Relationship between gain and feedback resistance R in transimpedance amplifier

魔双月壁

Relationship between gain and feedback resistance R in transimpedance amplifier [Copy link]

 

As shown in the figure, V0=Ip*R. Is the gain calculated like this? G=20lgR? I feel that the resistor R is the amplification factor, so it should be correct to calculate it this way. According to the gain-bandwidth product calculation GBP=BW*G, the resistor R=200K can be calculated as G=106 BW=15.09MHZ. However, it is a bit confusing to understand that BW=275MHZ and G=10 in this manual. In addition, my input pulse is a 2KHZ, and the pulse width is only about 150ns. Is the frequency calculated according to the frequency at 150ns or 2KHZ? Please help, looking forward to the answer, thank you.

maychang

"I don't quite understand the manual's statement BW=275MHZ and G=10." G=10 refers to the test conditions, i.e. 275MHz is measured under G=10.

maychang

"My input pulse is a narrow pulse of 2KHZ and the pulse width is only about 150ns. So should the frequency be calculated according to the frequency at 150ns or 2KHZ?" Neither. The pulse repetition frequency is 2kHz and the width is 150ns. Such a pulse contains abundant high-order harmonics (it is very likely that the harmonic content is higher than the fundamental wave). In order to amplify the pulse without distortion, the bandwidth of the amplifier needs to be infinite. In fact, of course, there is no amplifier with infinite bandwidth, but if the output waveform is close to the input waveform and the waveform is still "acceptable", the amplifier bandwidth must be at least dozens or even hundreds of times the reciprocal of 150ns.

maychang

"I feel like this resistance R is the amplification factor, so it should be correct to calculate it this way." It's true. Otherwise, how can it be called transimpedance amplification?

gmchen

The bandwidth calculation of the original poster is wrong. In the formula GBP=BW*G, G must be a dimensionless ratio. However, the gain of the transimpedance amplifier is calculated according to the ratio of the output voltage to the input current, so its gain is a value with resistance dimension. For this value with resistance dimension, it is not suitable to use decibel as the unit of measurement, so the algorithm G=20lgR is incorrect. The manual says BW=275MHz@G=10, which refers to the -3dB bandwidth when the gain is 10 times (note that it is not 10dB) as a voltage amplifier. This value is the best value obtained through experiments. Note that this value does not even satisfy GBP=BW*G. This is because the OPA657 op amp is not actually fully compensated, so when the closed-loop gain is small, its closed-loop frequency characteristics have overshoot at the high-frequency end. This overshoot causes the closed-loop BW>GBP/G. In the transimpedance amplifier, the distributed capacitance of the signal source, the feedback network, and the frequency response of the op amp itself affect the bandwidth. The following formula should be used to estimate the closed-loop bandwidth of the transimpedance amplifier: BW=sqrt[GBP/(2*pi*Rf*Cs)], where Rf is the feedback resistor and Cs is the distributed capacitance of the signal source.

xiaxingxing

maychang posted on 2018-4-12 15:56 "I feel that this resistance R is the amplification factor, so it should be correct to calculate it this way." It's not wrong. Otherwise, how can it be called transimpedance amplification?

Hello, isn't the amplification factor of this transimpedance amplifier the impedance value after R and C are connected in parallel? ? ? Please guide me, thank you!

maychang

xiaxingxing posted on 2018-4-15 17:52 Hello, isn't the gain of this transimpedance amplifier the impedance value after R and C are connected in parallel? ? ? Please guide me, thank you!

Isn't the gain of this transimpedance amplifier the impedance value after R and C are connected in parallel? Yes.

魔双月壁

gmchen posted on 2018-4-14 09:13 The bandwidth calculation of the original poster is wrong. In the formula GBP=BW*G, G must be a dimensionless ratio. But the gain of the transimpedance amplifier is calculated according to the input...

Thank you, I will try to calculate it. I am using OPA657 to amplify the photocurrent in laser ranging. Now I encounter the same distance light from different angles hitting the APD and find that the rising edge changes greatly. I wonder if it is a problem of bandwidth and gain, so I ask the question.

fickle

The so-called transimpendance amplifier is to convert the current signal into a voltage signal. The current flowing through the diode is very small, so a resistor (200k) is needed to amplify it and obtain a measurable voltage signal. The key to selecting an op amp is low bias current.

魔双月壁

Take the total capacitance as 45pf Rf=200K According to formula 2, we can calculate f=168.2K. I would like to ask if the frequency f here refers to the frequency of the sine wave, which is 168.2K? Thank you

gmchen

All references to cutoff frequency refer to the frequency of a sine wave. If the signal is not a sine wave, its decomposed spectrum must be considered.

魔双月壁

gmchen posted on 2018-4-20 15:42 All the references to cutoff frequency refer to the frequency of a sine wave. If the signal is not a sine wave, its decomposed spectrum must be considered

Does decomposition refer to Fourier expansion?

gmchen

Yes

xbhero

Posted by Mo Shuang Yue Bi on 2018-4-17 16:09 Thank you. I will try to calculate it. I am using OPA657 to amplify the photocurrent when doing laser ranging. Now I encounter the same distance light from different angles hitting A...

Was it finally resolved?