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Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
Can someone explain, if the inverter voltage is a square wave, why the inverter current is a triangle wave when one is above resonance (inductive load).
Also, when one is below resonance (capacitive) why is there ringing during inverter transitions?
Richie, I've seen your website, but maybe you or someone can give a more detailed reason with respect to the physics.
Registered Member #2463
Joined: Wed Nov 11 2009, 03:49AM
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Posts: 1546
When voltage is applied to L the current rises slowly because of back emf, as per Lenz law. Over the leading edge, rise time of current is retarded at rate of L(henry)/R(resistance of load). At flat of wave, voltage is constant, no back emf,current is not changing, is maintaining Bfield at Voltage/turn. At trailing edge of first half cycle ,voltage falls, back emf changes polarity and process repeats down (through zero) to rising leading edge of negative half cycle. Summary: Current follows the magnetic field, Process is integration, Works same in voltage mode with capacitors and resistors.
Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567
I look at this link and it says that the voltage falls exponentially with time
V = L*di/dt
So, when the voltage takes a step up the current is initially zero as the inductor resists a change in current. The current starts to climb, exponentially, until it reaches a steady state determined by the R of the RL circuit. Final current is V/R. So, how is it linear, unless it approximates a linear function if R is large compared to L_reactance?
Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567
Thanks, Peter. I am aware that the integral is a linear slope. I think my source of confusion is that the current is a linear slope, but the voltage is an exponential decay in a RL circuit subjected to a step input.
This linear response assumes the voltage is always constant across the inductor. In an RL circuit, as current rise, RI increases and Ldi/dt decreases in order to maintain the constant voltage, V = RI + Ldi/dt
I will have to set up a small breadboard and just scope the waveforms, and see what happens when I vary R.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
IamSmooth wrote ...
So, how is it linear, unless it approximates a linear function
That's exactly how it is linear. If you only see a small part of the exponential, then it looks like a straight line.
In Laplace speak, it's the difference between 1/s (a true integrator) and 1/(s+1) (a RC or LR lowpass filter) For values of s large compared to 1, which is to say frequencies high compared to the time constant of the filter, they both behave practically the same.
To be pedantic, we're dealing not with a LR or RC, but a LRC circuit. So the triangle wave you're seeing is a small part of a damped sinusoid, not an exponential.
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induction heater waveforms
