionality you actually want to reach. As said, your equations show no input, so a single run in some calculation system, for ex. in Excel or Matlab should be enough if the initial values are known.

ADD: I made a couple of numerical test runs to see something of the behaviour of your system. My parameters:

  • run time t=0...100, time unit = 1 second for convenience
  • sample rates 1kHz and 10kHz were both tried, no radical differences in 100 s run period
  • integration = trapezoidal

A plot of the state variables x,y,z and w:

enter image description here

NOTE: variable value transfer blocks were used to avoid too messy wiring

The problematic variables (due the exponentation) y and z vary widely being as well positive and negative. Making the same calculation with an analog circuit with 1 volt per numerical one needs exp(A) block which can produce hundreds of volts outputs. We can of course scale down say to 1% to keep everything below 10V, but then the smallest output would be well below one microvolt.

Antilog amps cannot handle bipolar input, something complex or clever is needed to either find equivalent equations which do not need negative input exp(A) or to design a bipolar input exponentation circuit. I do not know any simple such circuit.

Adding an offset to input A of exp(A) to and compensating it by attenuating the output is out of the question because the output is already high. But it's well possible that you can clip the inputs of the exponentials to some realizable range because the effects of the near zero exp values can be neglible for the integration. I haven't done this check, so it 's up to you to do it.

Then there's a need to show numerically that the result do not get spoiled if there's errors (noise, offset and scale errors) Without knowing the effect of the errors it's useless to try to make an analog circuit.

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  • \$\begingroup\$ seriously u r good .. but i m not . m still thinking what m i going to do. \$\endgroup\$ – zweo Nov 20 '19 at 16:33
  • \$\begingroup\$ hy. i have updated the full question with desire outputs along with the values of a b and c as mentioned by u .. no can u tell me how i am i going to set "exp" correctly.. \$\endgroup\$ – zweo Nov 30 '19 at 8:55
  • \$\begingroup\$ @zweo Now you should dig out how much error is allowed to your EXP function in this application. Let's assume your forthcoming circuit will calculate instead of EXP(t) function (1+K)(EXP(t))+J where K and J are scale and offset errors. Try to find numerically how far K and J can be from zero without making the circuit useless. \$\endgroup\$ – user287001 Dec 4 '19 at 20:46
  • \$\begingroup\$ i dont understand what u say? i dont know about K and J reference..! \$\endgroup\$ – zweo Dec 5 '19 at 20:19
  • \$\begingroup\$ multisim.com/help/components/abm-voltage/… @user287001 above link is for multisim mathematical functions. I found this link which i need to know how m i going to use this in multisim . \$\endgroup\$ – zweo Dec 5 '19 at 20:22

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