Question about circuit design

Loki008

I am new to this kind of stuff. I have a very basic understanding of how it all works, but most of my prior background comes from IT and simple programing. So i may need clarification and a little hand holding till i get everything figured out.

What i am trying to build is an programable firework ignition system. I plan on starting and stopping the squence via IR, and i think i have all that figured out from the playing i have been doing with the homework board i picked up. My main concern is the firing of the ignitors. I dont want to risk damaging the chip. What the ignitors are made of is a small peice of nicrome wire coated with a small bit of black powder.

http://www.fireworks.com/phantom_mall/photo.asp?pid=927

Below is the diagram is what i have come up with so far. I am sure that there is room for some refinement. Any sugestion would be appreciated.

Mike Green

1) You didn't show the polarity of the transistor. The emitter of an NPN transistor goes to (-).

2) You have to have a common ground connection. If you're using an NPN transistor, the negative supply gets connected to the BS2 ground (Vss).

3) I don't see a mechanical interlock here or a pilot light. If switching on the transistor can be dangerous, you need a mechanical interlock (switch) that you have to have in place for the ignitor to be connected in the circuit. In most hobby rocketry, the interlock is a largish plastic piece that has to be in place to close a contact for the ignitor circuit to connect to power. Whoever is serving as "safety officer" keeps that while someone else is working on the rocket and/or ignitor. The pilot light is just an LED (with a current limiting resistor) that attaches around the ignitor/interlock so, if the power is on, the LED will be on even if the interlock is off. That way, you know if the ignitor circuit is live before you insert the interlock and you can shut the whole thing down until you figure out what's wrong.

LilDi

I think its a tossup which is going to blow first. The fuse or the transistor. And you need a 10K resistor between the P0 pin and the base of the transistor to protect the Basic Stamp P0 pin from blowing too. Your going to need a very beefy transistor that can handle the amps before the fuse blows.

Then there is the issue of critical safety. The Basic Stamp is probably not recommended for critical safety applications and the transistor is going to take a big amperage surge each time you fire a fuse off, probably damaging the transistor any making the transistor unreliable. The transistor could short out.

It would help to know the voltage of the battery and the ohms resistance of the Nichrome fuse wire.

Loki008

It is NPN with C to E to ground.

As for safety controls, i have several peices in place. One is a keylock switch on the power to the stamp, second is that ignition program is started via an IR code. As for notification, when power is on, there is a multicolor LED that is green when power is on, but no IR code has been recieved as of yet, Once the IR code has been recieved, the LED turns red, and a buzzer starts beeping for 25 seconds followed by a 5 second continious tone. Anytime durring this application an IR abort code can be sent which will end the firing control.

This is just a sub section of the system that i am bringing up becasue of how i am basically causing an intentional short i am concerned of possably damaging the stamp.

As for the Common ground connection, assuming that this is the same battery that is powering the basic stamp the ground doesnt have to go through the stamp, just the same - terminal right?

Mike Green

A keylock switch on power to the Stamp is inadequate. The keylock has to be in the power lead to the ignitor, otherwise you have no way of knowing, for example, if the transistor has shorted out. Similarly, if you have an indicator that shows that the switching transistor is conducting, you don't want to turn on the keyswitch. You also want a keyswitch that must be open to remove the key. If the key is in the lock, you don't fiddle with the ignitor until you remove the key.

LilDi is correct that you need some kind of resistor in series with the base lead from the transistor. It should allow enough current at approximately 5V to saturate the transistor. How much current depends on the gain (hFE) of the transistor and the amount of current it has to conduct. The amount of base current has to be larger than hFE x max. current.

Loki008

Mike, how does this work for the safety issues



Before the fuses are attached, the unit is keyed on, assuming proper opperation no lights will light up, if there is a light then there is a failure.

Scratch that, duh, just connect the + of the LED prior to the switch.

Post Edited (Loki008) : 2/15/2008 5:58:43 PM GMT

Loki008

LilDi said...
I think its a tossup which is going to blow first. The fuse or the transistor. And you need a 10K resistor between the P0 pin and the base of the transistor to protect the Basic Stamp P0 pin from blowing too. Your going to need a very beefy transistor that can handle the amps before the fuse blows.

Then there is the issue of critical safety. The Basic Stamp is probably not recommended for critical safety applications and the transistor is going to take a big amperage surge each time you fire a fuse off, probably damaging the transistor any making the transistor unreliable. The transistor could short out.

It would help to know the voltage of the battery and the ohms resistance of the Nichrome fuse wire.

I will know this info this weekend. I need to pick up a pack of the fuses.

LilDi

The Diode is in backwards. I would use a high current relay or Silicon Control Rectifier instead of the transistor because they can handle high current better than a transistor. The relay idea will be less likely to damage components.

Post Edited (LilDi) : 2/15/2008 6:38:45 PM GMT

Mike Green

Actually, if you meant an LED for a diode, it's in the correct direction, just needs a current limiting resistor to limit the current to 20ma. If you're really using a 9V battery, that would be something like 330 ohms. Note that 9V radio batteries don't have much capacity and their peak current is not high enough for reliable use for something like an ignitor. At least use an AA cell battery pack.

RadioShack has a Darlington NPN transistor, the TIP120 which would work very nicely with a 220 ohm series base resistor and could handle currents of several Amperes.

LilDi

Sorry Mike. I didn't see the last sentence of Loki008's post (Scratch that, duh, just connect the + of the LED prior to the switch).

I think the idea is to have the voltage low enough to cause the Nichrome wire to glow red and ignite the fuse cord, but not so high as to burn out the nichrome wire which will then not ignite the fuse cord.

Is this what you mean Mike?
See attachment

Post Edited (LilDi) : 2/15/2008 10:01:23 PM GMT

allanlane5

Better, but you still have to connect the 'ground' of the circuit (the emitter of the transistor) to the Vss of the BS2. The way a transistor works, a little bit of current flowing into the base and out the emitter (and back into the ground of the driving device -- in this case, the BS2 Vss pin) controls a much larger amount of current flowing from the collector through the base and out the emitter (and back into the ground of the high-current loop -- in this case, the 9-volt battery).

Unless you're using the 9-volt across the Vin and Vss of the BS2, and you're just not showing that. In which case, this looks good to go.

Mike Green

I would suggest a smaller base resistor. A 10K resistor would allow maybe 1/2 mA base current for a regular transistor or maybe 1/4 mA for a Darlington (because of the higher base-emitter voltage). Even with the Darlington, that's not enough collector current for an ignitor. You should use no more than a 470 ohm resistor for a regular transistor or 220 ohm for a Darlington to ensure enough current (10mA) to saturate the transistor. You do also need the common ground connection (emitter to BS2 Vss).

Loki008

Ok, Like i said, i am fairly new to this stuff so thank you for bareing with me. I just wanted some clarification on some basics.

Right now i am using the homework board, it has its advantages where for the most part you can see the traces and i am understanding most of whats going on, but i am a bit confused with VSS VIN and VDD

VDD - +5v rail as supplied by the voltage regulator

VIN - + terminal of powersupply (9v battery on my homework board)

VSS - This is the one i am a bit confused, i guess somewhat since i can not see any traces on the homework board for this one. I am assuming that it is the - terminal of the powersupply (9v battery on my homework board)

So when you say that the emitter of the transister has to be connected to the VSS of the basic stamp, do you mean that it has to go through the stamp, or to the same ground point as the stamp. I am assuming that this means the same ground point as the stamp

Also in regards to the BS2 OEM module. When i am ready to assemble this project i will take the + terminal and attach it to VIN and - terminal to VSS. VDD will be +5v out from the OEM module for components that require the stable 5v current.

Like i said, this is a starter project after playing around with the homework board a bit. I figured its a good starter project due to the relative simplicity of the circuit as well as the variety of logic being applied with the IR, switches, LED's ect. Thanks for being so patient and giving me the advice you have given me so far.

Post Edited (Loki008) : 2/16/2008 12:58:15 AM GMT

allanlane5

Correct, Vss is the - terminal of the 9-volt battery.

Loki008

Thanks, thats what i thought. I was just a bit confused with all the questions about my diagram not terminating at Vss.

Loki008

2 more quick questions. I have decided to go with 8x AA batteries for the power supply as well as the recomended TIP120 Darlington transistor. My questions are these, will the homework board function at 12v rather than a 9v battery. Secondly does this change the resistor needed on the i/o pins?

Thank you all for your help

Mike Green

1) Yes, the homework board will function at 12V. The issue with 12V vs. 9V is that the regulator on the Homework Board has more power to dissipate if you draw the full amount of current through it. In your case, you're only supplying current to the transistor's base, not 3 or 4 LEDs as well. If you added 4 LEDs to the Stamp, that might be as much as 20mA each or 80mA. With an additional 3V drop in the regulator, that would be 3 x .08 = .24 Watts. A quarter Watt is not a huge amount of heat, but the regulator doesn't have a heatsink other than the printed circuit board and might get quite warm.

2) The resistor on the I/O pin is based on the use of the regulated 5V supply and isn't affected by the use of a 12V power supply.

Loki008

Thank you very much. I really appreciate that you take the time to explain your answers so that i can take something from them for future use.