I am quite new to op-amps but have looked at them in college and have done my own research on them. I am trying to use an op-amp to build a simple band-pass filter which has an in-band voltage gain of 5. The main goal is to have a signal which is received from an ultrasonic transducer enter this filter. The signal will be a sine-wave in the order of a couple hundred millivolts around 1.1MHz frequency. I want the band-pass filter to limit the noise that passes through. I also want the in-band gain to be around 5.

Let me go through what I have tried already. I have been looking at Texas Instrument’s TDC1000 which is an “Ultrasonic Sensing Analog Front End”.

I have been looking in particular at the Low Noise Amplifier in the receiver path. I am trying to replicate the circuit seen in Figure 16 on page 14:

LNA Circuit in TDC1000

According to the datasheet, the in-band gain is set by the capacitor ratio:

Gain_in-band = CIN/CF = 300pF/30pF = 10

The high-pass corner is set by the feedback resistor and capacitor:

FCH = 1/(2πRFCF) = 1/2π(9000)(30x10^-12) = 589.5kHz

The low-pass corner is set by the Gain Bandwidth Product and the gain:

FCL = GBP/Gain = 50MHz/10 = 5MHz

This all seemed to make sense, however, other websites showed different circuit configurations for band-pass filters. Some circuits included another resistor before CIN and there was no mention of the "Gain-Bandwidth Product".

I decided to build my own version of the Low Noise Amplifier circuit that was presented in the TDC1000. Picking out an op-amp was tricky enough but after doing some research, the LT1128 seemed like a good choice. The datasheet for this op-amp can be accessed below:


It is a low noise precision high speed op-amp with a Gain-Bandwidth product of 20MHz. I built the following circuit on a breadboard:

My LNA Circuit Built on Breadboard

Following the previous equations:

Gain_in-band = CIN/CF = 100pF/20pF = 5

High-pass corner:

FCH = 1/(2πRFCF) = 1/2π(10000)(20x10^-12) = 795.8kHz

Low-pass corner:

FCL = GBP/Gain = 20MHz/5 = 4MHz

To test out the circuit, I applied a 500mV peak-to-peak sine-wave from my signal generator and varied the frequency. The circuit was not operating as I had expected. Below are a few screenshots from my oscilloscope. The yellow channel is the input signal, and the blue channel is the output of the LT1128.


800kHz Input


1MHz Input

I am not too sure what is going on, changing the amplitude of the input signal did not change the amplitude of the output. However, varying the frequency of the input signal had an affect on the phase and amplitude of the output signal. What I had expected to happen was that within the frequencies from 795.8kHz to 4MHz the output signal would be inverted and 5 times greater than the input.

I decided to take a step back and just build an inverting amplifier with a gain of 4.7. Below is a schematic of the simple circuit that I built on a breadboard:

Inverting Amplifier Circuit

I applied a 1V peak-to-peak sine-wave and varied the frequency. At 300kHz the output signal is as expected, it is inverted and has a peak-to-peak voltage of 4.7V. Below is a screenshot from my oscilloscope, yellow channel is the input signal and the blue channel is the output:

Inverting Amplifier - Input 300kHz

As I increase the frequency, the phase and amplitude of the output signal changes and the amplitude of the input signal no longer has an effect on the output.


Inverting Amplifier - Input 700kHz

I am confused as to why this happens, I thought that the op-amp should continue to amplify the input signal at much higher frequencies.

I have done lots of research on the bandwidths of op-amps, but it seems that I am missing something.


  1. Is the op-amp itself that I picked not right for this application, is it one of the parameters of the LT1128 that I may have overlooked?
  2. In regards to the band-pass filter circuit, can anyone spot any mistakes with the schematic or can explain what is happening?
  3. In the inverting amplifier configuration, why does the output signal start to change above an input signal of 300kHz?

If it’s the fact that I need to do more reading, does anyone have any recommendations?

  • \$\begingroup\$ What filter BW min is needed? \$\endgroup\$ Commented Jul 29, 2021 at 16:31
  • \$\begingroup\$ What is the source impedance? \$\endgroup\$ Commented Jul 29, 2021 at 16:45

2 Answers 2


For an Op Amp to have low gain error at 1MHz you need about an open loop gain of >100 x the closed loop gain or 500 MHz= GBW.

A video amp would be better suited using discrete components.

Not knowing your source impedance is assumed to be 50 Ohms. Buffer with Emitter follower if higher or if lower adjust R according as Rb(eq)/Rin controls the closed loop gain of 5.

Proof of concept for a PN2222A

enter image description here

The reason why Op Amps in general are not well suited for amplifying in 1MHz band bandpass filters is the GBW product needed to overcome the Error in gain is also multiplied by Q^2.

Theory of Operation.

This is my classic common emitter 50 Ohm amp using negative feedback. Gains are traded-off to improve linearity and increase bandwidth. Base pull down adjusts collector Q point for low gain. In other situations if Vc is too high (low Ic) then a bigger pull-up R base to Vcc is used to increase Ic and thus reduce Rbe and increase gain.

There are better transistors but the old PN2222A has a minimum current GBW of 300MHz which beats any “old” Op Amp ! This is what makes it work here.

AC coupling the input signal is necessary so the 100nF is just to isolate DC from input to base bias with ~1 ohm at 1MHz. You can make this bigger which affects bias settling time after T=(50+50)C =10us rise time.

Rcb/Rs=500/50 ohms sets the closed loop gain maximum of 10 which is further reduced by other base loads.

The input impedance is further loaded by the 400R to ground which pulls up the collector DC voltage to midpoint and attenuates the gain. Re =2 plays an important role to set the open loop gain Rc/(Re+Rbe*)= 15 to 25 and at the same time raise the input impedance hFE*Re~200 ohms for hFE=100 which further attenuates the input just as the 400 Ohms does.

Rbe* = 25/Ic (mA) [ohms] is controlled by collector current , which makes it in the 1 ohm range thus reducing the open loop gain, so Re was chosen around the same or slightly higher for linearity vs gain reduction tradeoff.

Overall this is far better than an open loop H bias due to the excess gain used for error correction or negative feedback. Albeit only a small amount but sufficient.

If you need more linearity then more power can be used to achieve this with 5V and reduce Rbe from higher Ic and thus increase open loop gain for extra feedback from this excess open loop gain.

Rc =50 sets the open loop impedance and also open loop gain from Re is reduced by base to emitter load of hFE*Re.

The net result is the closed loop gain is 5x with an open loop gain of 20 to 50 so the excess gain, lowers both the base and collector impedance by negative feedback. This is similar to Op Amps effect with NFB except the excess gain being fed back is much less and there is no integrator as in the OA to limit the open loop gain, so it has much greater BW than your OA. (Depending on GBW of transistor)

  • \$\begingroup\$ Hi, thank you for taking the time to respond. The source impedance is actually 50ohms. I don't require a very specific bandwidth, I was thinking for a bandwidth between 600kHz and 1.5MHz. The goal is to attenuate the noise received from the piezo transducer and to also amplify the 1.1MHz signal that I am interested in without introducing a lot of noise, that is why I picked a low-noise amplifier. I assumed that my requirements could be met by using an op-amp as is done in the TDC1000 IC that I mentioned above of which the LNA only has a GBP of 50MHz. \$\endgroup\$
    – user291514
    Commented Jul 30, 2021 at 10:50
  • \$\begingroup\$ So would you recommend for me to build a circuit using a transistor similar to the one you have posted above instead? \$\endgroup\$
    – user291514
    Commented Jul 30, 2021 at 10:51
  • \$\begingroup\$ Yes with a 50 ohm source, the series R is not required. 3.3V limits the power dissipation and the Vc ought to be centred from feedback ratio. Use good filtering on Vcc and make it small with a good gnd. Don’t use excessive cable C loading on output. My sim. shows 3 different signal inputs that you click on Sw to change position. \$\endgroup\$ Commented Jul 30, 2021 at 12:46
  • \$\begingroup\$ Gain & Vcc may be increased to 5V with small changes to bias pulldown and increase to Rcb. \$\endgroup\$ Commented Jul 30, 2021 at 12:53
  • \$\begingroup\$ At 5V , total Pd = 1/4W.. others can speak up to confirm this solution 8>) \$\endgroup\$ Commented Jul 30, 2021 at 13:00

The maximum output amplitude of an op-amp, above a certain frequency, decreases with frequency. From the LT1128 datasheet we have this graph

enter image description here

What this graph tells us is that even though a particular op-amp may have a unity-gain bandwidth of, say, 10 MHz, that only applies to very weak signals.

  • \$\begingroup\$ Any comments on my solution? \$\endgroup\$ Commented Jul 30, 2021 at 16:36

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