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 Author Topic: RLC circuit discharge simulation  (Read 18280 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
dsieron
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 « Embed this message on: December 03, 2009, 05:30:55 am » posted from:Milford,Connecticut,United States

I have a closed series circuit consisting of a capacitor, inductor resistor and diode. There is no external voltage source.

Typical capacitor size is 0.02 mfd.  Inductor is 370 mhy, resistor is 9 ohms
Typical initial conditions (time zero) are capacitor volts = 20, inductor current = 120 ma.

Beginning clockwise around the loop, current flows clockwise out of the  inductor into the positive side of the capaciotr
Current is flowing in the forward direction of the diode (of course).
Need simulation to track voltage at any node after time zero
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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: December 03, 2009, 01:05:05 pm » posted from:Taipei,T'ai-pei,Taiwan

For example: Diode can be approximate as a small voltage (Vd=0.8V?) and a small resistor (25Ω/I(mA) ) when current flow in the forward direction.
When voltage is applied to a diode, it can be simulated as a very large resistor. What is the value for the resistance when the voltage is reversed?
This is similar to a RLC circuit, but with different R and a voltage source when current flow in different direction.
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dsieron
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 « Embed this message Reply #2 on: December 03, 2009, 08:15:29 pm » posted from:Milford,Connecticut,United States

Thank you for your attention.  Your approximation for the diode is good.
Because with a real diode, current can never reverse direction, my interest is confined to what is happening only while current is flowing in the forward direction of the diode.  In your suggested  simulation, I would therefore not be interested in what happens when current changes direction.
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Fu-Kwun Hwang
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 « Embed this message Reply #3 on: December 03, 2009, 09:54:51 pm » posted from:Taipei,T\'ai-pei,Taiwan

For a real diode, the current is very small (in the order of $uA$, but not zero) when the voltage is applied in reversed direction.
It means that the resistance of the diode is very large, but it is not infinite.

When the voltage is applied in forward direction, the result will be similar to standard RLC circuit ($V_V+V_R+V_C+0.8=0$).
However, when the voltage in in reversed direction, the change of the current will be limited by inductor, most of the voltage will be across the diode and the resistor. I will need to know the resistnce of the diode in order to calculate the voltage across the inductor.
i.e. $V_L= - (V_C+V_R+I*R_{diode})$. I need this value to calculate $T(t+dt)$ from $\frac{dI}{dt}= -V_L/L$
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dsieron
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 « Embed this message Reply #4 on: December 05, 2009, 09:28:23 pm » posted from:Milford,Connecticut,United States

Can you calculate only while the current is flowing in its original direstion?  I am not interested in what happens when current reverses direction.  If it is essential for you to know the resistance of the simulated diode, use a value of one ohm while it is conducting and a very high value while blocking, say one megohm.
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Fu-Kwun Hwang
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 « Embed this message Reply #5 on: December 06, 2009, 03:10:04 pm » posted from:Taipei,T\'ai-pei,Taiwan

For RLC and diode circuit:$V_C=V_{diode}+V_L+V_R$,
where $V_L=L\frac{dI}{dt}, V_R=IR,V_C=Q_C/C , I=-\frac{dQ_C}{dt}$

For example:
At t=0; V_L=V_C-I R-V_{diode}=20-0.12*(9+1)-0.8
this can be used to calculate $\frac{dI}{dt}=V_L/L$ which will give us I(t+dt).
With I(t) we can calculate $Q_C(t+dt)=Q_C(t=0)-I dt$
We can repeat the above step to calculate the charge Q(t+dt) and current I(t+dt) for each time step t.

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