Audio guru,

Thanks mate. A couple of points on what I am doing. The brightness is set by the AD8061 feedback circuit. The reason I am doing this is that the enable isn't really PWM. This LED array shines red light on to a microdisplay at a specific time. The brightness is not that important but it has to be balanced with the green and the blue LEDs. The brightness has to be controlled so that you can illuminate the display brighter. So the enable is really a timing pulse to switch on the LED.

You are absolutely on the money with 1us not giving a lot of light output. This is the smallest pulse segment applied to the LED. It is followed by another one, maybe a longer one and so on and so on. Its really a pulse train of timing pulses.

I agree that the AD8061 is causing a bit of the problem. I have found that its the cause of the continued resonance. It feeds back the inital pulse caused by the switch on of the LED and puts it through the NPN transistor. 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.

All you guys help is much appreciated.

Thanks in advance

Chris

 

Hi Chris,
Hurray, you fixed it! (Almost)
I didn't mention slowing-down the opamp because I thought there was a reason for using a 30MHz one so I had visions of a moving, video-modulated microdisplay like those pendulums or spinning thingies.
Snub the darn 30V spike.

 

Optikon, in one post, says "dampen any resonant behaviour" and in another "to dampen it out", and audioguru says "will certainly dampen it".

To dampen something is to make it slightly moist.

cchalmers has got it right with "before it is damped" and "it should damp down".

 

Hi David,
I see that you speekuh zee Queen's angrish.
If I make a towel wet, I dampen it. I'm left with a damp towel.
When I reduce the Q of a resonant circuit, I damp it. I'm left with a dampened circuit, not a damp one.
My angrish is awful. The Queen hasn't been over here for a while.

 

Ah, David! Picky, picky!
From Dictionary.com (see definition 2):
1. To make damp.
2. To deaden, restrain, or depress: “trade moves... aimed at dampening protectionist pressures in Congress” (Christian Science Monitor).
3. To soundproof.

Is it the tense that I'm not getting?

 

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.

 

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"

 

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?

 

Chris,

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

 

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

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

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

Ron

 

Edit:
Ramblings of a madman.

 

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.

 

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.

 

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