This article covers transimpedance amplifiers and how to stabilize them.

If you'd like to learn more, please check out our article on how to analyze stability in transimpedance amplifiers.

### What Is a Transimpedance Amplifier?

We begin by defining what a transimpedance amplifier is. For context, let's take a look at an example circuit.

*(a) (b)*

**Figure 1**. (a) Basic I-V converter, or transimpedance amplifier (TIA). (b) Practical implementation, showing the stray capacitance C_{n} associated with the op-amp’s inverting input pin.

**Figure 1**. (a) Basic I-V converter, or transimpedance amplifier (TIA). (b) Practical implementation, showing the stray capacitance C

_{n}associated with the op-amp’s inverting input pin.

The circuit of Figure *1(a)* accepts an input current *I _{i}* and converts it to an output voltage

*V*. Aptly called

_{o}*I-V converter*, it finds a variety of applications, two prominent ones being as

*photodiode preamplifier*and as a

*buffer for current-output digital-to-analog converters*(DACs). To find the relationship between

*V*and

_{o}*I*, we use Ohm’s law to write

_{i}*V*

_{o}– V_{n }=

*RI*, and the op-amp law to write

_{i}*V*=

_{o}*a(0 – V*=

_{n})*–aV*, where

_{n}*a*is the op-amp’s open-loop gain. Eliminating

*V*and solving for the ratio

_{n}*V*, we get the

_{o}/I_{i}*closed-loop gain*

#### $$A=\frac{V_{O}}{I_{i}}=\frac{R}{1+\frac{1}{a}}$$

*Equation 1*

*Equation 1*

In the ideal op-amp limit *a→∞*, we have *A → A _{ideal}* =

*R*. Since

*A*has the dimensions of volts/amperes, or ohms, which are the dimensions of impedance,

*A*is aptly called the

*transimpedance gain*, and the circuit is also known as a

*transimpedance amplifier*(TIA).

A real-life TIA, depicted in Figure*1(b)* includes also a stray capacitance *C _{n}*, consisting of the parasitic capacitances (discussed in a previous article on input capacitance in op-amps)

*plus*the parasitic capacitance of the circuit providing

*I*(typically, a photodiode or a current-output DAC). Depending on the application,

_{i}*C*is typically on the order of 10 pF to 100 pF or higher. Regardless, it is a common tenet that

_{n}*C*tends to destabilize the TIA, so it is the task of the designer to take suitable measures to stabilize the circuit.

_{n}

### Destabilization in Stray Capacitance: Rate of Closure

Let us investigate the destabilizing tendency of *C _{n}* using the

*rate-of-closure*(ROC). To this end, we set the input source to zero, break the loop as in Figure

*2(a)*, apply a test voltage

*V*and calculate the

_{t}*feedback factor*

*β(jƒ)*as

*(a) (b)*

**Figure 2**. (a) Finding the feedback factor β(jf). (b) Rate of closure (ROC) approaching 40 dB/dec. (Here a_{0} is the DC gain, ƒ_{b} the bandwidth, and ƒ_{t} the transition frequency).

**Figure 2**. (a) Finding the feedback factor β(jf). (b) Rate of closure (ROC) approaching 40 dB/dec. (Here a

_{0}is the DC gain, ƒ

_{b}the bandwidth, and ƒ

_{t}the transition frequency).

#### $$β(jf)=\frac{V_{n}}{V_{t}}=\frac{\frac{1}{j2πfC_{n}}}{\frac{1}{j2πfC_{n}}+R}=\frac{1}{1+j2πfRC}$$

**Equation 2**

**Equation 2**

This is readily put in the form

#### $$β(jf)=\frac{1}{1+\frac{jf}{f_{p}}}$$

**Equation 3**

**Equation 3**

where

#### $$f_{p}=\frac{1}{2πRC_{n}}$$

**Equation 4**

**Equation 4**

Physically, *C _{n}* and

*R*establish a

*pole frequency*within the feedback loop. Consequently, a signal traveling around the loop will have to contend with

*two poles*, one due to the op-amp and the other due to

*C*, with the risk of a phase shift approaching 180° and thus jeopardizing circuit stability. We can better visualize this in Figure

_{n}*2(b)*, which shows the plots of the open-loop gain |

*a*| and the reciprocal of the feedback factor |

*1/β(jƒ)*|, where

#### $$\frac{1}{β(jf)}=1+\frac{jf}{f_{p}}$$

**Equation 5**

**Equation 5**

The *pole frequency* *ƒ _{p}* of

*β(jƒ)*is a zero frequency of

*1/β(jƒ)*, indicating that the |

*1/β(jƒ)*| curve starts to rise at

*ƒ*. If

_{p}*ƒ*is low enough compared with the

_{p}*crossover frequency*

*ƒ*, the ROC will approach 40 dB/dec, indicating a phase-margin approaching zero.

_{x}A common cure for combating the *phase lag* due to *C _{n}* is to introduce

*phase lead*by means of a feedback capacitance

*C*across

_{ƒ}*R*, as depicted in Figure

*3(a)*.

*(a) (b)*

**Figure 3**. (a) Combating the phase lag due to C_{n} by means of the phase-lead introduced by C_{ƒ}. (b) Imposing ƒ_{z} = ƒ_{p} for a phase margin φ_{m} ≈ 45°.

**Figure 3**. (a) Combating the phase lag due to C

_{n}by means of the phase-lead introduced by C

_{ƒ}. (b) Imposing ƒ

_{z}= ƒ

_{p}for a phase margin φ

_{m}≈ 45°.

Equation (2) still holds, provided we replace *R* with *Z(jƒ)* = *R*||[*1/(j2πƒC _{ƒ})*]. This gives, after some algebraic manipulation,

#### $$\frac{1}{β(jf)}=\frac{1+\frac{jf}{f_{p}}}{1+\frac{jf}{f_{z}}}$$

*Equation 6*

*Equation 6*

where

#### $$f_{p}=\frac{1}{2πR(C_{n}+C_{f})}$$ $$f_{z}=\frac{1}{2πRC_{f}}$$

**Equation 7**

**Equation 7**

Note that *C _{ƒ}* creates a

*zero frequency*

*ƒ*for

_{z}*β(jƒ)*, while also lowering the existing pole frequency

*ƒ*somewhat (recall that a pole/zero for

_{p}*β*becomes a zero/pole for

*1/β*).

An easy-to-visualize technique specifies *C _{ƒ}* so as to position

*ƒ*right at

_{z}*ƒ*, as in Figure

_{x}*3(b)*. This reduces the ROC from about 40 dB/dec to about 30 dB/dec, thus ensuring a phase margin of about 45°. To find the required

*C*, we note from Figure

_{ƒ}*3(b)*that

*ƒ*equals the

_{z}*geometric mean*of

*ƒ*and ft, that is, ƒ

_{p}_{z}=

*(ƒ*. Using the expressions of Equation (7) and simplifying gives

_{p}×ƒ_{t})^{1/2}

#### $$C_{f}=\sqrt{\frac{C_{n}+C_{f}}{2πRf_{t}}}$$

*Equation 8*

*Equation 8*

This transcendental equation is readily solved by iterations, as shown next.

### A Photodiode Preamplifier Example

The circuit of Figure 4 typifies a photodiode preamplifier, such as those used in *light detection and ranging* (LiDAR).

**Figure 4**. Photodiode preamplifier with compensation for a phase margin φ_{m} ≈ 45°.

**Figure 4**. Photodiode preamplifier with compensation for a phase margin φ

_{m}≈ 45°.

Incident light causes the photodiode to draw a small current (up to a few microamperes), which the op-amp then converts to a useable voltage. The op-amp is assumed to have a constant gain-bandwidth product of 10 MHz, and the *total stray input capacitance* (sum of the diode’s junction capacitance and the stray capacitances of the op-amp, circuit components, and circuit traces) is assumed to be 50 nF. The value of *C _{ƒ}* is found via Equation (8). Start out assuming

*C*= 0 and get

_{ƒ}$$C_{f}=[\frac{(50+0)×10^{-12}}{(2π10^{6}×10^{7})}]^{1/2}=0.892pF$$

Iterate as *C _{ƒ}* = [(50 +

**0.892**)×10

^{–12}/(2π10

^{6}×10

^{7})]

^{1/2}=

**0.900 pF**. Another iteration gives again 0.900 pF, so we stop at this value.

Next, let us verify our findings via PSpice. Using the circuit of Figure *5(a)* we obtain the plots of Figure *5(b)*.

*(a) (b)*

**Figure 5**. (a) PSpice circuit to plot |a| and |1/β| for the TIA of Figure 4. (b) The |1/β| curves for the uncompensated (C_{ƒ} = 0) and the compensated (C_{ƒ} = 0.9 pF) cases.

**Figure 5**. (a) PSpice circuit to plot |a| and |1/β| for the TIA of Figure 4. (b) The |1/β| curves for the uncompensated (C

_{ƒ}= 0) and the compensated (C

_{ƒ}= 0.9 pF) cases.

For the uncompensated case we measure *ƒ _{x}* = 178.4 kHz, and the phase angles ph[

*a(jƒ*] ≈ –90° and ph[

_{x})*1/β(jƒ*] ≈ 89.0°, so the phase margin is

_{x})

#### $$φ_{m}=180º+ph[a(jf_{x})]-ph[\frac{1}{β(jf_{x})}]≈180-90-89=1º$$

*Equation 9*

*Equation 9*

indicating an almost oscillatory circuit.

For the compensated case we measure *ƒ _{x}* = 224.8 kHz, and the phase angles ph[

*a(jƒ*] ≈ –90° and ph[

_{x})*1/β(jƒ*] ≈ 37.4°, so the phase margin is now

_{x})*φ*= 180 – 90 – 37.4 = 52.6°, a bit better than the desired

_{m}*φ*= 45°. The above findings are confirmed by the closed-loop transient responses of Figure 6. Without compensation, the circuit gives a slow-decaying oscillation, whereas compensation tames the oscillation dramatically (what a 0.9 pF capacitor can do!).

_{m}

*(a) (b)*

**Figure 6**. (a) PSpice circuit to plot the step response of the TIA of Figure 4. (b) The |1/β| curves for the uncompensated (C_{ƒ} = 0) and compensated (C_{ƒ} = 0.9 pF) cases.

**Figure 6**. (a) PSpice circuit to plot the step response of the TIA of Figure 4. (b) The |1/β| curves for the uncompensated (C

_{ƒ}= 0) and compensated (C

_{ƒ}= 0.9 pF) cases.

The compensated response still exhibits some ringing, and the AC response (shown in Figure 8 below) exhibits some peaking. To eliminate peaking, *φ _{m}* must be raised to 65.5°, and to eliminate ringing it must be raised to 76.3°. (For this, I am referencing my book, Design with Operational Amplifiers and Analog Integrated Circuits, 4th Edition.)

Raising *φ _{m}* above 45° will result in the situation depicted in Figure 7.

**Figure 7**. – Linearized diagram showing the 1/β curve for φ_{m} > 45°.

**Figure 7**. – Linearized diagram showing the 1/β curve for φ

_{m}> 45°.

Using the PSpice circuit of Figure *5(a)*, we find by trial-and-error that the required values of *C _{ƒ}* are as follows:

For *φ _{m}* = 45.0° use

*C*= 0.738 pF and get

_{ƒ}*ƒ*= 209 kHz

_{x}For *φ _{m}* = 60.5° use

*C*= 1.098 pF and get

_{ƒ}*ƒ*= 248 kHz

_{x}For *φ _{m}* = 73.3° use

*C*= 1.606 pF and get

_{ƒ}*ƒ*= 326 kHz

_{x}The corresponding closed-loop responses are shown in Figure 8.

**Figure 8.** – Closed-loop (a) AC responses and (b) transient responses of the circuit of Figure 4.

**Figure 8.**– Closed-loop (a) AC responses and (b) transient responses of the circuit of Figure 4.

As usual, the price for an increased phase margin is a reduced AC bandwidth and a slower transient response.

### TIA Using a Current-Feedback Amplifier (CFA)

The stray inverting-input capacitance has a destabilizing effect also on TIAs based on current-feedback amplifiers (CFAs), as depicted in Figure 9.

*(a) (b)*

**Figure 9**. (a) CFA-based TIA, and (b) compensation for a phase margin of about 45.0°.

**Figure 9**. (a) CFA-based TIA, and (b) compensation for a phase margin of about 45.0°.

This topic is covered in a previous article on dual CFAs and composite amplifiers, where it is shown that the required feedback capacitance for *φ _{m}* ≈ 45.0° is

#### $$C_{f}=\frac{1}{R_{F}}\sqrt{\frac{r_{n}C_{n}}{2πf_{t}}}$$

**Equation 10**

**Equation 10**

### TIA Using a *T*-Network

As discussed in connection with Equation (1), the transconductance gain, in the limit *a →∞*, is *A _{ideal}* =

*R*. There are applications requiring much higher values of

*R*than 1 MΩ, values that may prove physically impractical. A popular trick around this conundrum is to interpose a

*voltage divider*

*R*between the op-amp output and the feedback resistance

_{1}-R_{2}*R*, as depicted in Figure

*10(a)*.

*(a) (b)*

**Figure 10**. (a) TIA with a T-network. (b) The voltage divider reduces the effective transition frequency from ƒ_{t} to ƒ_{t}/(1 + R_{2}/R_{1}).

**Figure 10**. (a) TIA with a T-network. (b) The voltage divider reduces the effective transition frequency from ƒ

_{t}to ƒ

_{t}/(1 + R

_{2}/R

_{1}).

The voltage at the node common to the three resistances is still, ideally, *RI _{i}*. The op-amp then magnifies this voltage according to the gain expression of the

*noninverting configuration*, in this case,

*1*+

*R*/(

_{2}*R*||

*R*), so

_{1}

#### $$A_{ideal}=(1+\frac{R_{2}}{R||R_{1}})R=mR$$

*Equation 11*

*Equation 11*

We are in effect witnessing a resistance multiplication by a factor of

#### $$m=1+\frac{R_{2}}{R_{1}}+\frac{R_{2}}{R}$$

**Equation 12**

**Equation 12**

With the component values shown, *m* = 1 + 9/1 + 9×10^{3}/10^{6} ≈ 10, so we are achieving *A _{ideal}* = 10

^{7}V/A with a physical resistance of only 10

^{6}Ω. As depicted in Figure

*10(b)*, the voltage divider shifts the baseline from 0 dB to +20 dB. Comparison with Figure

*3(b)*reveals that we are now dealing with an effective transition frequency of

*ƒ*/10, or 1 MHz. Equation (8) still holds, provided we use 1 MHz for

_{t}*ƒ*, so

_{t}*C*must be made 10

_{ƒ}^{1/2}times as large. For

*φ*≈ 45° we calculate

_{m}*C*= 0.900×10

_{ƒ}^{1/2}= 2.85 pF.

The more refined value of *C _{ƒ}* = 2.26 pF shown in Figure

*11(a)*is found by trial and error, as usual.

*(a) (b)*

**Figure 11**. Simulating a TIA with a T-network. Using trial-and-error to find C_{ƒ} for φ_{m} = 45°.

**Figure 11**. Simulating a TIA with a T-network. Using trial-and-error to find C

_{ƒ}for φ

_{m}= 45°.

### Practical considerations

The above examples indicate rather small values of *C _{ƒ}*, typically in the range of picofarads or even sub-picofarads. Such small values may prove physically impractical, so we start out with a more practical value, such as

*C*= 10 pF, and then we force the op-amp to drive

_{ƒ}*C*via a voltage divider to scale

_{ƒ}*C*down to the (smaller) desired value. This is depicted in Figure 12 for the case

_{ƒ}*φ*= 45.0°.

_{m}

*(a) (b)*

**Figure 12.** (a) Simulating a TIA with the capacitance divider R1-R2. (b) Plots of |a| and |1/β| after R_{2} has been fine-tuned for φ_{m} = 45°.

**Figure 12.**(a) Simulating a TIA with the capacitance divider R1-R2. (b) Plots of |a| and |1/β| after R

_{2}has been fine-tuned for φ

_{m}= 45°.

As seen earlier, this requires an effective capacitance of 0.738 pF, so we need to impose

$$0.738=\frac{R_{1}}{R_{1}+R_{2}}10$$

Letting *R _{1}* = 1 kΩ, we need

*R*= 12.6 kΩ. Starting out with this value, and then fine-tuning it by trial-and-error to achieve

_{2}*φ*= 45.0°, we end up with the value 11.4 kΩ, as shown in Figure 12. Clearly, the voltage divider provides the additional advantage of capacitance tuning via resistance tuning.

_{m}Figure *12b* reveals also a high-frequency rise of the |*1/β*| curve, but this is inconsequential if we manage to keep it sufficiently above *ƒ _{x}*. We achieve this by imposing

*R*||

_{1}*R*<<

_{2}*R.*

I hope this article has helped you gain a better understanding of how to stabilize transimpedance amplifiers.

If you'd like more articles like these, please let us know what you'd like to learn in the comments below.

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