LC ringing wave gen

I want to feed the squarewave output of my bench (audio) function generator through an LC network that will cause each pulse to ring at around 10MHz. IOW each pulse then appears as a damped wave, thus effectively applying 10MHz modulation to the original squarewave. The load varies experimentally between 100 Ohms to 10K Ohms. Can anyone tell me what calculation to use, and if possible, provide an example for me to follow of how the formula is used?

Thank you very much for any advice.

Bob Steiner

Reply to
Bob Steiner
Loading thread data ...

I'll take a stab at this one...

As I recall from a filter book: LC circuit Q is: Q = Q1*Q2/(Q1+Q2) The bigger the circuit Q, the longer the sustain.

Q1 = Xl/Rs Q2 = Rl/Xc

in--Rs--L-+---+---out | | C Rl | | com com

D from BC British Columbia Canada.

Reply to
D from BC

Oops..forget to mention...

L and C resonant as per f= __1____ ____ 2*pi*\\/ LC

D from BC British Columbia Canada.

Reply to
D from BC

Just two thoughts:

1/ Differentiate the square wave (light capacitive coupling from the function generator) 2/ Use a buffer amplifier to reduce the loading effects.
Reply to
Charles

--
Assuming that your edges are fast enough to cause a passive LC to
ring, you could do this: (View in Courier)

SQIN>--+
       |
      [L]
       |
       +-->OUT
       |
      [C]
       |
GND>---+-->GND

The frequency of the damped oscillation will be:

                1
     f = --------------
          sqrt (2piLC)

and, at resonance, with a fixed load, the output amplitude should
only vary with the amplitude of SQIN.
Reply to
John Fields

Does your audio generator have fast enough rise and fall times on the square wave output to excite a 10MHz circuit? If so, as others have shown, a series LC should do the trick. It's possible to get fancier, of course; you can build an LC bandpass filter that will have a different natural response.

The couple designs I saw posted fed the L from the generator, and have the C to ground. I'd be inclined to reverse that, and in fact, to measure the resonant frequency of a tank, or to measure an inductance using known capacitances, I form a high Q parallel tuned tank, one side grounded, and the other side excited through a relatively small capacitance to the square wave. That can give you more flexibility in adjusting the Q. But you may also find it advisable to put a buffer amplifier at the output of the LC circuit, whatever it is, because your 100 ohm load is VERY low to be using with practical LC values. Get LTSpice and simulate the circuit. In a few minutes, you can get a feel for how it all works. Be sure to include the source resistance of your generator in the model.

Cheers, Tom

Reply to
Tom Bruhns

These are the ideas that come to my mind. It also depends on your application, of which I have no clue.

there is this

Version 4 SHEET 1 880 680 WIRE 128 160 0 160 WIRE 272 160 192 160 WIRE 192 192 192 160 WIRE 272 240 272 224 WIRE 0 272 0 240 WIRE 192 320 192 272 FLAG 272 240 0 FLAG 0 272 0 FLAG 192 320 0 SYMBOL ind 176 176 R0 SYMATTR InstName L1 SYMATTR Value 1.2µh SYMBOL cap 256 160 R0 SYMATTR InstName C1 SYMATTR Value 208pf SYMBOL diode 128 176 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D2 SYMBOL voltage 0 144 R0 WINDOW 3 -221 -105 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value PULSE(0 1 0 0 0 .0005 .001 1) TEXT -16 112 Left 0 !.tran 0 .005 .000617

or this at first glance

Version 4 SHEET 1 880 680 WIRE 208 112 64 112 WIRE 240 144 224 144 WIRE 208 160 208 112 WIRE 224 160 224 144 WIRE 224 160 208 160 WIRE 240 160 224 160 WIRE 448 160 304 160 WIRE 224 176 224 160 WIRE 240 176 224 176 WIRE 240 192 224 192 WIRE 224 208 224 192 WIRE 224 208 208 208 WIRE 240 208 224 208 WIRE 256 224 256 208 WIRE 208 240 208 208 WIRE 208 240 64 240 FLAG 256 224 0 FLAG 64 192 0 FLAG 64 320 0 SYMBOL voltage 64 96 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value PULSE(0 5 0 0 0 .0005 .001 2) SYMBOL voltage 64 224 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V2 SYMATTR Value PULSE(0 5 0 0 0 .05us .1us 20000) SYMBOL Digital\\\\and 272 112 R0 SYMATTR InstName A1 TEXT 30 348 Left 0 !.tran .002

Bob

Reply to
sycochkn

--
Oops, LOL...

                1
      f = --------------
           2pi sqrt(LC)
Reply to
John Fields

What's the range of the audio input?

Why does the load vary 100:1?

How long do you want this ringing to last?

What output amplitude do you need?

How accurate do you want the 10MHz?

Reply to
Fred Bloggs

On Sun, 24 Feb 2008 18:31:34 -0800 (PST), Tom Bruhns wrote:

---

In the series resonant case, if the OP wants to keep the DC part of the square wave intact after the ringing has stopped, then the inductor must be connected to the hot side of the generator and the capacitor grounded:

Version 4 SHEET 1 940 740 WIRE -48 -128 -80 -128 WIRE 80 -128 32 -128 WIRE 448 -128 416 -128 WIRE 576 -128 528 -128 WIRE 80 -96 80 -128 WIRE 576 -80 576 -128 WIRE -80 -16 -80 -128 WIRE 416 -16 416 -128 WIRE 576 16 576 -16 WIRE 720 16 576 16 WIRE 80 32 80 -16 WIRE 224 32 80 32 WIRE 224 48 224 32 WIRE 576 48 576 16 WIRE 720 48 720 16 WIRE 80 64 80 32 WIRE -80 160 -80 64 WIRE 80 160 80 128 WIRE 80 160 -80 160 WIRE 224 160 224 128 WIRE 224 160 80 160 WIRE 416 160 416 64 WIRE 576 160 576 128 WIRE 576 160 416 160 WIRE 720 160 720 128 WIRE 720 160 576 160 WIRE -80 224 -80 160 WIRE 416 224 416 160 FLAG -80 224 0 FLAG 416 224 0 SYMBOL voltage -80 -32 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value PULSE(0 5 0 1e-9 1e-9 100e-6 200e-6) SYMATTR InstName V1 SYMBOL ind 64 -112 R0 SYMATTR InstName L1 SYMATTR Value 159e-6 SYMBOL cap 64 64 R0 SYMATTR InstName C1 SYMATTR Value 159e-12 SYMBOL res 208 32 R0 SYMATTR InstName R1 SYMATTR Value 10k SYMBOL res 704 32 R0 SYMATTR InstName R2 SYMATTR Value 10k SYMBOL ind 560 32 R0 SYMATTR InstName L2 SYMATTR Value 159e-6 SYMBOL cap 560 -80 R0 SYMATTR InstName C2 SYMATTR Value 159e-12 SYMBOL voltage 416 -32 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value PULSE(0 5 0 1e-9 1e-9 100e-6 200e-6) SYMATTR InstName V2 SYMBOL res 48 -144 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R3 SYMATTR Value 50 SYMBOL res 544 -144 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R4 SYMATTR Value 50 TEXT -72 184 Left 0 !.tran 300e-6

The parallel resonant way seems too complicated, so for simplicity's sake I'd use the series resonant circuit.

-- JF

Reply to
John Fields

I suspect the "experimentally" determined load impedance variation over

100:1 range is a significant reactance making the whole exercise the usual SED waste of time...
Reply to
Fred Bloggs

Yes, upon re-reading the OP's posting, I see he does want to keep the square wave. My original vision was that he just wanted 10MHz damped pulses; not sure why I missed the "keep the square wave" part. I'm not sure if anyone has mentioned it yet, but the result of the circuits/suggestions we've posted is NOT modulation of the square wave; it's simply linear filtering in the frequency domain. As F.B. notes, it's probably all an exercise in futility.

Of course, the circuit you posted works OK for the 10k load, nicely underdamped, but if you change the load over the range the OP wanted, it goes from the underdamped to a very overdamped case. I believe you will find that with the simple one inductor, one capacitor case, it's not possible to make a circuit that stays underdamped over the whole range of load values, 100-10k ohms, given a 50 ohm source. If amplitude is not an issue, one could simply add a resistor in series with the load, to limit the loading on the LC filter; or add an output buffer amplifier.

I suppose, without going through the math and proving it to myself, that the maximum loaded Q for the circuit you posted occurs when Xl=Xc=sqrt(R1*R3), and is equal to sqrt(R1/R3)/2. I suppose your values of 159pF and 159uH were picked to make the simulation output look prettier for an audio range square wave; substituting 1.125uH and

225.08pF to get 10MHz instead of 1MHz and to maximize the Q for the 100 ohm load case, you do see a quickly damped sinusoid for all loads. The Q for higher load resistance is limited by the source resistance, and the Q varies over much less range than if you use a higher L/C ratio.

Cheers, Tom

Reply to
Tom Bruhns

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.