# LTSpice AC Simulation giving unexpected output

I am currently working through AoE, 3rd Ed., and am on Exercise 3.7 (pg. 180). I am trying to find the -3dB point of the circuit pictured, and I am using LTSpice to check my answer.

LTSpice says the -3dB cutoff is ~882kHz. This does not agree with the answer I got on paper, which is ~398kHz. Can anyone point me in the right direction?

For reference, my quick math is:

• Z(C2||R2 @ 398kHz) = ~24.2k
• Voltage divider with Rser, V(out)/V(in) = 24.2k/34.2k =~ 0.7079 = -3dB
• What circuit variable are you looking for the 3-dB cut-off frequency of? Jun 21, 2020 at 22:03
• Ah, sorry, forgot to mention I am measuring V at the top of R2. Jun 21, 2020 at 22:18

Rser (10 kohm) is the dominant resistance so use that in your calculation. If you want a more precise calculation Use 10 kohm in parallel with R2 (about 9 kohm) to calculate the cut off frequency.

$$\boxed{f_C = \dfrac{ 1}{ 2\pi CR}}$$

So, if I do a simulation and use the DC gain as a reference I get this: -

The left cursor is positioned at an arbitrarily low frequency (1.14 kHz) and this shows that the DC gain is -0.827 dB as predicted in the comments. If I then move the right cursor so that the "delta" is 3.01 dB, the frequency is shown to be at 1.094 MHz. This tallies with the equation I wrote last night when you use the parallel resistance of the 10 kohm and the 100 kohm. The parallel resistance is 9.0909 kohm.

• I wish you'd explain it more, but it doesn't change the fact that this is square-on. The OP could, as I'm sure you already considered did but didn't spell out in so many words, just look at the two resistances as a Thevenin voltage divider. It simplifies the view, a lot, and gets the right answer. Anyway, +1. I just hope the OP "sees" things okay. (Another way to "see" is to realize that the 100k is eventually gradually bypassed as the capacitor starts to take over. When that happens, the 100k doesn't matter much and it's just the 10k and the cap left over -- as you say, it dominates.)
– jonk
Jun 22, 2020 at 2:55
• @jonk it was late last night and I was on my tablet. I think you've done a decent job of enhancing my answer so, if the OP "gets it" that's fine else I'll add a few more words. Jun 22, 2020 at 7:29
• Thank you both for the insight. After plugging in C=16e-12 and R=9e3 in the above equation, fc=1.1MHz. This is still different from the LTSpice output... am I missing something here? Thanks. Jun 22, 2020 at 10:22
• @Mason_Schellenberg the issue is simply one of choosing the correct 3 dB point. In your circuit with a 10 k resistor and a 100 k resistor, at DC, the gain is not unity but $\dfrac{100}{10 + 100}$ = 0.90909. That in dB terms is -0.828 dB so, you shouldn't be using the -3 dB point but the -3.828 dB point. Jun 22, 2020 at 10:37
• @Andyaka. Okay, that makes sense. I tried plugging in these new values to see at what frequency an RC lowpass filter would get at V(out)/V(in) = -3.828dB, R=10e3, and C=16e-12. I ended up with f=1.18 MHz. This is different from the 882 kHz estimated by LTSpice. Could you explain where I am going wrong? Thanks again. Jun 22, 2020 at 12:09

Since an answer has been selected (and a good one), I'll add some thoughts now.

Your circuit looks like this:

simulate this circuit – Schematic created using CircuitLab

It's pretty easy to see the voltage divider formed up by the two resistors, above. This has a Thevenin equivalent: $$\V_\text{TH}=V_\text{AC=1}\cdot\frac{R_2}{R_\text{SER}+R_2}\$$ and $$\R_\text{TH}=R_\text{SER}\,\mid\mid\,R_2\$$. The AC crossover will be at $$\f_{\text{c}}=\frac1{2\pi\,R_\text{TH}\,C_2}\$$. So you can just use the Thevenin equivalent and "get there," quickly.

A more formal approach is to use the Laplace equivalent. (I've written a little something on the topic, recently.) Here, a capacitor's impedance is described as $$\Z_C=\frac1{s\,C}\$$. (A resistor's is the obvious $$\Z_R=R\$$.) So long as you apply this idea fastidiously, it works out in the end. So let's do that.

Again, this is a voltage divider. But now, we treat $$\R_2\$$ and $$\C_2\$$ as a parallel pair and try and work out its impedance. Once we have that, we can apply it as we would with any other divider.

The parallel impedance of $$\R_2\$$ and $$\C_2\$$ is, let's say, $$\Z_2=Z_{\text{R}_2}\,\mid\mid\,Z_{\text{C}_2}=\frac{R_2\,\frac1{s\,C_2}}{R_2+\frac1{s\,C_2}}=\frac{R_2}{1+s\,R_2\,C_2}\$$. Same algebra, but weird symbols for now. Just live it for a moment.

Now, we apply this to our equivalent divider. Here, $$\V_\text{OUT}=V_\text{IN}\frac{Z_2}{R_\text{SER}+Z_2}\$$. Let's draw this out, painstakingly:

\begin{align*} V_\text{OUT}&=V_\text{IN}\cdot\frac{Z_2}{R_\text{SER}+Z_2}\\\\ &=V_\text{IN}\cdot\frac{\frac{R_2}{1+s\,R_2\,C_2}}{R_\text{SER}+\frac{R_2}{1+s\,R_2\,C_2}}\\\\ &=V_\text{IN}\cdot\frac{R_2}{R_\text{SER}\cdot\left(1+s\,R_2\,C_2\right)+R_2}\\\\ &=V_\text{IN}\cdot\frac{R_2}{R_\text{SER}+s\,R_\text{SER}\,R_2\,C_2+R_2}\\\\ &=V_\text{IN}\cdot\frac1{1+ \frac{R_\text{SER}}{R_2}+s\,R_\text{SER}\,C_2}\\\\ &=V_\text{IN}\cdot\frac{\frac1{\left[1+ \frac{R_\text{SER}}{R_2}\right]}}{1+s\,\frac{R_\text{SER}\,C_2}{\left[1+ \frac{R_\text{SER}}{R_2}\right]}}=V_\text{IN}\cdot\frac{K}{1+\tau\,s} \end{align*}

This is the standard form for a low pass filter. The numerator, $$\K\$$, is the gain. And, as you can see, this is slightly less than 1. The denominator has been arranged so that it is 1 plus "something times $$\s\$$." (Having a "1 +" in the denominator is a requirement and taking the trouble to massage the denominator to achieve it causes the numerator, sometimes, to no longer be 1, as in this case.) This "something" (times $$\s\$$) is also known as (aka) the tau ($$\\tau\$$.) (The angular frequency is $$\\omega_{_0}=\frac1{\tau}\$$.)

Let's look at $$\\tau\$$:

$$\tau = \frac{R_\text{SER}\,C_2}{\left[1+ \frac{R_\text{SER}}{R_2}\right]}$$

Here, you can see that this is dominated by the value of $$\R_\text{SER}\,C_2\$$. But that it is modified, a little, by the near-1 value of $$\1+ \frac{R_\text{SER}}{R_2}\$$ that appears in the denominator. The denominator, in your case, is $$\1.1\$$. This acts to slightly reduce the value of $$\R_\text{SER}\$$ before it is combined with $$\C_2\$$ to compute $$\\tau\$$. But the upshot is that the value is primarily determined by $$\R_\text{SER}\,C_2\$$ and then only slightly modified by the value of $$\R_2\$$.

That's why someone can say that $$\R_\text{SER}\$$ is the dominant resistance value in determining the cross-over frequency. Yes, it is modified by $$\R_2\$$. But as you can see, still larger values of $$\R_2\$$ will have diminishing impact, too.