Crazy Resonance Problem

[David Bridgen]

Hi David,

..... When I reduce the Q of a resonant circuit, I damp it. I'm left with a dampened circuit .....

No. You are left with a damped one.

And hello to you, audioguru. I've only just discovered this forum. Been a fairly regular contributor to others.

Regards.

 

[David Bridgen]

[quote="Ron H]

2. To deaden, restrain, or depress: “trade moves... aimed at dampening protectionist pressures in Congress” (Christian Science Monitor).

[/quote]


Hi Ron. I think you were a regular contributor to the sadly defunct PopTronics forum?

The Christain Science folk have it wrong too. It should be "damping ...pressures"

 

[Optikon]

Those inductors you have drawn are really transmission lines, and they will resonate and cause your transistor to oscillate. Try putting the resistor in series with the other end of the cable (the +8 volt end). This is counter-intuitive, but it should terminate the cable in something near its characteristic impedance, reducing reflections and hence ringing/oscillation. Putting the resistor near the collector will dampen the resonance, but it is a poor termination.

I think the original author has his problem under control now but I'd like to hear some elaboration on this point. I feel like you are talking about source termination of his "wire" transmission line in this case. My undersanding is that if you have a mismatch at the load, you will get at least one reflection. As you suggest, source terminating will stop that reflection and prevent a second one from occuring. But in general, the only way to eliminate any reflections is both source and load termination.

1) Source and load terminated (BEST)
2) Source mismatched, but load terminated (Better)
3) Source terminated, but load mismatched (Better)

Between choices 2 and 3, which is better? and why?

 

[David Bridgen]

Chris,

have you tried damping in parallel with the line, either across the L1 section or the L2?

 

[Nigel Goodwin]

Has this actually been built? - or is it just simulated?.

 

[Roff]

2. To deaden, restrain, or depress: “trade moves... aimed at dampening protectionist pressures in Congress” (Christian Science Monitor).

Hi Ron. I think you were a regular contributor to the sadly defunct PopTronics forum?

The Christain Science folk have it wrong too. It should be "damping ...pressures"

David, I did participate in that forum. Actually, there are still vestiges of it remaining:
https://www.poptronics.com/forums/electronic_bench/
https://www.poptronics.com/forums/resource_bin/

Ron

 

[Roff]

Edit:
Ramblings of a madman.

 

[FRIED]

The amplifier is too damn fast. I have stopped the resonance going up to 30V by putting a 1nF cap between the output of the AD8061 and the inverting output (inplace of the 15pF cap that was already there). All I am left with is one 30V spike when the FET switches. Any ideas on how to get rid of this? Or is it just a snubber circuit. Its just that the PCB is designed and I don't want to do any crazy stuff.

Use a slower amp. Edit: I also tried this in a simulator and it fails shortly after the fet switches with no added inductance. Changing to a slower amp fixed.

 

[Roff]

Those inductors you have drawn are really transmission lines, and they will resonate and cause your transistor to oscillate. Try putting the resistor in series with the other end of the cable (the +8 volt end). This is counter-intuitive, but it should terminate the cable in something near its characteristic impedance, reducing reflections and hence ringing/oscillation. Putting the resistor near the collector will dampen the resonance, but it is a poor termination.

I think the original author has his problem under control now but I'd like to hear some elaboration on this point. I feel like you are talking about source termination of his "wire" transmission line in this case. My undersanding is that if you have a mismatch at the load, you will get at least one reflection. As you suggest, source terminating will stop that reflection and prevent a second one from occuring. But in general, the only way to eliminate any reflections is both source and load termination.

1) Source and load terminated (BEST)
2) Source mismatched, but load terminated (Better)
3) Source terminated, but load mismatched (Better)

Between choices 2 and 3, which is better? and why?

The collector is the (current) source. The 8 volt battery looks like a short circuit at the load end. If you have a line which is shorted at the far end, the short will cause 100% reflection. If the source is an open circuit (current source, as we have here), then the wave reflected from the short will reflect again. Not a good thing unless you want to generate ringing. The resistor I suggested adding would be a load termination.

If you have a tline with no parasitics (L and/or C) at the load end, then a perfect load termination (R=Z0) will not generate a reflection, so source termination is not necessary, regardless of the impedance of the source. The only effect of a source termination would be 50% signal amplitude at the load. If you need to place taps (receivers) at intermediate points on the line, source and load termination is necessary.

If you have a tline with no parasitics (L and/or C) at the load end, the absence of a load termination will cause 100% reflection,but a perfectly matched source termination will absorb the reflection completely, so the waveform at the load will still be perfect. Intermediate points will still look bad, because the reflected wave interferes with the incident wave.

If you have a line with source parasitics, but no load parasitics (as we have in my suggestion), then load termination will still not generate any reflection.

Having said all that, nearly all practical applications have parasitics on both ends, I think that, if you need to preserve signal amplitude, load termination is generally preferable to source termination for signal integrity, partly because proper source termination must include the impedance of the driver, which frequently is nonlinear and varies from device to device. Receivers, on the other hand, generally have input impedance much higher than the transmission line (which will never be more than about 120 ohms), and so are easier to terminate. One disadvantage of load termination is that it dissipates more power than source termination. As you said, for best signal integrity, source andload termination is best.

 

[cchalmers]

resonace

Thanks for all your replies. Although I think a whole new thread is developing.

Sorry I have been away for last couple of days trying things. I have now got a spice simulation running of it and I am following through.

The circuit was simulated before hand (not by me) but by someone who is no longer contactable. I breadboarded this before I actually made the circuit and although there was a wee bit of ringing around the FET switching edges on the breadboard, this was small and thought just to be because of the inducatance in the wire interconnects. This would disappear when on a PCB. However, the ringing turned in to a 30V pulse.

I slowed the amp down by replacing the 15pF cap with a 1nF cap. I also added a 100ohm gate resistor to the FET. The only thing I am left with is a ~15V pulse on the FET switch on and off. I am analaysing the spice simulation at the moment.

FRIED, you said you simulated the AD8061. What did you mean that the amp fell over without any inductance? In my simulation, without any inductance, there are spikes but the amp doesn't go in to resonace.

Thanks all

Chris

 

[fredsm]

In an ideal world, the source, line, and load impedances
would be perfectly matched. Since this is almost never
achievable, the best one can do is get as close as possible.
And since most power sources are of a low-impedance
nature, matching networks or the equivalent are generally
necessary. Note that in a mis-matched transmission line,
there will be standing waves (current and voltage peaks)
at sub-multiples of the wavelength, always to the detriment
of maximum power transfer. In an environment where it is
difficult to control the line impedance (circuit board traces,
connectors, etc.) it is preferable to terminate both ends of
the line in a low impedance, to swamp the effects of the
variances in the line. It is preferable to lose some signal
amplitude, rather than to compromise the signal waveshape
(especially in digital signals, which are subject to the
effects of the transmission line causing signal distortion.

As to the question of whether it is preferable to match
the source or load ends, the load termination will prevent
reflections from that end only, and the source termination
will tend to reduce the effects of the reflections on the
source (harmonic generation, intermodulation, and
other nasty types of signal distortion). Either one will
reduce signal amplitude.

<als>

 

[Roff]

Perhaps you should consider getting rid of the FET switch and doing the switching in front of the op amp. See sim results below. As you know, the hardware may behave differently.
I show switching with an analog switch. You could do the switching digitally in front of the DAC, which would be preferable if your DAC is fast enough.