Other Parts Discussed in Thread: TINA-TI, , LMH6645
Hello everyone,
I need help to calculate the gain of the amplifier circuit below.
Thanks,
Hai
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Hi Lai,
I would run a TINA-TI simulation. It's free and can be downloaded here:
http://www.ti.com/tool/TINA-TI
Kai
Hello Hai,
Running a TINA-TI circuit simulation as Kai suggested would allow you to see what the gain looks like across frequency. It will change with frequency due to the addition of the various RC networks and limited gain-bandwidths of the op amps.
If you are looking for the dc and low-frequency ac gain it can be determined as follows:
U1 - Inverting amplifier, G1 = -R2/R1 = 196 k/499 = -392.8 V/V
U2 - Variable gain non-inverting amplifier, G2min = 1 + VR1min/R5 = +1 V/V, when VR1 = 0 Ohms, and G2max = 1 + VR1max/R5 = 1 + 25 k/1 k = +26 V/V
U3 - Sallen-key low-pass filter, G3 = 1 + R9/R8 = 1 + 1 k/1 k = +2 V/V
Gmin = G1 x G2min x G3 = (-392.8 V/V) (+1 V/V) (+2 V/V) = -785.6 V/V
Gmax = G1 x G2max x G3 = (-392.8 V/V) (+26 V/V) (+2 V/V) = -20,426 V/V
Therefore, the gain is dependent on the VR1 setting.
I hope this is what you needed.
Regards, Thomas
Precision Amplifiers Applications Engineering
Thomas Kuehl said:Hello Hai,
Running a TINA-TI circuit simulation as Kai suggested would allow you to see what the gain looks like across frequency. It will change with frequency due to the addition of the various RC networks and limited gain-bandwidths of the op amps.
If you are looking for the dc and low-frequency ac gain it can be determined as follows:
U1 - Inverting amplifier, G1 = -R2/R1 = 196 k/499 = -392.8 V/V
U2 - Variable gain non-inverting amplifier, G2min = 1 + VR1min/R5 = +1 V/V, when VR1 = 0 Ohms, and G2max = 1 + VR1max/R5 = 1 + 25 k/1 k = +26 V/V
U3 - Sallen-key low-pass filter, G3 = 1 + R9/R8 = 1 + 1 k/1 k = +2 V/V
Gmin = G1 x G2min x G3 = (-392.8 V/V) (+1 V/V) (+2 V/V) = -785.6 V/V
Gmax = G1 x G2max x G3 = (-392.8 V/V) (+26 V/V) (+2 V/V) = -20,426 V/V
Therefore, the gain is dependent on the VR1 setting.
I hope this is what you needed.
Regards, Thomas
Precision Amplifiers Applications Engineering
Dear Thomas,
Thanks for your answer. This is a very detailed answer.
But I still want to know more about the roles of C2, R3, R4 and C3, C4, C5, R6, R7 and C6.
Thank you very much.
Hai
kai klaas69 said:Hi Lai,
I would run a TINA-TI simulation. It's free and can be downloaded here:
http://www.ti.com/tool/TINA-TI
Kai
Dear Kai,
Thanks for your answer.
I spent time with TINA but it was still too new for me.
Can you help me set up the simulation?
Thanks,
Hai
Hello Hai,
Determining the frequency responses contributed by the various components in your application circuit is somewhat outside of the realm of the assistance we provide. It has less to do with the actual OPA140 and LMH6645 op amps and more to do with network analysis.
I will provide some general comments about your question, "I still want to know more about the roles of C2, R3, R4 and C3, C4, C5, R6, R7 and C6."
A mathematical analysis of each stage can be accomplished and their math product would reveal the overall frequency response of the complete amplifier. Doing so requires a good understanding of complex circuit analysis, and the effort can be time consuming.
Kai's suggestion to apply TINA-TI and do a frequency sweep will provide the correct response over frequency and should take less time than attempting a mathematical analysis - even if you have to learn how to use TINA-TI. TINA-TI is very intuitive and easy to learn and Kai has pointed you to online training.
Regards, Thomas
Thomas Kuehl said:Hello Hai,
Determining the frequency responses contributed by the various components in your application circuit is somewhat outside of the realm of the assistance we provide. It has less to do with the actual OPA140 and LMH6645 op amps and more to do with network analysis.
I will provide some general comments about your question, "I still want to know more about the roles of C2, R3, R4 and C3, C4, C5, R6, R7 and C6."
- C1 shunts R2 and as frequency is increased the capacitive reactance of C1 decreases. This causes the gain to roll of with increased frequency. This introduces a first-order low-pass filter response (-20 dB/dec).
- R3 and R4 form a voltage divider at the output of U1, and C3 shunts R3. As frequency increases C3's capacitive reactance decreases shunting R3 in the divider. This has the effect of increasing the voltage divider output voltage as frequency is increased, which results in a first-order high-pass response (+20 dB/dec).
- C4, C5, R6, R7 are the network components associated with a Sallen-Key low-pass filter topology. The circuits produces a second-order low-pass filter response (-40 dB/dec). The math of the second-order stage is reasonably complex and if you need to know more I suggest Googling the Sallen-Key filter topology. There are many good resources that delve into the theory of the filter.
- C6 modifies the low-frequency response of the Sallen-Key low-pass filter stage. C6 in conjunction with R8 adds a low-frequency, first-order high-pass response to the filter stage (+20 dB/dec).
A mathematical analysis of each stage can be accomplished and their math product would reveal the overall frequency response of the complete amplifier. Doing so requires a good understanding of complex circuit analysis, and the effort can be time consuming.
Kai's suggestion to apply TINA-TI and do a frequency sweep will provide the correct response over frequency and should take less time than attempting a mathematical analysis - even if you have to learn how to use TINA-TI. TINA-TI is very intuitive and easy to learn and Kai has pointed you to online training.
Regards, Thomas
Dear Thomas,
I have set up simulation file for this circuit.8446.amplifier.TSC
Now I need to analyze what factors, what characteristics of the circuit, to understand how this circuit works?
Hai,