# What is the peak of the short circuit current

I have this problem:

Consider the circuit below:

$$\V_{in}\$$ has an amplitude of 17V and a frequency of 125 kHz.

What is the peak of the short circuit current between the terminals $$\a\$$ and $$\b\$$?

Okay so this is how I thought about solving the problem. First finding all the impedances with the formulas:

$$\X_C = \frac{1}{j\omega C}\$$

$$\X_L = j\omega L\$$

Now creating mesh current equations, all currents going clockwise:

$$\-V_{in}+X_{C1} \cdot i_1+X_L(i_1-i_2) = 0\$$

$$\ X_L(i_2-i1)+i_2\cdot R+X_{C2}(i_2-i_3)=0\$$

$$\X_{C2}(i_3-i_2)=0\$$

Inserting all the values, setting $$\V_{in}=17 \mbox{V}\$$ to get the peak value and solving the equations with Maple we find that:

$$\i_3=-0.029289+0.263783\cdot j\$$

Taking the absolute vales we get: $$\i_3=265.4042 \mbox{mA}\$$, which I assumed was the correct answer. However, double checking with Lt-spice I get a different result for the peak value.

Who should I trust? My calculations or the simulation tool?

I hope someone can help me out.

EDIT

Another problem I have with the simulation tool, is that doesn't give me a consistent result of the peak value, as seen below.

• Just throwing an idea: if you reduce the time step in the simulation, you’ll get a result that’s closer to ideal. Mar 22, 2020 at 18:58
• Assuming your model was correct, you should trust the tool. For a simple ideal L-C circuit Its far less likely that the software made a mistake compared to hand solved equations. Mar 22, 2020 at 18:58
• @user4574 I made an edit to the post. Maybe you can help me out?
– Carl
Mar 22, 2020 at 19:11
• So far as i can see, your results match with a micro-ampere accuracy. I couldn’t see a problem in this. I don’t think high accuracy is important in this circuit. Mar 23, 2020 at 5:02
• @BenFM Although it's good accuracy, I need to give an answer in mA with 2 decimals precision. So I'm in a little bit of trouble here.
– Carl
Mar 23, 2020 at 8:17

If you want precise measurements you should not rely on the cursors: they are handy, but only for quick references, as you can see in the time values of the two readings. Instead, use .meas for an optional interval:

.meas Imax max I(V2) from 0 to 100n


Here I chose 100n as the final time because I only wanted to measure the initial peak current.

You should also know your tool before you use it: inductors have a default series resistance of 1mΩ, which can be seen when right-clicking on the inductor (message at the bottom of the settings window). Also, there is a default Rpar, to allow damping, setting which can be altered in Tools > Control Panel > Hacks! > Supply a min inductor damping.... It seems to default to gmin*L = 1e12*L, so if your inductance is small enough, it may become influential enough. Directly setting Rpar to zero is much quicker and doesn't influence the rest of the schematic(s), since the Control Panel setting is remembered.

The capacitors also have the hidden Rpar, which seems to default to gmin, which isn't that much problematic. Still, since your example relies on ideal elements, you should probably take time to set to zero the builtin parasitic elements for both L and C. It's enough to do it once, for one, and then simply copy-paste for the rest.

Finally, you may want to impose a tighter timestep, for better resolution, and also disable waveform compression with .opt plotwinsize=0, since the default 300 points is not that much precise.

Considering your other question, it seems you need to rely on some simulator or another for your learning, so it would do you good to deal with the friendly, but rather spartan manual when things seem to not work properly. The undocumented LTspice can provide additional help.

– Carl
Mar 23, 2020 at 9:53

Well, let's solve this mathematically. We have the following circuit:

simulate this circuit – Schematic created using CircuitLab

The input voltage is given by:

$$\text{V}_\text{in}\left(t\right)=\hat{\text{u}}_\text{in}\cos\left(\omega t+\varphi\right)\tag1$$

In the complex form this input voltage can be written as:

$$\underline{\text{V}}_{\space\text{in}}=\hat{\text{u}}_\text{in}\exp\left(\varphi\text{j}\right)\tag2$$

Where $$\\text{j}^2=-1\$$.

Now, the input impedance is given by:

$$\underline{\text{Z}}_{\space\text{in}}=\frac{1}{\text{j}\omega\text{C}_1}+\frac{\text{j}\omega\text{L}\left(\text{R}_1+\frac{1}{\text{j}\omega\text{C}_2}\text{||}\text{R}_2\right)}{\text{j}\omega\text{L}+\text{R}_1+\frac{1}{\text{j}\omega\text{C}_2}\text{||}\text{R}_2}=\frac{1}{\text{j}\omega\text{C}_1}+\frac{\text{j}\omega\text{L}\left(\text{R}_1+\frac{\text{R}_2}{1+\omega\text{C}_2\text{R}_2\text{j}}\right)}{\text{j}\omega\text{L}+\text{R}_1+\frac{\text{R}_2}{1+\omega\text{C}_2\text{R}_2\text{j}}}\tag3$$

Now, the input current is given by:

$$\underline{\text{I}}_{\space\text{in}}=\frac{\underline{\text{V}}_{\space\text{in}}}{\underline{\text{Z}}_{\space\text{in}}}\tag4$$

And the current trough $$\\text{R}_2\$$ can be found using the current divider formula:

$$\underline{\text{I}}_{\space\text{R}_2}=\underline{\text{I}}_{\space\text{in}}\cdot\frac{\frac{1}{\text{j}\omega\text{C}_2}}{\frac{1}{\text{j}\omega\text{C}_2}+\text{R}_2}\cdot\frac{\text{j}\omega\text{L}}{\text{j}\omega\text{L}+\text{R}_1+\frac{\text{R}_2}{1+\omega\text{C}_2\text{R}_2\text{j}}}\tag5$$

Now, we can find $$\\left|\underline{\text{I}}_{\space\text{R}_2}\right|\$$ when $$\\text{R}_2\to0\$$. Because you've already found an answer I used Mathematica to find it:

In[1]:=R1 = 3/10;
C1 = 20*10^(-9);
C2 = 330*10^(-9);
L = (33/10)*10^(-6);
w = 2*Pi*125*1000;
uin = 17;
\[Phi] = 0; FullSimplify[
Limit[Abs[(((I*w*
L))/(((I*w*
L)) + (R1 + (((R2)*((1/(I*w*C2))))/(((1/(I*w*
C2))) + (R2))))))*((1/(I*w*C2))/((1/(I*w*C2)) +
R2))*((uin*
Exp[\[Phi]*
I])/((1/(I*w*
C1)) + (((I*w*
L)*((R1 + ((((1/(I*w*C2)))*(R2))/(((1/(I*w*
C2))) + (R2))))))/((I*w*
L) + ((R1 + ((((1/(I*w*C2)))*(R2))/(((1/(I*w*
C2))) + (R2)))))))))], R2 -> 0]]

Out[1]=(1870 \[Pi]^2)/Sqrt[64000000 + 483472000 \[Pi]^2 + 1089 \[Pi]^4]

In[2]:=N[%1]

Out[2]=0.265404


So, the exact expression is:

$$\lim_{\text{R}_2\to0}\left|\underline{\text{I}}_{\space\text{R}_2}\right|=\frac{1870\pi^2}{\sqrt{64000000+483472000\pi^2+1089\pi^4}}\approx0.265404\space\text{A}\tag6$$

• Great, so you also get an answer of 265.40mA, that's good to see. This is also a really clear and well-written answer. I really appreciate the effort!
– Carl
Mar 23, 2020 at 11:47
• @Carl You're welcome, I happy that I could be of any help. Mar 23, 2020 at 12:23