# Driving to Resonance

A parallel LC filter has impedance that's theoretically infinite at its resonant frequency. You can exploit this characteristic to decouple a voltage generator from its load. Doing so lets the generator drive any load.

The resonant frequency of a LC filter can be obtained by equating the denominator of its impedance, shown in Figure 1, to zero.

Figure 1 LC filter. The equation for the resonant impedance is below. Usually the resonant frequency is centered at the working frequency, but why?

Consider, for example, the simple circuit of Figure 2. A driver with output impedance Rs (typically 50Ω) has to feed a small load, let’s say 50Ω.

Figure 2 12 V generator with a 50Ω series resistance and 50Ω load.

The maximum voltage that the driver can provide to the load is 6 V, half of its nominal value. This loss of power supplied to the load is caused by the current in the load, which creates a voltage drop of 6 V that represents a loss. To solve this problem we use a buffer with high input impedance and low output impedance, as shown in Figure 3.

Figure 3 Circuit from Figure 2 with a decoupling buffer.

A high-impedance circuit won't create an additional voltage drop. The voltage applied to the load is equal to the input voltage, so there is no power loss. An LC parallel resonant circuit has high input impedance, as the buffer in Figure 3 shows. You can use the circuit to decouple the driver from the load.

A few months ago, I measured the small-signal parameters of a military-grade power switch. The measurements had to match the requirements of the guideline MIL-STD-750. One of the methods described in this standard is for small-signal measurements. This method contains an LC resonant block to drive the switch by decoupling the driver signal generator from the switch. (See Figure 4.)

Figure 4 One of the MIL-STD-750 test methods to measure a small signal parameter, the short circuit input impedance of a radiation hard power switch.

To perform my task, initially I had to test an electrical circuit assembled according to the diagram in Figure 4. I expected that the LC filter would supply the input voltage to the DUT. The reality, however, was different. I have found in the laboratory some waveforms very much more attenuated than at the entrance. So, I simulated the circuit with LTspice.

Figure 5 LTspice simulation of the circuit of Figure 4.

The problem was that at DC, the inductor behaved like a short circuit by altering the working point of the power switch, in this case the power BJT, but the same happened with a power MOSFET.

I tried to insert a decoupling capacitor, but this solution affected the frequency response of the system around the resonant frequency, which had been designed to 1 MHz, the frequency at which I wanted to measure the parameters of small signal of the device according to MIL-STD-750.

I solved the problem by modifying the circuit. I assembled a first prototype yarn and then a PCB using a bias resistor bridge. The result of the simulation is shown in Figure 6.

Figure 6 Updated circuit from Figure 4 (here used in constant-current configuration).

Using the circuit in Figure 6, I measured in the laboratory parameters of small-signal device and fixed the decoupling problem. Have you ever used the LC resonant filters for driving loads? Have you had any problems related to the operating point of the circuit because of the short in DC of the inductor?

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## 4 comments on “Driving to Resonance”

1. Scott Elder
October 25, 2013

Hi Paolo,

I think the circuit of figure 4 should work.  Certainly the base of the PNP is pulled (shorted) to 0V, but since the emitter is driven by a current source, and there is no alternative dc current path but through the emitter, the emitter would simply lift up from 0V to about +0.7V. [This makes the physical circuit design a bit tougher because now you need another power supply voltage to build a current source compliant to ~+1V, but LT SPICE would let you get away without that requirement).

Under the above condition, and with such a large capacitor on the emitter, the emitter is essentially grounded for ac.  So as your oscillograph shows, the ac voltage at the emitter is essentially 0V.

The base voltage does not swing +/- 1V because (a) the hie forms a signal divider with the 3kohm input resistance and (b) the LC circuit Q is not infinity and is shunted by the hie.

Looking at the oscillograph that you state is working, it isn't enabling a measurement of hie, but instead showing that the voltage on the emitter is following the voltage on the base.  That could never be the case when the emitter is ac shorted to the supply through 5000uF, but certainly works when the emitter is essentially open (i.e. 100 kohm).  I am a bit puzzled though that the oscillograph dc voltages seem to be the same when I would expect a ~0.7V difference.

I'm also wondering about the 1MHz signal test.  The oscillographs show ~1Khz which is also near the resonant frequency of the large L and large C.  Perhaps that's a misprint?

Perhaps I'm missing something, but it seems like you should have been able to see this work on LT SPICE.  What am I missing?

2. D Feucht
October 27, 2013

I concur with Scott on his analysis and agree that this is a rather odd way of designing such a circuit.

Also, Fig 3 is somewhat puzzling too because 50 ohms arises typically when transmission lines (interconnect cables) are to be used to connect amplifier outputs to loads. In this case, the buffer amplifier defeats the purpose of back-terminating the amplifier.

There are better ways of “isolating” amplifiers from reactive loads than using resonant circuits. See my book, Designing Dynamic Circuit Response , chapter 2, “Output Load Isolation” (www.scitechpub.com/r/category.php?cPath=5) for more on this.

3. etnapowers
October 28, 2013

@D_Feucht: that's correct the circuit of figure 3 was not properly designed, due to the insulation of the load was missing, your book is very interesting.

4. etnapowers
October 28, 2013

@Scott: I agree too with your analisys, the circuit of figure 4 worked properly because it was an update of the circuit of figure 3 (not working due to missing insulation)

The frequency was 1MHz, corresponding to a period of 1us, many periods are considered to see a steady state condition of the circuit.

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