What is a Transimpedance Amplifier?
A Transimpedance Amplifier, or TIA for short, is an electronic amplifier circuit who’s job is to convert an input current source into a proportional output voltage. That is the transimpedance amplifier operates as a current-to-voltage converter (I-V converter).
The previous operational amplifier circuits have all used voltage as their primary input signal. But many active (self-generating) sensors like photodiodes (in photovoltaic mode) generate a current, not a voltage, in response to some external or environmental change.
Then we need to be able to convert this self-generated current into a voltage and the transimpedance amplifier allows us to do just that.
Converting Current Into A Voltage Drop
Ohm’s Law tells us that when an electric current (I) flows through a simple fixed resistor (R), a voltage (V) is developed or dropped across it with Ohm’s law taking the simple form of: V = I x R.
Because this Ohm’s Law relationship is linear, the voltage developed across the resistor is directly proportional and linear to the current flowing through it as shown by this 10Ω graph.
Therefore if we change the value of the current, the I*R voltage drop across the resistor proportionally changes by the exact same factor.
Note that this strict proportionality only holds true as long as the resistance, (R) remains constant and is actively flowing. Thus resistors are Ohmic devices.
However, the problem with different Types of Resistors is that they can quickly heat up. This is due to their I2R power loss if too much current passes through them causing them to change their resistance. Then the voltage drop will begin to change and the I-V relationship will lose its property of proportionality and therefore become non-linear.
Thus, using a single fixed resistor as our current-to-voltage device, can create various issues. Since if a large value resistor is used it presents a large input impedance to a particular circuit when we really want our current-to-voltage converter to have a low (near zero) input impedance for maximum signal transfer.
One way around this heat loss problem is to use an Inverting Operational Amplifier which allows us to control both the input impedance and output impedance of the circuit creating a much improved current-to-voltage converter.
The Transimpedance Amplifier
Transimpedance Amplifiers allows the current at one place in the circuit to become a voltage source elsewhere in the same circuit, rather than having the voltage drop affect of a resistor. This makes the amplifier circuit useful in converting a current to a voltage drop.
But how does a transimpedance amplifier work, and why do we need to build an amplifier circuit to accomplish the same thing as a fixed resistor?
From Inverting Amp to Transimpedance Amp
The basic op-amp transimpedance amplifier can be constructed using an inverting operational amplifier. When used in a closed-loop configuration with negative-feedback, its current to voltage gain is based on the amount of resistive feedback.
Then the basic op-amp transimpedance amplifier has a current source connected to the op-amp’s inverting (–) input with a feedback resistor Rf between its inverting input and its output as shown:
Basic Transimpedance Amplifier Circuit
So how does the circuit work? The analysis of this transimpedance amplifier circuit is similar to that for the previous inverting op-amp amplifier. The difference here is that we have removed the input resistor RIN and the input voltage VIN and replaced them with a current source IIN. Since we are not using the positive non-inverting input this is connected to a common ground or zero.
Remember that when dealing with operational amplifiers there are two very important rules. These are: “No current flows into either input terminal”, and that “V- always equals V+”. This is because the junction of the input and feedback signal is at the same voltage potential as the positive (+) non-inverting input producing a “Virtual Earth” condition.
Since the inverting input of the op-amp is a virtual earth summing point, the voltage at this point (V-) must be the same as V+, that is zero. Thus any currents flowing into this virtual earth point must sum to zero. So the input current to the op-amp is zero. That is:
IIN + IR = 0
Since the inverting op-amp both amplifies and inverts its input signal by 180o. This inversion means that a positive input signal will produce a negative output and vice-versa. Therefore:
IIN = –IR
As input V- is zero due to the virtual earth summing point, current IR can be defined as:
–IR = Vout ÷ R
Then we can correctly say that:
IIN = –IR = Vout ÷ R
Rearranging above produces the following transimpedance amplifier formula.
Transimpedance Amplifier Formula
Vout = –IIN × R
Note that the output voltage (Vout) is directly proportional to the input current (IIN. Thus an ideal transimpedance amplifier is a current-controlled voltage source with an infinite transimpedance gain (-R) and is sometimes referred to as a Current Feedback Op-amp Circuit or simply a Current-to-Voltage Converter.
Remember that the minus sign indicates that the input current flows in the direction of the arrow producing a negative output voltage. Thus the gain is -R and is negative.
Choosing a Suitable Feedback Resistor
As stated above, the transimpedance amplifier is a “current-to-voltage converter”, that consists of a shunt feedback across a high-gain voltage amplifier. As such it has a transfer ratio of: A = Vout/Iin. That is, it has the dimension of V/I or Resistance, which is normally expressed in V/A, or V/mA.
So when designing a transimpedance amplifier circuit we have only one component value to consider. The feedback resistor, Rƒ itself. Then as we can see, the amplifiers transimpedance gain is simply the value of Rƒ. Note that sometime the transimpedance gain is presented as “–R” showing negative resistance because it uses an inverting amplifier.
So how do we determine the value of the feedback resistor? Considering DC conditions, if Rƒ is too large, then the input current signal can saturate the op-amp’s output at either its positive or negative supply rail limits, causing clipping of the output signal.
Likewise, if R– is too small, then the output voltage signal may also be too small to be useful and therefore unable to detect small changes or variations in input current. So we need to make the feedback resistor, Rƒ sufficiently large enough to be able to detect small changes in input current, Iin without causing Vout to saturate.
Transimpedance Amplifier Worked Example No1
Let us assume we have an operational amplifier powered from a ±10 volt supply, and we want to measure input currents up to ±100 uA. Then the maximum value for the feedback resistor will be:
R = V ÷ I = 10 ÷ (100 x 10-6) = 100kΩ or 100V/A
That is: Rƒ = 100kΩ. Therefore if we have an input current, Iin or say, 50uA, then the output voltage will be:
V = I x R = 50 x 10-6 x 100000 = 5 volts
Remember that in this simple example, the op-amps output will saturate if the input current exceeds 100uA.
Photodiodes Convert Light into Electrical Current
One very common application of the transimpedance amplifier is to use a photodiode as its current source for use in optoelectronic circuits.
Typical Silicon Photodiode
Photodiodes are solid-state semiconductor devices that convert light energy into electrical current through the photovoltaic effect.
The basic photodiode consists of a p-n junction formed by doping silicon in the same way as for signal diodes and bipolar transistors.
When photons of light with sufficient energy strike the semiconductor p-n junction, a photocurrent proportional to the incident light intensity is generated.
Then we can use photodiodes with our high-input transimpedance amplifier to convert the small photocurrent of light into a usable output voltage signal which scales linearly with the light intensity for use in light meters or optical encoders in positional sensing applications.
A Photodiode Transimpedance Amplifier
As we can see, the anode terminal of the photodiode is connected to the inverting input of the op-amp. That is, the photodiode is connected directly across both the inverting and non-inverting inputs of the op-amp (with non-inverting terminal grounded).
The photodiode current, IPD has the same value as the current flowing through the feedback resistor, Rƒ. This resistive feedback forces the amplifier to convert the diodes small photocurrent without there being any other external voltage being present at its input.
The amplifiers gain measures the diodes photocurrent, IPD and converts it into an output voltage per microampere input, (V/mA) equal to the diode current times the feedback resistance, Rƒ. Thus as before, the gain of the photodiode amplifier is determined by the resistive value of the feedback resistor Rƒ. That is: A = Rƒ.
The magnitude of the amplifiers gain can also be thought of as its “sensitivity of conversion” because it gives an amount of voltage output change for a given input current change. For instance, for a sensitivity of 1 V/mA we may need Rƒ = 1kΩ. While a sensitivity of 1 V/µA we may need: R = 1MΩ, and so on.
Then selecting the correct feedback resistance can be a compromise between sensitivity and output voltage range. So in order to use our transimpedance amplifier over a larger range of input photocurrents. We can use switches to select different resistors for Rƒ to achieve different levels of transimpedance as shown.
Adjustable Photodiode Transimpedance Amplifier
The example transimpedance circuit offers a wide range of sensitivities depending upon the position of the 4-way rotary switch ranging from 100Ω for position 1 to 100kΩ for position 4. Clearly then for a given photocurrent, switch position 4 offers one thousand times the amplification of switch position 1.
We could take this photodiode transimpedance amplifier circuit one step further by replacing the rotary switch and fixed value resistors, R1 to R4 with a 100kΩ potentiometer giving us a fully adjustable gain stage to change light sensitivity.
Transimpedance Amplifier Worked Example No2
Suppose we have a silicon PIN photodiode that passes a maximum reverse current of 5 mA when fully illuminated across a spectral range of 350 nm to 1100 nm. What is the maximum value for the feedback resistor if the amplifier is fed from a single 5V supply.
Rƒ = Vout (max) ÷ Iphoto (max) = 5 ÷ 0.005 = 1000 Ω (1 kΩ)
Thus a 1kΩ resistor (or potentiometer) is the maximum value of the transimpedance gain we can use to prevent the output from saturating.
Transimpedance Amplifier Tutorial Summary
We have seen here that a Transimpedance Amplifier (TIA) is a current-to-voltage converter built around a high-impedance inverting operational amplifier. Thus a transimpedance amplifier can be used to convert a sensor’s current-controlled output, I into a usable voltage, V.
This output voltage Vout is proportional to the current Iin generated at the input with the transimpedance open-loop gain determined by: Vout/Iin expressed in volts per ampere or ohms (kΩ).
The relationship between Vout and Iin and therefore its conversion gain can be accurately controlled using a feedback resistor, Rƒ with typical values ranging between thousands of ohms (kΩ) to tens of mega-ohms (MΩ).
The transimpedance amplifier circuit finds many applications in optoelectronics since it can be used to measure the very small current flowing through a silicon photodiode and convert it into a useable output voltage. Phototransistors can also provide current amplification of the detected light signals but require biasing first.
However, despite the above circuits simplicity, the current-to-voltage conversion of the transimpedance amplifier acting as a photodiode amplifier, can restrict its ac bandwidth due to the in-built parasitic capacitance across the photodiodes p-n junction.
Because any voltage developed across the photodiode reacts with its internal junction capacitance, shunting away part of the diodes photocurrent when operated at high switching frequencies. Thereby reducing its upper bandwidth.
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Credit- Basic Electronics Tutorials. Distributed by Department of EEE, ADBU.
Curated by Jesif Ahmed.
