Does anyone know how to explain non-inverting amps?

 

I redraw your diagram

And as you can see this diagram is almost a classic diagram.

R2, R4 , R4 provide DC bias voltage. Because every "active device" to work properly as an amplifier supplied from a single source need proper bias circuit. When you use BJT as a CE amplifier, you use a voltage divider to bias the active device somewhere in the "linear region". In case of single supply op amp you have to do the same think. We need to bias the op amp somewhere in the middle of his "linear region".

C2 ensure that the DC voltage gain is equal to 1.

But AC gain is equal to
Av = 1+ R6/R5 = 101

As for this 0.584V it is a measurement error.
Becaues R4 has 100M and you use a multimetr which has 10M internal resistance.
So you form a voltage divider, solve for Va in this circuit

Va ≈ 6V * 10M/(10M + 100M) ≈ 6* 10/110 ≈ 0.5454V

 

Hi Jony130

Wow thank you so much this makes way more sense now! I just realized I should have drawn a better schematic like you did, sorry about that. I didn't realize the voltage reading error was due to the internal resistance of 10 MegaOhms from the multimeter.



Is the D3 diode for reverse polarity protection, or does it also serve another purpose?
I'm guessing the C3 capacitor is for filtering?

 

Is the R4 resistor of 100 MegaOhms meant to limit the current?

This resistor determined the input resistance for AC signals
Rin = R2||R3 + R4 ≈ R4

Is the D3 diode for reverse polarity protection, or does it also serve another purpose?

Only reverse polarity protection.

I'm guessing the C3 capacitor is for filtering

C2 and C3 determine the amplifier frequency response.
See the example for the ideal op amp case


But for your op amp OPA637 the gain bandwidth product is equal to 80MHz
So the gain will start to drop at 235KHz at rate 6dB/octave and at Fc = 80MHz/101 = 790KHz gain start drop even faster at rate 12dB/octave. Google bode plot.

So C3 prevent circuit form osculation. To be sure the full bode plot is needed.

 

Hi Jony130,

Thanks so much for the response!

 

What is the purpose of the R1 resistor which has a value of 200 ohms?

R1 and R2 are simply input/output protection resistor.

Also, I see at the output of the op amp that the C4 capacitor removes the DC offset from the signal. Is the R7 and R8 resistors for additional filtering or why are they needed?

R7 is pull down resistor. Prevents output from floating.

 

This circuit is a non-inverting op amp running off a single +12V battery. It seems to be biased at the halfway on the non-inverting input although I don't understand why there is a 100Mohm resistor between the bias point and +IN pin.

Because the piezo transducer must feed into a 100M extremely high resistance. Therefore R4 is 100M. It separates the input of the opamp and also the piezo from the 50k total resistance of the R2 and R3 voltage divider.

I also have no idea why I measure 0.588 volts DC at the input where the transducer connects, 0.583 volts DC at the op amp +IN pin, 5.93 DC volts at the -IN pin.

Because your multimeter loads down the 100M resistance which causes the voltage measurement to be too low.

.... 5.96V DC at the output pin.

Because the R2 and R3 voltage divider causes the input of the opamp to be +6V and since the opamp has a DC voltage gain of 1 then its output is also +6V.

The RF resistor is 68.1kohms and the Rg resistor is 681 ohms giving a gain of 101.
If the +IN pin is at 0.583 volts DC I don't see how this should give a 5.96V DC at the output unless the gain was only 10. Someone told me I am likely measuring a diode drop at the +IN pin because the op amp has a virtual ground but it doesn't make sense to me.

I repeat, your meter is loading down the input DC voltage. The DC input is +6V. Therefore the output DC voltage is also +6V. C2 causes the DC gain to be 1.

 

Thanks for the explanations.

 

As for C3 I made a stupid mistake.
F3 is not equal to 23.5MHz It should be
F3 = 0.16/ ( C3* R6 ) = 235KHz

 

Thanks for the explanations.

I'm still not sure why the C3 capacitor would help the op amp behave more stable in the unity gain configuration?

Check out page 13 of the datasheet http://www.ti.com/lit/ds/symlink/opa637.pdf

They indicate layout as being a factor (stray capacitance), but the real hooker is the low, but non-specified input capacitance that adds phase shift to the feedback.

 

Thanks jimkeith for your response. Its still not easy for me to understand how that capacitor interacts with the non-specified input capacitance but I'm guessing its hard to without looking at a bode plot.

 

Thanks Jony130.

Is there an easy analogy or explanation for computing the F1, F2, F3, and F4 cutoff frequencies of the amplifier response?

I cannot find any of those computations in the design references that I have for op amps. For instance, why is F2 computed without R6 and C3 in the equation? I don't understand why F1 doesn't use C3, and why F3 doesn't use R5 and C2, and why F4 doesn't use C2 in the equations.

Thanks.

 

Thanks jimkeith for your response. Its still not easy for me to understand how that capacitor interacts with the non-specified input capacitance but I'm guessing its hard to without looking at a bode plot.

Read this form page 13
http://www.analogzone.com/acqt0704.pdf
And you will see how C3 helps.

 

Is there an easy analogy or explanation for computing the F1, F2, F3, and F4 cutoff frequencies of the amplifier response?

I cannot find any of those computations in the design references that I have for op amps. For instance, why is F2 computed without R6 and C3 in the equation? I don't understand why F1 doesn't use C3, and why F3 doesn't use R5 and C2, and why F4 doesn't use C2 in the equations.

You need to find the transfer function.
And by knowing transfer function you can find pols and zeros.
And since C3<<C2 we can analyze the circuit much easier.
Simply if we analyze C2 influence we remove C3 from the circuit, and vice versa, we remove C2 when we analyze C3 influence.
http://www.electro-tech-online.com/general-electronics-chat/120949-cutoff-frequency.html#post997029

 

Thanks again Jony130. I will read the material in the links you attached and try and learn about both the op amp frequency response and stability.