Does anyone know how to explain non-inverting amps?
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This resistor determined the input resistance for AC signalsIs the R4 resistor of 100 MegaOhms meant to limit the current?
Only reverse polarity protection.Is the D3 diode for reverse polarity protection, or does it also serve another purpose?
C2 and C3 determine the amplifier frequency response.I'm guessing the C3 capacitor is for filtering
R1 and R2 are simply input/output protection resistor.What is the purpose of the R1 resistor which has a value of 200 ohms?
R7 is pull down resistor. Prevents output from floating.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?
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.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 your multimeter loads down the 100M resistance which causes the voltage measurement to be too low.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 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..... 5.96V DC at the output pin.
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.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.
Check out page 13 of the datasheet http://www.ti.com/lit/ds/symlink/opa637.pdfThanks 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?
Read this form page 13Thanks 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.
You need to find the transfer function.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.