# Ltspice helpp

#### arthurwinchester

Joined Jun 1, 2023
5
Hello guys, Can anyone help me put this into the software and find V0 in the software???

#### djsfantasi

Joined Apr 11, 2010
9,080
Why do you need this information? It wouldn’t be schoolwork, would it? We need this information in order to help.

#### DickCappels

Joined Aug 21, 2008
10,067
What do you mean by "V0"?
By software, do you mean find it in LTspice, or by other means?

#### arthurwinchester

Joined Jun 1, 2023
5
I'm trying to learn how to use the software, this is an example of a circuit that is more advanced and I need to insert in LTSPICE, and V0 is the tension on the 2 ohm resistor

#### arthurwinchester

Joined Jun 1, 2023
5
Why do you need this information? It wouldn’t be schoolwork, would it? We need this information in order to help.
schoolwork is to find it manually using tension node, my teacher said we can check our answers in the software(if we know how to use it) since his slides don't have solutions

#### DickCappels

Joined Aug 21, 2008
10,067
Have you tried entering the circuit into LTspice, running a simulation and "looking" at the voltage?

#### crutschow

Joined Mar 14, 2008
32,914
Since this is homework we can't give you the complete answer but we can help.
Start by entering the schematic into LTspice (free download from Analog Devices).
From the Edit/Component drop-down menu, use the source "e" in for the 2Vs source, voltage for the voltage source, and current for the current source.
Then, if you have any questions, post the LTspice schematic here.

#### WBahn

Joined Mar 31, 2012
29,165
schoolwork is to find it manually using tension node, my teacher said we can check our answers in the software(if we know how to use it) since his slides don't have solutions
By "using tension node", I'm going to infer that you are to find the answer using what we normally call "node voltage" or "nodal" analysis.

So what answer did you get manually, since you need that before you can check it with a simulation.

If you haven't gotten an answer, then that is the far-more-important issue that you need help with. Don't worry about checking it with a simulator until you have something to check.

Once you get to that point, there are a number of circuit simulators out there -- the dominant one used by members here is LTSpice, which is decent and pretty good... and free. There are lots of tutorials out there to get you started and this circuit doesn't have any hidden gotchas in it. One key thing to remember is that you MUST define one of the nodes as Node 0 (commonly called "ground") for the simulator to work properly.

#### Papabravo

Joined Feb 24, 2006
20,392
First use the pencil icon to draw all the wires like this:

Then use the resistor symbol to place the resistors. When you place a component on top of a wire it removes the short across the terminals of the component. When you grab a symbol to place, the Ctrl-R key combination will rotate the symbol and Ctrl-E will flip the symbol side-to-side.

Place a voltage source on the left-hand side using the letter V or select it from the components menu, which is between the diode symbol and the hand symbol. Also place the current source on the right-hand side.

Lastly you want to add a behavioral voltage source whose name in the component directory is "bv"

Once you get this far then we have to show you how to edit the values, construct a simulation command, and run it.

#### crutschow

Joined Mar 14, 2008
32,914
Lastly you want to add a behavioral voltage source whose name in the component directory is "bv"
I think the "e" voltage-dependent voltage-source would likely be easier to use for that function.

#### Alec_t

Joined Sep 17, 2013
13,788
There are lots of tutorials out there
.... including those in the Help available in LTspice itself.

#### Papabravo

Joined Feb 24, 2006
20,392
I think the "e" voltage-dependent voltage-source would likely be easier to use for that function.
There are two problems with that:
1. The main E-source has 4 pins and requires a control voltage which in this context complicates the situation.
2. The E-source with only two pins is a bit harder to find and is called EPOLY and located in the "misc" folder.
I don't remember if LTspice IV had the B-source.

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#### arthurwinchester

Joined Jun 1, 2023
5
First use the pencil icon to draw all the wires like this:
View attachment 295518
Then use the resistor symbol to place the resistors. When you place a component on top of a wire it removes the short across the terminals of the component. When you grab a symbol to place, the Ctrl-R key combination will rotate the symbol and Ctrl-E will flip the symbol side-to-side.
View attachment 295520
Place a voltage source on the left-hand side using the letter V or select it from the components menu, which is between the diode symbol and the hand symbol. Also place the current source on the right-hand side.
View attachment 295523
Lastly you want to add a behavioral voltage source whose name in the component directory is "bv"
View attachment 295524
Once you get this far then we have to show you how to edit the values, construct a simulation command, and run it.
I did that, and I put the ground point already. I'm stuck now in the dependent voltage source

#### Irving

Joined Jan 30, 2016
3,496
Use F4 to label the R1/R2/R3 junction as V0
Then enter the formula V=2*V(V0) for the behavioural voltage source
Then right-click on the background, choose RUN from the menu and select the DC Op Pnt (DC Operating point) tab

#### ericgibbs

Joined Jan 29, 2010
18,032
Hi,
This is a homework assignment, please no complete solutions.

Moderation.

#### Papabravo

Joined Feb 24, 2006
20,392
I did that, and I put the ground point already. I'm stuck now in the dependent voltage source View attachment 295571
OK - very good. First, place a Net Name on the junction of R1, R2, and R3. There is an icon showing the letter "A" between the GROUND icon and the resistor icon. That is what you use for Net Names. Call it something like "Vx"

The expression underneath the behavioral source is a hint that it wants a functional expression for the voltage of
Right-Click on the text "V=F(...)" and in the small dialog, make its say "V=2*V(Vx)". This means make it equal to 2 times the voltage at node "Vx"

Lastly go up to the menu bar, click on Simulate and choose from the sub-menu "Edit Simulation Command". Choose "DC op pnt" and this will put the text ".op" onto the schematic.

Make it look like this:

Net names are in brown color. I did change my color palette so don't worry about that for now.

#### Papabravo

Joined Feb 24, 2006
20,392
Hi,
This is a homework assignment, please no complete solutions.

Moderation.
It is fairly clear that the homework is not to show a solution in LTspice, but to solve the problem. He did most of the work on his own and this is just a bit of guidance. He still has to be able to justify and verify the results. That is something the simulator won't help with. It is why on an exam you still have to "show your work". I believe the TS understands that and that is why I did not finish my answer in my first post on this thread.

I should also point out that this solution is sensitive to the way the resistors are placed, because they do have an orientation with respect to the signs of currents. The magnitudes of the currents will be correct, but the signs may be confusing. I assure that KCL has not been violated.

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#### ericgibbs

Joined Jan 29, 2010
18,032
Hi,
I agree, it was a cautionary note only, that's why I did not delete any posts.

Mod

#### crutschow

Joined Mar 14, 2008
32,914
The main E-source has 4 pins and requires a control voltage which in this context complicates the situation.
I don't see why.
The control voltage is the Vs voltage across the 2Ω resistor so the "e" control terminals are just connected across that resistor.
The only entry is then to give "e" a gain of 2.

#### Papabravo

Joined Feb 24, 2006
20,392
I don't see why.
The control voltage is the Vs voltage across the 2Ω resistor so the "e" control terminals are just connected across that resistor.
The only entry is then to give "e" a gain of 2.
OK, there is no argument that it works, but I was trying for a schematic that looked as much like the original problem as possible.