Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

op-amp stray-capacitance

Status
Not open for further replies.

RS_Vimperk

New Member
Hello,

i posted the problem as png file.

The first question is to determine the transfer function for the circuit taking into account the stray capacitance. Using voltage-devision rule, i dervied the tf and it was correct, same as the answer at the end of book.

\[ \frac{v_{0}(s)}{v_{in}(s)}= 3.5(1+3.17*10^{-6}s) \]

The second question is to determine R1 and R2 so the DC gain of the circuit is unchanged. What i understand is when DC, then capacitor is open so we get only resistances and since s=0 the transfer function has the value 3.5, and according to this rewriting the formula, we have

R2=2.5 R1

which is also good, the answers at the end of book show that R1=60.2 k, R2=150.6 k

so it's right that R2=2.5 R1

now i'm not sure about that part that says that the capacitance affects the gain by no more than 3 dB at 100 kHz. Can you guide me?

R2 is the feedback resistor.

Thanks.
 

Attachments

  • op-amp stray-capacitance.png
    op-amp stray-capacitance.png
    62 KB · Views: 287
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top