R-L Circuit RMS current

[Misterbenn]

Hello,

I'm hoping you can help me with some working here. I'm trying to find the rms current in a freewheeling (flyback) circuit with a 2.2mH inductor and a 0.1ohm resistor.

Ok i've now had time to attach a picture of my simulation. So you can see we have a current source in series with an IGBT switch & snubber with a load inductor and freewheeling thyristor/diode/resistor network. The thyristor is constantly triggered.
At time t=0 to t=0.001 the source supplies 1500A and the IGBT is closed. so the current flows in the inductor and the freewheeling components are reverse biased.
At t=0.001 to 0.02 The source current is zero and the IGBT is open breaking the current path (switching the sunbber into the circuit to eliminate large voltage spikes). The current circulating in the inductor is comutated to the freewheeling path. - i want to calculate the freewheeling current in the thyristor as this is a selection parameter.

a graph of my simulation results is attached, showing current in the inductor (IL), current in the freewheeling path (Ifreewheeling) and the rms current (mislabled as I2t). The rms current measurment has a time constant of 20ms (henze the initial delay)


So i'm trying to work out my rms current - simulations on PSIM are saying 620 A(rms) - but i want to show this mathmaticaly (or disprove if thats the case)

So:

To work out rms current:
[LATEX]
\mathcal{I}_{rms} = \sqrt{I^2}
[/LATEX]

Current in an inductor at time t:
[LATEX]
I(t) = I(0)e^{-\frac{t}{\tau}}
[/LATEX]

Calculating the time constant
[LATEX]
\tau = L/R = 22ms
[/LATEX]
[LATEX]
I(0) = 1500
[/LATEX]


so intergrating the current over a time 0-0.02s, finding the mean and then rooting this gives:
[LATEX]
\therefore \mathcal{I}_{rms} = \sqrt{\frac{1}{0.02 - 0} \int_{0}^{0.019} {(1500 e^{-\frac{t}{0.022}}})^2 \,dt}
[/LATEX]

now i'm a bit lazy so i let the wonder of the modern world do some of the work and wolframAlpha www.wolframalpha.com calculates my intergral as:
[LATEX]
\mathcal{I}_{rms} = \sqrt{\frac{1}{0.02} [-24750 e^{-90.9091\times 0.019}] - [-24750 e^{-90.9091\times 0}]}
[/LATEX]

[LATEX]
\mathcal{I}_{rms} = \sqrt{\frac{-24750}{0.02} [0.177768] - [1]}
[/LATEX]


[LATEX]
\therefore \mathcal{I}_{rms} = 1008.71 A [/LATEX]
Which is different to my simulated result

So can anyone confirm my calculation? or find a reason for the discrepancy? It may be due to a snubber C value of 1.98 mF, all components in parallel, source, inductor, resistor and capacitor

 

[Misterbenn]

[LATEX]
\begin{align}
\frac{dV}{dt} &= \frac{1500}{1.98\times 10^{-3}}
\frac{dV}{dt} &= \frac{25}{33}\frac{A}{us}
\mathcal{V}_{pk} &= I\sqrt{\frac{L}{C}}
\mathcal{V}_{pk} &= 1581 V
\end{align}
[/LATEX]

 

[Misterbenn]

[LATEX]\mathcal{I}_{rms} = \sqrt{\frac{1}{t2-t1} \int I(0) e^{-\frac{t}{\tau}}}
[/LATEX] and again

 

[Ratchit]

Misterbenn,

What is the value of the simulation? Inquiring minds would like to know. You should use 0.021 as the upper limit of the integral, not 0.020. That gives a different value.

Ratch

 

[Misterbenn]

Ratch,

The rms Value from my simulation is 620 A(rms).
actualy is should use 0.019 as the upper limit of my intergral as this is the time between swiches - the final 1ms of the 20ms period has the freewheeling part in reverse bias so no current

 

[MrAl]

Hi Misterbenn,

You had your calculation idea right i think and you integrated correctly too. Perhaps you need to adjust your time periods.

When you want to find the RMS value you integrate over the period from 0 to 0.021 as Ratch said, but when you integrate over a period where part of the wave is zero you only have to integrate over the period that is non zero. Thus when you integrate from 0 to 0.021 it's the same as integrating from 0.020 except integration from 0 to 0.020 allows you to define the wave the way you did. If you try to integrate from 0 to 0.021 then you have to find a way to make the exponential come out to zero for that short time 0.001 where the curved triangle wave no longer exists. Thus, integrate the exponential from 0 to 0.020 not 0 to 0.021 unless of course the time is really 0.019. The averaging period however is still 0.021 so you would still divide by 0.021.

If you have two parts to the wave you have to do them independently and add the results quadratically to get the total RMS value of the wave. Thus a pulse plus triangle would mean two different calculations followed by a quadratic summation (sqrt of the sum of squares).

Example:
exponential part goes from 0 to 0.010 and the total period is 0.020.
You would integrate from 0 to 0.010 but divide by 0.020. That's because during the time you dont integrate you assume the wave is zero and that is part of the cycle too but doesnt add anything to the rms value.

 

[Misterbenn]

Thanks guys, the calculation looks about right to me. I'm missing the capacative effect of my snubber circuit but i think that should have only a small effect on the Irms.

Anyway i'm going for a thyristor with 1800 Irms so should be ok. Just would be interesting to know why the simulated and calculated values dont match up. I bet it has somthing to do with the simulation program rather than the hand calculation.

 

[MrAl]

Hi again Misterbenn,

I just noticed you posted a waveform. That waveform is not just an exponential, but is both an exponential for part of the time and a pulse for the rest of the time (short time).
If that's the right waveform then you need to calculate the rms of the pulse too and add that to the rms of the exponential as the square root of the sum of squares. That will be the total RMS value of the wave.

Interesting that wave looks like it dips a little too low too for 20ms. You can check by using your exponential formula and calculate the lower value of current after 20ms (or whatever time period you are using) with a starting current of 1500 amps. Using the values you gave for L and R and 20ms time period it comes out to around 604 amps at the lowest point.

 

[Misterbenn]

MrAl,

It's the rms current in the flywheel circuit i'm trying to calculate - i think you were looking at the inductor current?? The pulse you discribe relates to a current zero in the freewheeling circuit- this is why i'm only intergrating for 19ms not the full 20ms duty cycle

 

[MrAl]

Hello,

Yes that's what i thought. So if you integrate from 0 to 19ms and divide by 20ms you should come out with a decent approximation. If the circuit is not as perfect as in theory though there are going to be some differences. If you are worried, create a more theoretical circuit and work with that first to see if you are doing it right.

 

[Misterbenn]

sadly can't do that as the system is already in place. My thyristor is an addition to solve an issue we're currently experiencing. I'm giving it a healthy margin for error by selecting a thyristor with Irms = 1800A this is higher than the maximum system output so i don't foresee any problems.

I may contact PSim thought and try to find out the reason for the discrepancy in our calculations as its not insignificant

 

[Misterbenn]

I've another query, trying to calculate dv/dt now in A/us units.

[LATEX]
\frac{dV}{dt} = \frac{I}{C} \frac{V}{s}
[/LATEX]

maximum at I(max) which is 1500

[LATEX]
\frac{dV}{dt} = \frac{1500}{1.98 \times 10^ {-3}}
[/LATEX]

[LATEX]
\frac{dV}{dt} = \frac{25}{33}\frac{V}{us}
[/LATEX]

Correct? the values seems quite low - though i do have a large snubber C, but again simulations are showing a few volts per us

 

[MrAl]

Hi,

You mean V/us right?

 

[Misterbenn]

Yes sorry, silly mistake