Electronics experts help needed....

[Level42]

Hi there, I know this is the monitor-video section, but I also know there are some really good electronics experts here.

I need your help. I have tried to improve a schematic (make it safer) for triggering a knocker that was originaly used in the Q*bert cab.

The original schematic can be found here:


However, I didn't like the fact that it connects the grounds of the I-pac/J-pac and the external 30 Volts supply. If you wire something incorrectly, or the transistor blows, you could very well damage the I-pac/J-pac.

So I figured this needed an optoisolator.

This is my schematic:


I think I got the schematic right, but I wonder if I should use a resistor between the emitter of the optoisolator and the base of the darlington transistor (2N6044). I guess so, because else the current might be too high ?

Any help very much appreceated. This is the thread where we are discussing this:
http://forum.arcadecontrols.com/index.php?topic=54147.0

[MonMotha]

Personally, I think the original one is fine.  A failure condition of a short from base to collector is fairly uncommon in my (admitadly limited) experience, and most digital outputs could handle the other failure short condition (base to emitter, which is grounded) for a little while (enough to probably notice things are broken) without sustaining damage.  If you're concerned, you could chuck a low Vf diode (with high reverse breakdown) in front of the base of the darlington.

However, if you want to use the optoisolator, you'll need to check a few things.  One, make sure your opto can handle VCE of 30V.  Some are only good to 25.  Also, if you get the output transistor of the optoisolater "good and on" you'll have very little voltage drop across it (VCEsat is usually around 0.1-0.5V), which would put upwards of 29V across VBE of the darlington.  That's way above spec and will probably result in excessive base current.  Figure from the datasheet you'll get at most 4.5V across VBE of the darlington (before it blows up) and you can put up to 120mA through that base.  30-4.5-0.5 = 25V.  R = V/I = 25/120m = 208ohms.  So start with a 270-330ohm and decrease until it works reliably.

You probably won't need all that base current, since the gain of a darlington is rather huge. Watch the power dissipation on your resistor, as well. (P=V^2/R or I^2*R) Depending on the gain of your device (which can vary wildly for darlingtons) and the current required for your load, a much higher resistance (resulting in lower current and lower power dissipation in the resistor, all of which are good) is probably possible.

[Level42]

Thanks, excellent info !

Yes I had thought about that the optoisolator should be able to handle the 30V, it was my first concern, but I've already found a couple that can handle that easily.

Indeed, almost all of the 30V will go onto the base of the darlington, so that's why I already expected that a resistor would be needed, escpecialy to limit the current. Yes a darlington needs very little to be triggered, so I'll start off with a 1K resistor or something like that and see what happens. I'll also measure the current, but it will be pretty hard, because the triggering pulse lasts very shorly only, probably way to fast to measure with a regular multimeter.

Still waiting for the knocker to arrive and I will go and get the other parts and try things out.

Thanks again for the help !

[Level42]

Ow, I just had another look on Brad's schematic and take a look at it here:


You can see he wired up a led in series with the base and a 5k6 resistor. His schematic is confirmed to work fine and is basicaly the same as John's. So, at 5v, a 5k6 is still triggering the darlington transistor without a problem. So I figure I should use something in the order of 34k to reach about the same base current on 30 V.

The 120mA base current is an absolute maximum rating, there's some diagrams showing that the "normal" base current is 10mA max.

When I do some calcultions on Brad's diagram I get this: I=U/R => 5V/5600 Ohms= 0,0009 A = 0,9 mA (about !) and his diagram works, so this darlington needs very little to trigger.
So I think I should have a resistor around that 34K value to get the same amount of current. Please let me know if I make some stupig mistake here.

(actualy, the current is even lower at Brad's diagram because of the LED he has in series with the 5k6 resistor.)

[Level42]

So this would be my "final" design: