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I'm trying to build a reactive load box, so I can run a guitar amplifier without a speaker. It's basically a device that simulates the impedance/frequency curve of a relatively high power speaker.

I need 4Ω nominal impedance, but my local electronics store does not sell 4Ω 100W resistors, so I though of getting four 16Ω resistors and put them in parallel.

LTSpice impedance/frequency curve

On the right it's the correct impedance curve using a single 4Ω resistor, and on the left, the same circuit but using four parallel 16Ω resistors.

Why are the simulation results different? I thought these circuits were supposed to be equivalent.

Edit by Steve G.: The following circuit gives the same results as the 4 x 16Ω resistor circuit (note the floating wire):

Upload by Steve G

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    \$\begingroup\$ what are the parameters for the resistors you use ? Ideal resistors ? \$\endgroup\$
    – Marla
    Commented Jan 18, 2017 at 14:39
  • \$\begingroup\$ @Marla, I'm sorry but I'm not sure I can understand your question. I just grabbed resistors from LTSpice menu and set the resistance values. I'm guessing that since this is a this simulation they are "ideal"? \$\endgroup\$
    – Telmo Marques
    Commented Jan 18, 2017 at 14:43
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    \$\begingroup\$ Likely the resistors you chose are ideal (standard resistor). But the question had to be asked (never know what others do in spice). . Also, don't assume that spice models are ideal models. \$\endgroup\$
    – Marla
    Commented Jan 18, 2017 at 14:54
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    \$\begingroup\$ Not believing it, I tried to replicate, and I have the same behavior. Strange. I wonder what the explanation is. Have an upvote, at least you'll have that going for you. \$\endgroup\$
    – dim
    Commented Jan 18, 2017 at 14:57
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    \$\begingroup\$ There is clearly something going wrong with Ltspice here. Using the single 4 Ohm circuit, if I add a connection to the left of R1, going upwards and to the right (as if you were about to place another resistor) but leaving the connection floating then Ltspice gives the single peak at 70Hz without the upward sweep at 1kHz. \$\endgroup\$
    – Steve G
    Commented Jan 18, 2017 at 15:23

1 Answer 1

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I tried to replicate, and I saw the same behavior. @SteveG did, too. After a few questions regarding the sanity of each of us, I think I found why, and it is pretty simple:

Are you absolutely sure you are actually plotting the right voltage node? In both of your plots, it shows "V(n003)/I(V1)". But I actually realized that if I mess up just a bit with the schematics, the node names change. In my case, n003 became n001 when I added the resistors, or just changed the wires layout.

To overcome this, you can explicitly name your nets by putting labels.

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  • \$\begingroup\$ +1. You're right. I didn't notice. In fact, adding the horizontal floating connection changes the net from n003 to n001. \$\endgroup\$
    – Steve G
    Commented Jan 18, 2017 at 15:45
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    \$\begingroup\$ Is this a lesson in specifying node names (net labels)? Eg: "Always label your nets!" \$\endgroup\$
    – Bort
    Commented Jan 18, 2017 at 16:05
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    \$\begingroup\$ I'm so happy the laws of physics still apply! \$\endgroup\$
    – Telmo Marques
    Commented Jan 18, 2017 at 16:35
  • \$\begingroup\$ lol @ dynamic node renaming ... \$\endgroup\$
    – jbord39
    Commented Jan 18, 2017 at 18:31
  • \$\begingroup\$ To overcome this, only work with .cir files :P \$\endgroup\$
    – Vladimir Cravero
    Commented Jan 18, 2017 at 21:21

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