New answers tagged

1

However, in Pspice it looks like the center frequency is almost 36 KHz. Did I use the wrong equations to calculate the center frequency? Your calculations assume an ideal op amp. In practice the center frequency should be lower due to the LM741's limited bandwidth. However in your case it's higher because you have the inputs swapped around. In AC analysis ...


1

Try doing a transient analysis. You'll probably find your output slammed in to one of the rails since the opamp inputs are flipped (positive input goes to ground).


2

Here is the result if you follow my comment above: I'm not going to check your calculations (at least right now) but maybe try simulating it with a better op-amp model and higher supply voltages. 30kHz is pretty high for a 741 (look at all that required gain!) and +/-5V is right at the minimum. Something like an AD8034 off +/-10V. And look at the output in ...


1

What you need is a Linkwitz-Riley filter, it's derived from the Butterworth. But, while the Butterworth has a corner frequency at \$\frac{1}{\sqrt{2}}\$, these have it at \$\frac12\$, which means that summing two filters, lowpass and highpass, with the same \$f_p\$, will result in a magnitude of \$2\cdot\frac{1}{\sqrt{2}}\$ for the Butterworth, and \$2\cdot\...


1

While your expansion is correct, the issue is your 1st plot is in Hz while the Matlab plot is in Radians, ~ 10x different num = [1 0]; den = [(1/(4000000*pi*pi)) 101/(20000*pi) 1]; G = tf(num,den); bode(G) However, a couple of extra commands can switch the units to Hz (you can do it from the UI as well) num = [1 0]; den = [(1/(4000000*pi*pi)) 101/(20000*pi) ...


-1

Well, using the voltage divider formula we can see that: $$\text{v}_+\left(\text{s}\right)=\frac{\text{R}_1}{\text{R}_1+\frac{1}{\text{sC}}}\cdot\text{v}_\text{i}\left(\text{s}\right)\tag1$$ $$\text{v}_-\left(\text{s}\right)=\frac{\text{R}_2}{\text{R}_2+\text{R}_3}\cdot\text{v}_\text{o}\left(\text{s}\right)\tag2$$ For an non-ideal opamp we know that \$\...


1

The CR highpass filter has one pole and one zero. The opamp itself has one pole (it doesn't have infinite bandwidth) as indicated by LTSpice:- So the transfer function will be in the form \$H(s) = \frac{b_1s}{s^2+a_1s+a_0} \$ which clearly is a transfer function for a bandpass filter. If you want to avoid this you can give your opamp more bandwidth in ...


Top 50 recent answers are included