Line Driver and Oscillating Circuit

1. Jun 25, 2009

lankan_ice_405

So I have this line driver using an IR2110 to amplify a square function tuned for a specific frequency. Its used to cause oscillations in an LC circuit (pulsed at the same frequency, 1/root LC).

Experimentally though I can't find that resonance frequency. As I go above the theoretical resonance freuqency, Vinductor also increases 2-3kHz then it has a sudden fall. (I think this is because the Line Driver is not fit to go that much above its tuned frequency)

If I take this same circuit, attach a resistor onto it and pulse it using a function generator, there is a visible peak of V inductor...that's not visible when using the line driver in between.

What's going on here when I use the line driver?

2. Jun 25, 2009

vk6kro

A line driver has a low output impedance and if you drive a parallel resonant circuit directly with it, this low impedance will become part of the circuit and destroy the resonance you are trying to observe.

or look for "tank circuits" on the current PF page.
and scroll down to see the diagrams by Bob S.
They include ideal setups for observing resonance effects.

3. Jun 25, 2009

lankan_ice_405

Its actually a series LC Circuit but I think I face the same problem.

Anyways, do you think you could explain Bob S's drawings, and how it relates to the issue I'm facing?

Last edited: Jun 25, 2009
4. Jun 25, 2009

vk6kro

Can you supply your exact circuit with component values?

As you are using a series tuned circuit, you should be able to put your function generator in sinewave mode and sweep it across the frequency you expect the circuit to be resonant at.

It will need some series resistance (as shown in Bob S's diagram, above the graph) to show the drop in impedance of the circuit at resonance. You can also observe the increase in voltage across the inductor or the capacitor at resonance.

A line driver has a sufficiently low output impedance to generate the same voltage regardless of load (within limits). So, you would not see any change in voltage despite an increase in current at resonance.

If your assignment is to use a square wave, you would be aware that square wave generators produce harmonics at odd multiples of the fundamental frequency. So, a square wave at 15 KHz will have output at 45 KHz as well as 15 KHz but these outputs are actually sinewaves. It is only the total waveform that appears as a square wave.

5. Jun 26, 2009

lankan_ice_405

I would except the line driver is using the IR2110 High/Low Side Driver. I cannot input a sine waveform, only square. The output of this is also a square wave, which goes to a 0.2uF cap and ~310uH inductor.

Last edited: Jun 26, 2009
6. Jun 26, 2009

vk6kro

That tuned circuit should resonate at 20212 Hz.

If you have a square wave out, one way to drive the tuned circuit is to turn it into a Pi network.

You would need a capacitor of about 1 or 2 uF (not an electrolytic) across the line driver low side. Then the inductor from the LO output to the capacitor, then the capacitor to the common side of the line driver low side.

Watch the voltage across the capacitor with an oscilloscope. It should peak at about 20.2 KHz.