Re: International standards

is

I was thinking that for a high current load, the veroboard traces wouldn't be thick enough to carry it. And the cable is too thick to put through one of the veroboard holes.

OK. I'm surprised it would matter so much.

I'll think about it - thanks.

No, that wouldn't work.

As shown the RC time constant is long relative to the typical ramp up of the power supply. So pin 2 remains below required 1/3 threshold for a short time, triggering the mono. Then it stays +ve so that it's still high when mono ends, avoiding re-triggering.

--- Sure they would [be thick enough]. The only thing which would make the traces fail is if they melted or vaporized with 12 amps going through them, and even if they were only 1/2 oz. copper, with a nice coating of solder they should be fine. Remember too, there's only going to current flowing for about a second.

---

--- It _might_ not, but it's better to eliminate the source of a possible problem than to have to deal with it later. An ounce of prevention...

-- John Fields

In message , James writes

A circuit that regulates the power dissipated by the laser diode by monitoring the output from the laser diode's internal photodiode. Many laser pointers are just a 'chip' with a bare bond wire and have no kind of drive circuit, they are very easy to destroy with too much current.

How would such an EMF cause this type of damage and how would your suggested solution solve the problem ? What is a TVS ? What would be the connection of this resistor/TVS - from shield to GND or some other form of connection (The signals on the bus are a differential pair).

it connected?

This was going to be one of the solutions. However, I wanted to try and come up with some sort of theory of what was causing the fault. I was also worried that someone had suggested that if we do short the shield to GND then we may cause large currents to flow, doing more damage than good (even though the devices with the jumper in it did not show this fault !)

Is there any way to measure these phenomena you mentioned above ? We do not have access to the system that the devices were placed in so we can't replicate the precise circumstances under which the fault occurred (we are not able to measure the "current due to ground imbalance" and I was only suggestion that this might be a cause of the fault). Is there any way within an "artificial" labratory environment for us to narrow down what the source of the fault is (and thus the remedy ?)

What a Rip-off Price. That laser pen in B.C. Canada, sells for $1.00 Canadian, thats about $.75 US. Also that particular product has a BAD Failure Rate. There is a Better brand out there that is as cheap and more relilable.

Those particular lasers only use a Series Resistor for current regulation. Add another resistor in series of an appropriate value and you can use any voltage adapter.

See my Laser Alarm Project, It Might help you.

Take care.........Gary

that link didn't work.

How can you have "no luck" with plated tips? I have some tips that I have been using on and off for over a decade. Plated tips are without a doubt the best. But, if you are grinding, sanding, and filing your tips, then you deserve the "no luck" you get. You have immediately destroyed the life-lengthening properties of the tip. The only thing you should use to clean a soldering tip is a damp sponge for constant use, and maybe once a month you might wire brush it and then tin it again. If you are doing something that dirties your tip up any worse, you are not using it properly. If you are doing something that bends or deforms your tip, you are not using it properly. If you are doing anything other than plain soldering with minimal force and on clean surfaces, then you might as well revert to thick leather gloves to go with the glass pliers and blowtorch.

Cheers!

Sir Charles W. Shults III, K. B. B. Xenotech Research

Was there a diode across the coil to deal with spike voltages? If not, those probably killed the FET. You need something like that:

o + Ub | |-------- | | | ) ____ |( /\ 1N4148 coil | ) / \ |( ---- | ) | |--------- | to FET

I know I will be boo'd off the stage but we have long since stopped using wet sponges to wipe the soldering iron tips. We use folded terrycloth wiper towels conveniently clipped onto a clip board. Try it before you laugh.

I just use paper towel, not even damp.

Decades ago, there was an article that suggested it, though I can't even remember if they say to keep it damp. It doesn't really seem to matter, though I suppose if one had a hot enough iron, perhaps the issue of keeping the sponge damp is to avoid setting it on fire.

I just can't be bothered keeping a sponge damp, and if it's not damp, it tends to wear out due to burning. It's more expensive than paper towel, especially if using those specifically made for the soldering iron station, so one may feel compelled to keep it damp.

Michael

First, no LED has a "voltage rating". The have a "current rating" and a *typical* forward voltage drop (at the rated current).

Second, *some* current limiting method is more or less always used. Either passive (resistor) or active (like in the white LED drivers) with current sensing feedback into the IC.

It's very hard to find a *voltage source* that has exactly the correct voltage. And if the LED heats a little, the neg tempco will lower the forward voltage drop of the LED, and you'll end up with a current rush and a dead LED...

Jan-Erik.

I've heard that the thermal shock from wiping on a damp sponge is somehow good for the tip, or its plating or tin coating or something. I have no idea what it could do, but I've always used a wet wiper, whether sponge, rag, or paper towel.

Cheers! Rich

Thanks all for your help!

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Here is the original circuit.

I built this and found that the compressor doesn't work very well. Of course it has only a volt and a half to work with. But it does amplify.

You should check the voltages, and if your earphones have about .2VDC across them then the output stage is feeding it a current of about 6 mA, which is most of the current for the whole amplifier.

You might disconnect D1 and the base of Q3 from C3 and P1. This will shut off the compressor circuit. Then see what happens. BTW, I found that for R2, 1 megohm was a bit high, I think I used 560k or so. But it depends on the gain of the preamp transistor Q1.

I built mine into the handset of a telephone. I sawed off the mouthpiece. The DC resistance of the handset is more like 150 ohms.

The 2k and 1k resistor are loading down the high impedance output of your signal source. You need 2 gain stages... first one needs to be non inverting (and hi impedance) with gain of 10 or 20, 2nd stage can make up the rest of the gain.

Resonate against known value cap. f=1/2*pi*sqrt(LC)