How do I use a transimpedance amplifier to measure small currents through biased component?

Question

In the simplified example, I have a component (here a 10000 ohm resistor) which I am biasing and trying to measure the current through. In my real application, it is a more complicated component. The problem is that changing its resistance should change the current through the resistor, but the output of the TIA isnt reflecting that. How do I rearrange my parts so that the output voltage reflects the current going through the test resistor?

If I put the test resistor in series with the voltage supply it seems to work ok. I just have to make sure that the voltage bias across my test device is what I think it is

Answer 1

In an inverting voltage amplifier, or a transimpedance amplifier (indeed any amplifier with feedback to the op-amp's inverting input), the op-amp takes control of the potential of its inverting input, and you must not interfere with that. Your first design interferes with that, rather brutally:

^{simulate this circuit – Schematic created using CircuitLab}

The op-amp wants to adjust its output to whatever potential is necessary to equalise its two inputs P and Q. Since \$V_P=0V\$, it's trying to maintain \$V_Q = V_P = 0V\$. With V4 connected to ground (0V) on one side and Q on the other, it is directly controlling \$V_Q\$, and preventing the op-amp from doing that. What you have built is just a comparator, comparing V4 with 0V, and R1 and Rf are disregarded.

A transimpedance amplifier (TIA) is no different, in that you must not interfere with feedback at Q. This would work:

^{simulate this circuit}

Since Q is being held at ground potential (0V) by the op-amp, not you, you must not touch it. What's being measured (amplified) here is input current \$I\$:

$$ V_{OUT} = -I \times R_F $$

It measures current \$I\$ into the TIA via the device under test (DUT, R1 here), but you have no access to current coming out of it at Q, since the op-amp needs control of Q's potential. With the op-amp holding Q at 0V, this circuit cannot be used to measure current through a DUT where both ends may have arbitrary potentials. You are constrained to the condition that the DUT's right end (Q) will always be 0V.

Answer 2

The problem

In the simplified example, I have a component (here a 10000 ohm resistor) which I am biasing and trying to measure the current through.

Your task seems to be reduced to measuring the resistance of the DUT (Device Under Test), or simply put, to create an ohmmeter. Another possibility is that you are asked to measure the IV curve of the DUT, that is, to create a curve tracer. I am not sure which one it is, so I will check out both options.

Basic idea

The relationship between voltage, current, and resistance is given by Ohm's law (I = V/I or V = I.R).

If your task is to make an "ohmmeter", resistance is the input variable. Depending on which variable is the output, we have two ways to solve the problem:

• Keep the voltage constant and measure the current as a function of the resistance (Iout = Vref/Rin).

• Keep the current constant and measure the voltage as a function of the resistance (Vout = Iref.Rin).

If you have to make a curve tracer, resistance is a parameter. Depending on which variable is the input and which the output, we have two more ways to solve the problem:

• Keep the resistance constant and measure the current as a function of the voltage (Iout = Vin/Rdut).

• Keep the resistance constant and measure the voltage as a function of the current (Vout = Iin.Rdut).

All these conversions can be implemented by an inverting amplifier topology. At its core, this configuration comprises two devices in series (cascaded): a voltage/resistance-to-current converter (R1) followed by a current/resistance-to-voltage converter (R2). The operational amplifier serves the crucial role of establishing ideal operating conditions, specifically a virtual ground at the junction between the two converters.

^{simulate this circuit – Schematic created using CircuitLab}

The circuit solutions that implement these ideas are simple and straightforward, but there is a lot of underlying philosophy worth exploring. The best way to do this is to build and examine them step by step using simulations.

Resistance-to-current conversion

Passive resistance-to-current converter

If I put the test resistor in series with the voltage supply it seems to work ok.

This is essentially the simplest 19th-century Ohmic circuit, where we apply a constant voltage Vref across a variable resistor Rin and measure the current Iout flowing through it. Metaphorically speaking, in this way the resistance is as if "converted" into current.

^{simulate this circuit}

By DC sweeping the IV characteristic of this converter (resistor) through a CircuitLab DC Sweep simulation, we see that it is highly nonlinear (hyperbolic), which is undesirable.

Adding an output current-to-voltage converter

However, we want voltage, not current. That is why, we connect another resistor R in series and take the voltage drop across it as an output voltage Vout.

^{simulate this circuit}

Now, another problem arises - the voltage drop across the resistor R is subtracted from the input voltage (resistance R is added to Rin) and the current, accordingly the output voltage, decreases.

Adding a compensating voltage source...

But we are inventive enough to find a remedy - we connect another but negative voltage source Vout in series so that its voltage compensates for the VR drop (travelling the loop, it is added to the input voltage and thus it supports the input source). The result is amazing - the old output voltage becomes zero and the current depends only on Vref and Rin. We use the compensating voltage as a new (grounded and inverted) output.

^{simulate this circuit}

The DC sweep graph is a mirror copy of that below Schematic 1.1 above.

... or equivalent negative resistance

Figuratively speaking, the compensating source plays the role of a negative resistor with resistance -R which annihilates R.

^{simulate this circuit}

Equivalent circuit

The result is the same - a "piece of wire".

^{simulate this circuit}

Op-amp implementation

In your circuit, an op-amp output voltage plays the role of the compensating voltage (the circuit is actually an inverting amplifier).

^{simulate this circuit}

Resistance-to-voltage conversion

Passive resistance-to-voltage converter

With the same success, we can pass a constant current Iref through the variable resistor Rin and measure the voltage Vout across it. Now we can say that the resistance is "converted" into voltage.

^{simulate this circuit}

A pleasant surprise is that the voltage depends linearly on the resistance.

Adding an input voltage-to-current converter

We can make the reference current source with a reference voltage source Vref and a constant resistor R.

^{simulate this circuit}

However, the variable resistor Rin disturbs the operation of this simple current source as R in Schematic 1.2 above.

Adding a compensating voltage source...

The remedy is the same - we connect another voltage source Vout in series to Rin so that its voltage compensates for the VRin drop. The result is the same - the old output voltage becomes zero and the current depends only on Vref and R. The compensating voltage serves as a new (grounded and inverted) output.

^{simulate this circuit}

The output voltage is linear and inverted.

Equivalent circuit

As a result, the right part of the circuit behaves as a short connection.

^{simulate this circuit}

Op-amp implementation

It is the same as Schematic 1.4 but with swapped resistors (the circuit is actually an inverting amplifier). A disadvantage is that Rin is not grounded.

^{simulate this circuit}

The output voltage is linear and inverted.

Voltage-to-current conversion

If your task us to measure the IV curve of a DUT, then you have to vary the voltage across or the current through it and to measure the current through or the voltage across it.

Linear V-to-I DUT

If the DUT is linear (resistor), it can replace either R1 or R2 in the inverting amplifier configuration. Let's first put it in for R1 here.

^{simulate this circuit}

The IV curve is linear (Ohm's law).

Non-linear V-to-I DUT

However, if the DUT is nonlinear, its placement depends on the shape of its I-V curve. If it is current stabilizing, then it should replace R1; otherwise, a conflict between two current sources will occur. As an example, let's examine the output characteristic of a bipolar transistor or a family of characteristics.

^{simulate this circuit}

As you can see, the partial curves have an almost horizontal region.

Current-to-voltage conversion

Linear I-to-V DUT

If the DUT is linear (resistor), it can also replace R2 in the inverting amplifier configuration.

^{simulate this circuit}

The IV curve is the same as the one under Schematic 3.1.1.

Non-linear I-to-V DUT

If the DUT is nonlinear voltage stabilizing, then it should replace R2; otherwise, a conflict between two voltage sources will occur. As an example, let's examine the IV curve of a diode.

^{simulate this circuit}

Now the curve has an almost vertical part.

Answer 3

You somehow mixed everything up in your circuit - so it didn't work.

Take a look at this: R_dut is your device under test, and R_f defines the transimpedance of the amplifier:

Note: the output voltage is inverted.