0
\$\begingroup\$

I need a circuit that senses a current and outputs a voltage, but in inverse proportion.

In my case: When the current is around 0A, the voltage source should be around 18V (my Vcc). And when the current is around 8mA, the voltage source should be 5V.

Can this be done ? Opamps ? Transistors ?

It's just for a simulation. No worry about part count. But generic parts are preferred.

Is it even possible ?

All I can do is thanks. No reputation to upvote.

The following is a directly proportional current controlled voltage source. A small illustration of the idea. All I need is to reverse the proportion.

This is a directly proportional current controlled voltage source

Edit 1: To make this thread short, I removed the part where I explain the impedance requirements. This can still be seen in the edit history.

The user Transistor understood correctly my necessities.

Edit 2:

I "ommitted" the following information to make the opening thread short, but that choice is creating gaps:

This whole circuit is a brief and partial model of the LM7805.

This IC has a quiescent current (as the datasheet calls) of 8mA leaving the control pin towards the GND. In this scenario, the regulator does its job and outputs clean 5V.

Now, if I interrupt this control current, the output approaches the Vcc of the supply.

I intend to control this current with a transistor, and this control should take into account the output voltage (so not to break the closed loop).

A voltage divider quite does the job, but the output voltage is much below the input Vcc. I know that a transistor will increase the minimum voltage much above 5V, but that's not a problem.

I don't have available LM317, by the way.

To "generate" this 8mA in the simulator I'll not use a current source, but a Vcc and a resistor.

| improve this question | |
\$\endgroup\$
  • \$\begingroup\$ If your source is a current why are you worried about input impedance? In fact, with high input impedance you'll need a high voltage source. e.g. For 1 MΩ input impedance and 8 mA you're going to need an 8,000 V supply on the source signal. You want a low input impedance. Similarly a high output impedance means that your output voltage will collapse if you attach a load. I think an edit is required. \$\endgroup\$ – Transistor Jul 21 '19 at 22:32
  • \$\begingroup\$ Your requirements are not consistent. How can the circuit be "high impedance" and still have 8mA going through it with 5V across it? \$\endgroup\$ – Elliot Alderson Jul 21 '19 at 22:32
  • \$\begingroup\$ @Elliot, I think the OP means 8 mA in and the signal conditioner produces 5 V out. Similarly with 0 mA in the conditioner gives 18 V out. \$\endgroup\$ – Transistor Jul 21 '19 at 22:33
  • \$\begingroup\$ @Transistor Possibly. But I would really like the OP to clarify the requirements themselves. We could waste a lot of time giving suggestions that are completely off track. \$\endgroup\$ – Elliot Alderson Jul 21 '19 at 22:38
  • \$\begingroup\$ Edited to clarify. I can clarify further if needed. I won't be able to assemble the suggestions today. \$\endgroup\$ – Lucas BS Jul 21 '19 at 23:21
0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Instrumentation amplifier has very big input impedance. It measures the voltage on current sensing resitor (low impedance), then it is amplified G=1+2*Rf/Rg. Bias of 18V is applied, then you have to calculate Rf, RG so that amplified voltage is subrtracted from 18V to get 5V at xyz current.

The output impedance is low, but you can add a series resitor of your choice to match hi-impedance output.

\$V_{out}=18V-R_{sh}\cdot I\cdot (1+\dfrac{2\cdot R_f}{Rg})\$

\$R_{sh}=10; V_{sh}=10\cdot 8mA=80mV\$

\$\dfrac{dV_{out}}{dV_{in}}=\dfrac{18V-5V}{80mV}=162.5\$

\$(1+\dfrac{2\cdot R_f}{Rg})=162.5\$

\$R_f=2.2k\$

\$R_g=\dfrac{2\cdot R_f}{161.5} = 27.2\$

\$R=500\$

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Perfect man. It works. The structure, the values, the behavior... Everything. Thank you. \$\endgroup\$ – Lucas BS Jul 22 '19 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.