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Hi,
Real quick, your "Off State" drawing does not look correct. You show the inductor and resistor RL voltages backwards from what they should be at that point in time. Once the switch opens, the inductor voltage will reverse and that will also reverse the voltage across the resistor. That may or may not be the source of your error in making your equations.
Also a side note, when showing current flow it is usually better to show the current as INSIDE a wire not outside.
For example in your diagram you show iL as OUTSIDE of both the inductor and RL and the arrow resembles the kind of arrow we use for voltage polarity so it can get confusing. This drawing is not as bad as some i've seen though
Hello again,Hi,
Well the voltage reverses numerically, but if your polarity signs are showing the symbolic currents and voltages then it may be ok.
Now I understand what did you mean. Decreasing current will turn the voltage to negative, ok.
I noticed now that you have the capacitor current arrow (if that is what it is) as pointing UP. is that a current arrow? If so then in the off state if the resistor current is UP then the cap current must be DOWN.
Yes, it's a current arrow, and I thought that to, but I took the sample from here (All About Circuits page)
When it's in ON State, the Capacitor discharges on to the load and OFF State is charging UP, so shouldn't the arrow current point up in OFF State?
Anyway, I've tried to change the Capacitor Sign Voltage and the result goes wrong again. Ouput Voltage is now 6.25 V.
I wrote the equations now and used a numerical method to solve the ODE's and i get a voltage that is closer to 3.57 volts.
I'll post the equations later once i clean them up.
Did you get the result adding the Parasital Resistance too? I would like to see that.
The capacitor is 10 mili farad and the Inductor is 600 micro Henry.
Thank you very much!
The voltage polarity across the resistor does not change.Once the switch opens, the inductor voltage will reverse and that will also reverse the voltage across the resistor.
Hi,The voltage polarity across the resistor does not change.
The inductor voltage reverses because the inductance now acts like a voltage currentsource to keep the current flowing in the same direction, but the voltage polarity across the resistor is unchanged, since the current direction does not change.
Hello again,
Well if you are doing it symbolically then the cap current must stay the same.
This is a conflict that comes up now and then, whether to show the cap voltage as the symbolic value or numerical value. I believe if you show the cap current numerically then you will find that it changes direction, but when we say it symbolically it will always be the same and that is because the numerical solution will naturally produce a negative result so we dont have to try to force that symbolically. For example:
0=x+1
we dont have to change x to x we just have to solve for it and then it comes out to a negative value.
Anyway, here are my two sets of ODEs for this circuit, and the drawing that goes with it.
Note the symbolic cap voltage arrow is always up, and the symbolic cap current arrow is always down.
Note that i consider ground to be the bottom of the cap and inductor too so that is zero volts for my solutions and that makes the cap voltage numerically negative.
The top of the cap is considered symbolic positive always.
Also note the second circuit to the right is for comparison, it's a circuit you can use if you still have trouble with the first one. If you can do that circuit with the input switching from 0 to 5v you may get more insight into these circuits.
The inductor current is 'i' and the cap voltage is 'v' below..
switch on:
L*di/dt=Ei*RL
C*dv/dt=v/R
(notice here iC=C*dv/dt the cap current is down so v must be negative because 'v' the cap voltage is positive symbolic with respect to ground and ground is the bottom of the cap)
switch off:
L*di/dt=vi*RL
C*dv/dt=iv/R
Vavg during the last on/off cycle came out to be 3.565232376 volts.
Here is the output of the numerical solution after many cycles using C=1 and L=1 and your R and RL values...
[organization: 't' is time in seconds, {inductor current, cap voltage}]
t=99.810000 {0.5470675577,3.577704524,1} (so this is time 99.81 seconds and iL=0.547 amps and vC=3.578 volts)
t=99.820000 {0.5913749758,3.574128607,1}
t=99.830000 {0.6352415277,3.570556265,1}
t=99.840000 {0.6786716002,3.566987494,1}
t=99.850000 {0.7216695362,3.563422289,1}
t=99.860000 {0.7642396355,3.559860648,1}
t=99.870000 {0.8063861554,3.556302567,1}
t=99.880000 {0.8481133103,3.552748042,1}
t=99.890000 {0.8894252731,3.549197069,1}
t=99.900000 {0.930326175,3.545649646,1}
t=99.910000 {0.885761531,3.551181306,0}
t=99.920000 {0.8415875129,3.556263966,0}
t=99.930000 {0.7978046796,3.560901981,0}
t=99.940000 {0.7544135416,3.565099708,0}
t=99.950000 {0.7114145605,3.568861505,0}
t=99.960000 {0.6688081503,3.572191729,0}
t=99.970000 {0.6265946772,3.575094737,0}
t=99.980000 {0.5847744607,3.577574885,0}
t=99.990000 {0.5433477736,3.579636527,0}
t=100.000000 {0.5023148426,3.581284018,0}
Just to note, the Laplace solutions come out pretty complicated because we have to include both the initial current for the inductor and initial voltage for the cap.
In the following, 'v' is initial cap voltage and 'i' is initial inductor current at start of half cycle...
switch on:
iL(s)=(i*s*L+E)/(s*RL+s^2*L)
vC(s)=(v*C*R)/(s*C*R+1)
switch off:
iL(s)=(L*(i*s*C*R+i)+v*C*R)/((s*C*R+1)*RL+L*(s^2*C*R+s)+R)
vC(s)=(v*C*R*RL+L*(s*v*C*Ri*R))/((s*C*R+1)*RL+L*(s^2*C*R+s)+R)
Of course the full time solutions are quite a handful also, but they would allow calculating the current and voltage exactly for some check point.
Do you mean the development of the equations are too long or the result it's too long?Long equations are no longer from my favourite things.
You should know that Wikipedia is not always true. There are a lot of false things written there, and Wikipedia is a nonprofit, not maintained organization, which does not give certainty of its posts' accuracy.
Okay.The voltage polarity across the inductor either changes or does not change according to how we view that very voltage.
Hi again,
Hi again,Okay.
But I was referring to your comment about the voltage across the resistor, not the inductor, which are not the same.
The voltage across the resistor does not change.
Hi again,
The values in the ODE's are as follows:
'v' is the symbolic cap voltage, positive on top negative on bottom, and that of course means numerically v comes out negative.
'i' is the inductor current flowing down through the inductor, and that means that when the switch is 'on' the voltage at the top of the inductor is numerically positive and when 'off' it is numerically negative.
Yes if we have a cap voltage increment that goes up then it must come down the same amount otherwise we are not yet in steady state. That means we only see that in steady state. HOWEVER, the sign of the increase is opposite to that of the decrease so dv/dt is positive one way and negative the other way.
I am not entirely sure i understand what you mean by the "off state current" in your previous post.
I'll take a closer look at this again later tonight.
Can I see the development steps ?Hello again,
Ok i got the same result for the average inductor current:
Iavg=Vo/((1D)*R)
Hi,Can I see the development steps ?
Did you redone the adding of cap voltage increments equal to zero, or another steps ?
Hello again,
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