# How do I get a low impedance current output from this circuit?

I am learning about the VI characteristics of diodes, specifically nonlinear resistor applications. The circuit in this question is used as an example, but the question applies to any high to low impedance current buffer requirement.

I simulated a circuit to "brute force" an approximation to the equation y=(2^(x-1) )/100000 as a way to create a linear voltage to exponential current converter where 0 < x < 5 volts and 0 < y < 1.6 milliAmps. The blue signal represents the triangle voltage source 'x' between 0-5volts, and the red signal represents a current 'y' (0-1.6milliAmps) taken between the triangle voltage source and the rest of the circuit. The non-linear properties of the current are determined by the resistive attenuator circuit. Here is the circuit below working as expected:

Question: How do I buffer the current 'y' (red signal) in order to drive other circuits (such as the bias current pin of an LM13700) or simply convert that non-linear current to gain a non-linear voltage? I tried to add a resistor to ground to convert the non-linear current 'y' to a non-linear voltage along the top path of the circuit, but that removed the non-linear properties of the current. I think that creating a buffered current would solve the problem. I could then connect a resistor to ground and convert the current to voltage.

I've gotten the hang of op amp voltage followers, but don't have a strong understanding yet of current buffer/followers. I tried employing a current buffer using a BJT in common base configuration based on some research googling "current buffer circuit". I don't think I am using it correctly though (see circuit shown below with current buffer added).

If anyone else is interested in V-I characteristics of diodes, I have been following the IIT Madras lecture series on Analog Electronics. This is one of the first videos in the series on diodes: https://youtu.be/qEg3DrBe-3U

• Two problems with your question: 1. When you make a current source, you want it to have high output impedance, not low. Low output impedance means the voltage stays constant even if the current varies a lot. You want the current to stay (nearly) constant even if the voltage of the load varies. Commented Oct 26, 2015 at 0:35
• 2. It's not at all obvious to me where the "red signal" is found in your schematic, and where you'd be able to break your circuit in order to sense the current. When designing a current source circuit we also need to know whether the load is going to be connected from our source to ground, to a voltage supply (high or low) or is floating. Commented Oct 26, 2015 at 0:37

I'm making a guess about what you're trying to do:

simulate this circuit – Schematic created using CircuitLab

You would place your circuit in place of the "Nonlinear Resistor" box, treating node VG as the ground for your circuit (VG is a virtual ground node).

VOUT will have a voltage proportional to the current through your nonlinear resistor.

Alternately, the current through R1 will be equal to the current through the nonlinear resistor, so if you want to drive a floating load, you could just place it in place of R1 (but then the output voltage might not be linearly proportional to the nonlinear resistor current).

If you want to drive a ground or supply-referenced load, you could use VOUT to drive your choice of voltage to current conversion circuit. For example, this is a common configuration:

• Thanks for the response. I edited the question to try and be a little more clear. Looking at your answer I don't think that an inverting amplifier will help me here as it doesn't really buffer the input signal. Also the second circuit, a voltage controlled current sink won't help me either because I would need to start out with a current. Is there a way to buffer the current through my circuit so that I can use it in other ways like voltage conversions or driving other parts of a circuit? In my circuit above, any resistors I add to the network affects that red current signal. Commented Oct 26, 2015 at 14:53

You could mirror the current and feed the mirrored current into a transimpedance amplifier.

This should not load the current you have produced and leaves you free to use the output of the TIA as you wish.