[SOLVED] How to calculate the total current in a series RL circuit with pulse train wave input

[iamakda]

Hi everyone,

I have a series circuit. Inductor of 10uH is in series with 100 ohm resistor with pulse train voltage of 6.59 V with a duty cycle of 0.31 and frequency of 180 kHz. I want to calculate the total current flowing in this series circuit. Can anyone help me finding this? I will be really grateful.

Thank you.

 

[godfreyl]

Peak current is very close to 6.59V/100Ω = 65.9mA
Average current = 0.31*6.59V/100Ω = 20.429mA

The waveform looks like this:

 

[iamakda]

Hi godfreyl,

First of all thanks for the reply. May I ask why you have taken only R in consideration while calculating the current? Are you sure you have calculated it correctly? And what is 0.31 factor while calculating averange current?

Thanks.

 

[godfreyl]

May I ask why you have taken only R in consideration while calculating the current?

The inductance doesn't affect the average (DC) current.

And what is 0.31 factor while calculating averange current?

0.31 is the duty cycle.
I.E. The voltage is 6.59V for 31% of the time, and 0V for 69% of the time. Thus the average voltage is 0.31* 6.59V.

 

[iamakda]

The inductance doesn't affect the average (DC) current.


But when we will calculate the peak current so even at that time we will not consider the effect of XL?

One more thing, if I excited series RL circuit with Sine wave is same as if the excite the same circuit with pulse train wave?

 

[godfreyl]

But when we will calculate the peak current so even at that time we will not consider the effect of XL?

Yes, we should consider it. If the frequency or inductance is much higher, or the resistance is much lower, then it will make a big difference. In this example it only makes a very small difference.

XL of 10uH at 180KHz = 2*pi*F*L = 11.31Ω
Thus the impedance of the inductor is much lower than the 100Ω resistance.

Also, you can see that the waveform looks almost square, only the leading corners are rounded, but most of the top and bottom looks quite flat.

One more thing, if I excited series RL circuit with Sine wave is same as if the excite the same circuit with pulse train wave?

The current waveform will be different - it will be a sin wave, but phase shifted from the voltage waveform.

First work out the DC current as I(DC) = V(DC)/R
Then work out the AC current as I(AC) = V(AC)/Z where Z = R+XL
Finally add and subtract the DC and AC values to get the minimum and maximum current.

 

[iamakda]

Yes, we should consider it. If the frequency or inductance is much higher, or the resistance is much lower, then it will make a big difference. In this example it only makes a very small difference.

XL of 10uH at 180KHz = 2*pi*F*L = 11.31Ω
Thus the impedance of the inductor is much lower than the 100Ω resistance.

Also, you can see that the waveform looks almost square, only the leading corners are rounded, but most of the top and bottom looks quite flat.

The current waveform will be different - it will be a sin wave, but phase shifted from the voltage waveform.

First work out the DC current as I(DC) = V(DC)/R
Then work out the AC current as I(AC) = V(AC)/Z where Z = R+XL
Finally add and subtract the DC and AC values to get the minimum and maximum current.

Thanks a lot godfreyl for your kind help

 

[kobeismygod]

Yes, we should consider it. If the frequency or inductance is much higher, or the resistance is much lower, then it will make a big difference. In this example it only makes a very small difference.

XL of 10uH at 180KHz = 2*pi*F*L = 11.31Ω
Thus the impedance of the inductor is much lower than the 100Ω resistance.

Also, you can see that the waveform looks almost square, only the leading corners are rounded, but most of the top and bottom looks quite flat.


The current waveform will be different - it will be a sin wave, but phase shifted from the voltage waveform.

First work out the DC current as I(DC) = V(DC)/R
Then work out the AC current as I(AC) = V(AC)/Z where Z = R+XL
Finally add and subtract the DC and AC values to get the minimum and maximum current.

i have a question，dose the XL=10uH at 180KHz = 2*pi*F*L = 11.31Ω will affect the average current or peak current?

 

[iamakda]

i have a question，dose the XL=10uH at 180KHz = 2*pi*F*L = 11.31Ω will affect the average current or peak current?

In this case it will not effect average and peak current much because of the very small value of XL as compared to the R.

I hope this will help.