# Understanding a parallel resonant circuit

Discussion in 'The Projects Forum' started by wayneh, Nov 2, 2011.

1. ### wayneh Thread Starter Expert

Sep 9, 2010
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In the attached circuit for a pulse generator, there's a small capacitor in parallel to the transformer output. I'd like to know how to choose and optimize the value of that capacitor to increase the pulse intensity into the load, which I don't have a good description of except that it has resistive (1-10K?) and capacitive (100pF?) components and probably no inductance.

I have a general understanding of the parallel resonance provided by an LC-tank. There's a good description on this website here. Choosing the right capacitor makes the LC tank "disappear" (appear open) to the AC frequency passed to it.

In my case, I'm generating brief pulses at ~100Hz, on the rising edge of each input clock wave. Without an oscilloscope, how can I choose a capacitor that will maximize - or at least improve - the output?

2. ### crutschow Expert

Mar 14, 2008
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I believe that circuit is a flyback type. The peak output voltage is determined by the inductance of the transformer and the output capacitance. The transistor causes a current to flow in the primary inductance of the transformer. The energy in that inductive current (1/2 LI^2) is transferred to the secondary when the transistor shuts off. So the smaller the output capacitance, the larger the output pulse.

3. ### SgtWookie Expert

Jul 17, 2007
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Your output will occur when you turn OFF the MOSFET, so it will be at the falling edge of your input.

The diode will result in a very reduced output. You need a snubber instead; or you could omit the diode and let the MOSFETs' built-in 100V Zener clamp it. Make sure you have a heatsink on the MOSFET.

4. ### crutschow Expert

Mar 14, 2008
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Yes, I missed that diode across the primary, which will indeed absorb most of the voltage and energy. You only use a diode across an inductor when you are trying to suppress the transient and absorb the inductive energy. Here you have the opposite. You want all the inductive energy transferred to the secondary. The diode must be removed.

5. ### SgtWookie Expert

Jul 17, 2007
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What the diode does is prevent the rapid collapse of the magnetic field around the primary; which is how the energy gets transferred to the secondary winding.

In a relay or solenoid where you don't care about speed or transferring energy to a secondary or tertiary winding, a regular diode works OK. If you start to care about speed, then adding a resistor in series with a diode makes the field collapse more quickly. A Zener in series with a diode makes the current flow stop yet more quickly.

Jun 5, 2009
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I don't meen to go of topic but how I don't understand how adding a resistor in series with the diode make the field collapse more quickly?

7. ### wayneh Thread Starter Expert

Sep 9, 2010
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I don't get that. After the clock goes high, the voltage at the gate drops to near ground after about 3 times the 1/RC time constant, a couple hundred microseconds in this case and long before the next rising edge. Any DC component is filtered out by the 0.1µF cap, which then discharges any remaining charge "instantly" when the clock goes low. So when the clock goes to ground, how could that do anything to the gate voltage, which is already at (or very near) ground? To my thinking, it just stays at ground most of the time until another rising edge comes along.

Since the "on" surge is as fast as possible and the "off" is slowed by the RC tank, I believe the "on" pulse in the transformer is much larger than the reverse, as the field collapses. The RC tank makes sure that the FET turns off more slowly than it was turned on.

If that is correct, then the diode is mostly irrelevant but does prevent any reverse EMF damage to the FET as the current through the coil is tapered off by the falling gate voltage. I did use the circuit for a while without the diode and added it for good measure. Maybe it's useless.

One thing I can report is that the pulse generator is definitely working as shown, with no capacitor on the output. Using a cap-and-diode peak detector, I see over 100V at the output. I'm just trying to get a bit more bang for the buck by adding a small capacitor.

8. ### SgtWookie Expert

Jul 17, 2007
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If it's just a diode conducting the current, the only things that will slow the flow of current is the resistance of the coil, the wiring, and the Vf of the diode.

If there is a resistor in the current path, current flow through the resistor will cause a voltage drop across the resistor. The voltage drop will be at its' peak when the current flow through the coil is first cut off. It's like applying brakes.

If there is NO path for the current, it's sort of like slamming a car headfirst into a huge boulder; LOTS of energy gets released.

9. ### SgtWookie Expert

Jul 17, 2007
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You're showing a 5k pot, but not what the setting of the pot is.
This is a rather poor arrangement.
It's OK for the MOSFET to turn on slowly, as at T=0, there will be no current flow in the inductor. However, you want the turn OFF time to be as short as possible, as that is when the energy transfer from the primary to the secondary occurs.

Current flow through the inductor builds as a function of voltage, time, and inductance. You need to turn off the current before the inductor saturates, which is a function of the core material. If it is an air coil, it won't saturate.

The gate goes negative, which isn't really a good thing. As long as it doesn't go below ~-15v or so, it probably won't hurt anything.

That's not how it works. You get a larger energy transfer when the current through the inductor gets stopped rapidly.

You'll either help it or hurt it, depending on the PRF of your pulse, the MOSFET on-time, and the resonant frequency. If the transformer pulse isn't in phase with the tank circuit, the existing oscillation will have to be stopped and re-started.

Try running some simulations of it.

10. ### wayneh Thread Starter Expert

Sep 9, 2010
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The pulse becomes quite strong as the pot setting rises above 1K or so. Higher ohms gives higher pulse intensity, since the gate voltage is held high longer.
Again, I'm pretty sure this arrangement is producing a pulse from the high dI/dt on the rising edge. The drop in coil current, as the 0.1µF cap charges, occurs over ≥100µsec which gives a much slower -dI/dt and I doubt there is much reverse pulse from this. One reason for the arrangement is undoubtedly to avoid developing a large current in the coil for any length of time, as this would require a lot more power from the battery.
Good point. It's brief, but the gate voltage must indeed go negative, not merely to ground as I was thinking.
I'll do some actual experimentation and see if I can tell whether the cap helps. Since the pulses are already adequate at just ~20% rotation of the dial on the 5K pot, frankly I'm not sure I really need still more intensity.