but even an ideal inductor is supposed to have inductance, right? like VL = di/dt there's supposed to be some opposition so i don't get how it arrives at the 1kA number. <edit>and it STARTS at 1kA. i backed up the time scale to ns, to ps. it's starting at that value</edit>Welcome to ETO!
Add some resistance in series with the inductor. 1000A is a tad unrealistic!
i don't understand. voltage V1 is supposed to start at 5 and the voltage at Vcap is supposed to start at 0. so spice is just not plotting what i'm telling it to plot. but when you change the voltage source to Pulse, then it interprets the circuit at face value and plots it correctly. WTF?Changed 1F cap to 1uF.
Changed your voltage source from a battery to a pulse source.
"Initial condition" is one of the problems. The program saw the voltage on C1 was going to charge to 5V so it started out with the voltage at 5V.
By using a pulse source, I started out with 0V, at time zero moved to 5V, then at time 10mS moved back to 0V.
No.voltage V1 is supposed to start at 5 and the voltage at Vcap is supposed to start at 0
.is there a more direct way to specify initial condition?
.IC -- Set Initial Conditions
I have not used this in a while but I think if you have this on the schematic somewhere:The .ic directive allows initial conditions for transient analysis to be specified. Node voltages and inductor currents may be specified. A DC solution is performed using the initial conditions as constraints. Note that although inductors are normally treated as short circuits in the DC solution in other SPICE programs, if an initial current is specified, they are treated as infinite-impedance current sources in LTspice.
Syntax: .ic [V(<n1>)=<voltage>] [I(<inductor>)=<current>]
Example: .ic V(in)=2 V(out)=5 V(vc)=1.8 I(L1)=300m
No resistance, no problem with calculating the current. The current for no resistance is Vo*t/L , where Vo is a step function. The curve a straight line starting from zero and goes as high as you can name if the voltage source can supply it, and you wait long enough. Notice that as L becomes smaller and smaller, the given formula produces output current values that are similiar to a dead short.
In LT Spice the voltage source has a place to add a resistance. Right click on voltage source, "series resistance". If not filled in there is a default value.and goes as high as you can name
Perhaps so, but I was calculating the current equation for a nonexistent inductor with no resistance, and energized with a step voltage. That appears to be what was wanted in post #1.In LT Spice the voltage source has a place to add a resistance. Right click on voltage source, "series resistance". If not filled in there is a default value.
Same thing for inductors. "series resistance" For real parts, there is a resistance of the wire.
In the real world, a car battery and wire might have 0.01 ohms. (don't really know) and a inductor might have 0.5 ohms. In the real world, when the current hits 24 amps the wire burns out. You don't really reach 1000 amps.
You need a step function because Spice normally calculates the steady-state DC bias conditions before it does the transient response.yes it was, just to see how long to saturate an L of 4.7uH so i could make a buck circuit. but why would i need a step voltage function i don't get it.
Well, you have to have some kind of voltage curve don't you? A common step function is what you get when you switch on DC. If you want me to calculate a pulse, square, triangle, or whatever, then specify what you need.yes it was, just to see how long to saturate an L of 4.7uH so i could make a buck circuit. but why would i need a step voltage function i don't get it.
A 4.7hH "spice" coil will never saturate. It is not a real world coil.just to see how long to saturate an L of 4.7uH
THANK YOU i didn't know what the initial operating point solution was