LPR Idea: 12V Rocket Launch System Schematic Using Transistors Configured as Darlington Pair; Also MOSFET-based Circuit

[brockrwood]

YES!!!!!! You need a mechanical switch to disconnect the igniter from battery. You never, ever want B+ voltage on an igniter lead ALL THE TIME just waiting for a command or accidentally provided source to ground.

Incidentally your design would not conform to NAR safety code without this manual disconnect.

See my post about the battery being connected to the system by two lamp wires and two alligator clips. Just yanking one of the alligator clips off of the battery in a scared panic is not adequate as a manual “break the circuit” feature?

No biggie. I will put a big old honking on/off household type light switch on the “pad box” right where the 12 volt wire from the battery comes in. That should do it!

 

[sghioto]

YES!!!!!! You need a mechanical switch to disconnect the igniter from battery. You never, ever want B+ voltage on an igniter lead ALL THE TIME just waiting for a command or accidentally provided source to ground.

Incidentally your design would not conform to NAR safety code without this manual disconnect.

What difference does it make if it's OPEN high side or low side?

 

[jderimig]

What difference does it make if it's OPEN high side or low side?

If the switch disconnects battery ground directly from circuit ground that makes it 1-fault tolerant. As shown the circuit is 0-fault tolerant. Or you can add a large resistor at the battery + or - minus that gets shunted by an arm switch but lets enough current leak through for continuity check.

 

[sghioto]

What is the nifty software or other tool you use to draw these schematics? I enjoy hand-drawing my Forrest Mims style schematics, but changes are easier to make using software!

Windows Paint.

Man, I have to read the datasheets for these MOSFET’s closely. Key parameters can be quite different from MOSFET to MOSFET.

Yes that's true. Adding a zener diode from the gate to ground will limit the max voltage to the mosfet.

 

[jderimig]

If the switch disconnects battery ground directly from circuit ground that makes it 1-fault tolerant. As shown the circuit is 0-fault tolerant. Or you can add a large resistor at the battery + or - minus that gets shunted by an arm switch but lets enough current leak through for continuity check.

You can take one of the two switches that drives the gate of mosfet and use it as power or shunt switch.

 

[brockrwood]

What difference does it make if it's OPEN high side or low side?

With a battery it should not make a difference I would think, because the negative side of the battery IS ground. With mains powered circuits it might matter because, somewhere in the circuit you could get a short to true ground (Earth). If the switch is on the low side, then the high side keeps on feeding current to the place where the circuit is shorted to true ground.

My understanding is that this is why mains plugs and sockets are polarized, so that the “hot” wire coming into the device is the one that gets broken by fuses and switches, so that, say if a short to a metal enclosure of the device is somehow connected to ground, flipping the switch will kill the current to the short.

The idea, I suppose, is that if the positive gets shorted to the metal enclosure of a device, and you touch the enclosure, YOU may become the path to “ground”. So, by making the plug polarized, the manufacturer can put all of the fuses, switches, and safety mechanisms on the “hot” wire, to make it possible to cut the power before it gets to the fault area.

For a model tocket controller I guess it is possible for something similar to happen. What if the leads to the igniter fall to the ground, the ground is wet, and that connects the positive side of the battery to true ground (earth)?

 

[brockrwood]

Windows Paint.

Yes that's true. Adding a zener diode from the gate to ground will limit the max voltage to the mosfet.
View attachment 613703

Windows Paint comes with electronic schematic symbols?

 

[sghioto]

Windows Paint comes with electronic schematic symbols?

No. You have set up a library file in Pictures and import schematic symbols into it. Then it's simply copy and paste.
Here's a portion of my file as an example.

 

[sghioto]

No biggie. I will put a big old honking on/off household type light switch on the “pad box” right where the 12 volt wire from the battery comes in. That should do it!

Not needed as per NAR model rocket safety code:
3. Ignition SystemI will launch my rockets with an electrical launch system and electrical motor igniters. My launch system will have a safety interlock in series with the launch switch (Sw1) and will use a launch switch (Sw2 and Sw3) that returns to the “off” position when released.

 

[brockrwood]

Here is the updated, MOSFET-based circuit, using the basic design of @sghioto with ideas from @jderimig and @brockrwood .

I will probably edit the title of this thread as it now contains both a Darlington Pair based circuit and the MOSFET-based circuit.

Nope. Just saw an error in the schematic. Back to the mechanical pencil...

OK, here it is.

Design compromise: I have the red LED, LED1 connected across the positive and negative rails. It is connected to the positive rail after the fuse and the big, master power switch (SW1). I want to be able to see LED1 glowing if the master power switch, SW1, is feeding battery power to the system. Basically, if battery is hooked up to the system, I want to know it. Resistor R3, 2.2K ohms, limits current through the "Power On" LED1, to about 6mA.

There is also another LED, a green one, LED2. This one is also current limited by a 2.2K ohm resistor, R4. This one, however, is connected to the positive rail THROUGH the igniter (about .711 ohm). It also has a push button in series with it, SW4. When power is on at the master switch, and an igniter has been clipped to the ignition wires, you push SW4 briefly to see if there is "continuity" through the igniter. If the green LED glows, then you have the igniter wired properly and you are ready for launch.

This presents an issue: When you press SW4 to check "continuity", you still have the "power on" LED glowing. You now have two LED's wired in parallel (and two current limiting resistors wired in parallel). I am really nervous about an any portion of the circuit that will draw current through the igniter before it is time to launch. So I have increased the value of the current limiting resistors on LED1 and LED2 to 2.2K ohms. This should make no more than about 6mA flow through either LED. Yes, that might make the LED's a bit dim. I might have to hold my hand over the LED's, in bright sunshine, to see if they are glowing.

Am I being too nervous here? I know it takes about 1 amp to set off an Estes-type igniter. Still, the idea of 10mA to 20mA flowing through the igniter, just to check it, makes me nervous. So I am going very conservative with the current limiting resistors on the LED's.

Also, hooking up LED's in parallel is trickier than hooking them up in series.

One solution: Put a push button in series with the "power on" LED, LED1. That way, you can "check to see if power is on" by just pushing the button. Once you have satisfied yourself that power is on to the system, you can release the button.

View attachment mosfet_based_12V_launch_controller.jpg

And another thing: Yes, I could have dispensed with the long wire in the schematic showing the connection to the negative terminal of the battery. Instead, I could have just used ground symbols where needed. But, the point of this controller is to separate the "controller box" and its switches from the rest of the circuit (the "pad box" and its innards and the battery connected to the pad box and the wires going to the igniter from the pad box). The idea is to be able to stand a considerable distance away from the pad (20 feet) while holding a light controller box that is connected to the pad box with light gauge wire. I wanted to show that in the schematic.

 

[John Brohm]

Forgive me if I'm not understanding your circuit properly, but as I see it, if you were to consider a couple of changes, you would be able to boost the current to your LED indicators (making them brighter) while limiting the continuity check current to just a few safe milliamps.

In the circuit provided below, I've moved your MOSFET to the high side, and use a PNP BJT as the check current source. It's wired in parallel to the active launch element (the MOSFET, which is normally OFF) and the current source is only active when the continuity check pushbutton SW4 is pressed. I've inserted R5 to make sure the PNP transistor is fully turned off when not active. This precludes any leakage current across the junction from activating the check current LED2. With this arrangement, LED2 enjoys a good 20mA of current, and the check current is limited to about 5 mA.

Hope this might help.

John

 

[sghioto]

Am I being too nervous here?

Yes! 20 ma is nothing to an Estes igniter. However depending on the LED used 6 ma may be more than enough.

Also, hooking up LED's in parallel is trickier than hooking them up in series.

I don't understand that statement.
I would highly recommend adding LED3 as shown in post #24.
I wouldn't use Sw1 as shown in your latest schematic unless the battery is concealed in the pad box. Just not necessary but it's your project.

 

[brockrwood]

Yes! 20 ma is nothing to an Estes igniter. However depending on the LED used 6 ma may be more than enough.

I don't understand that statement.
I would highly recommend adding LED3 as shown in post #24.
I wouldn't use Sw1 as shown in your latest schematic unless the battery is concealed in the pad box. Just not necessary but it's your project.

Series LED’s: Just add up the voltage drops of the LED’s and the resistances of the resistors. There you are. Current easy to calculate and control.

Parallel LED’s: How do I account for the voltage drop of the LED on each of the parallel paths? Just subtract it from the input voltage applied to all parallel paths?

Tricky for me because I am not skilled enough at circuit analysis.

 

[sghioto]

If the LEDs are perfectly matched then the voltage drop is the same as a single LED. Calculate the resistor value for twice the current required.
Better practice for each LED to have its own resistor but often seen in some products.

 

[brockrwood]

I would highly recommend adding LED3 as shown in post #24.

First: Live and learn: I did not know that the post number was in the upper right corner of each post. Doh!

Second: So, the "Fault" LED would show that there is current (at least a few milliamps) going through the MOSFET when there shouldn't be?

If the "continuity" push button is pressed, there is path to ground for both the "fault" LED and the "continuity" LED. Both of them should glow.

If the continuity push button is not pressed, but the launch push buttons are pressed, then, the "fault" LED has a path to ground through the MOSFET, no? Won't it glow when the launch buttons are pressed and the large current flows through the igniter? It doesn't really matter if it glows when the launch buttons are pressed, but it should, right?

If the launch buttons are not pressed and the continuity button is not pressed and the fault LED STILL glows, then something is wrong - probably a shorted out MOSFET. There should not be a way to ground for the fault LED if the continuity button or the launch buttons are not pressed. Correct?

 

[brockrwood]

I wouldn't use Sw1 as shown in your latest schematic unless the battery is concealed in the pad box. Just not necessary but it's your project.

Well, in deference to @jderimig , I thought a "master" on/off switch that could completely disconnect the battery from the launch system could not hurt anything. It might be unnecessary, but it did not seem harmful. Also (sheepish grin) the thought of a big, old A/C wall switch sticking up out of the "Pad Box" just seemed nerdy, fun, and cool looking to me. ;-)

Maybe, just maybe, there is a situation where there is a short that doesn't blow the fuse (or somehow gets around the fuse) and causes the wires to the battery to get super hot. Without a master on/off switch, the only way to disconnect the battery is to grab one of the hot wires and yank one off of the battery. (Or maybe kick the battery until one of them comes off.) In that admittedly unlikely situation, wouldn't an on/off switch come in handy?

How about this big, red one? Come on, that would be fun, don't you think? ;-)

 

[brockrwood]

Forgive me if I'm not understanding your circuit properly, but as I see it, if you were to consider a couple of changes, you would be able to boost the current to your LED indicators (making them brighter) while limiting the continuity check current to just a few safe milliamps.

In the circuit provided below, I've moved your MOSFET to the high side, and use a PNP BJT as the check current source. It's wired in parallel to the active launch element (the MOSFET, which is normally OFF) and the current source is only active when the continuity check pushbutton SW4 is pressed. I've inserted R5 to make sure the PNP transistor is fully turned off when not active. This precludes any leakage current across the junction from activating the check current LED2. With this arrangement, LED2 enjoys a good 20mA of current, and the check current is limited to about 5 mA.

Hope this might help.

John


View attachment 613971

First, are you J. R. Brohm, the author of this very helpful article on Igniter Continuity Tests?

https://www.rocketshoppe.com/info/Igniter_Continuity_Tests.pdf

If so, thank you so much! Great work! It is very helpful!

Second, let me make sure I understand your circuit idea to get around the "there is not much current flowing through the check continuity LED" problem.

Looking at your schematic, when SW4 is pressed, that creates negative voltage on the base of the PNP transistor (the venerable 2N2907). The current flows from the battery (+12V) through the emitter of the PNP transistor, through the base, through R6 (4.7K), through the igniter, to ground. That flow turns on the much bigger current flow from emitter to collector through the PNP transistor, through R4 (a mere 510 ohms resistor), through the continuity check LED (D2 in your schematic) to ground.

The continuity check switch lets current flow through R6 (and through R5) through the base of the transistor and through the igniter to ground. But that current is is a mere pittance. The two resistors in parallel are equivalent to a 2.35K ohms resistor. With 12 volts flowing through them (ignoring the voltage drop through the base of the transistor), the current getting to the igniter is 12V / 2,350 ohms or about .005 amp (5mA).

But the current getting to the LED is limited only by R4, the low value 510 ohms resistor. The current through the LED is about 23mA. That will make the LED glow nice and bright!

Did my amateur, hobbyist brain say it back to you correctly?

This is awesome!

The only major change I need to make is to put the MOSFET on the "high" side of the igniter rather than on the "low" side. This makes the current flow through the MOSFET first, then through the igniter, when the launch buttons are pressed.

Of course, as @sghioto says, running 12mA or even 20mA through an Estes-type, nichrome wire igniter will not get anywhere NEAR the 1 amp or so needed to fire the igniter. Still, this is a very elegant way to separate the current going through the "check continuity LED" from the current going through the igniter, while still having the fact that current, even a small of current, going through the igniter being the event that triggers the LED to glow - hence a "check of igniter continuity".

Thanks!

 

[sghioto]

The only major change I need to make is to put the MOSFET on the "high" side of the igniter rather than on the "low" side.

One problem using a N channel mosfet as a high side switch is the voltage loss across the mosfet. As shown now with the two 10K resistors the voltage out to the igniter is only appx 4 volts max. Connecting the gate directly to 12 volts expect appx 9 volts.
Keeping Q1 as a low side switch and using mosfet Q3 will accomplish the same with practically zero current through the igniter for the paranoid inclined.

EDIT: Removed Fault Indicator circuit.

 

[brockrwood]

One problem using a N channel mosfet as a high side switch is the voltage loss across the mosfet. As shown now with the two 10K resistors the voltage out to the igniter is only appx 4 volts max. Connecting the gate directly to 12 volts expect appx 9 volts.

Hmm. So, how do I determine the voltage drop across the MOSFET? Do I look at these graphs in the datasheet?







I need to go to MOSFET school.

It looks like if I feed just a volt or two more to the gate, the Drain to Source voltage stays lower. Maybe adjust the voltage divider on the gate to feed about 7 volts to the gate? Of course, per the graph, it doesn't matter until we get to about 40 amps on the drain?

 

[brockrwood]

T-I transfer curves? My brain hurts…

;-)

 

[Paul Howard]

Wow! This is a great thread! I've recently had thoughts of doing something similar to add a continuity LED to my home-made launch controller in 2 Altoids cans (soldered together with a flap-open side for easy access and modification, plus it's good place to store the safety-key made from a short-wired electric guitar plug.) for a housing and powered by 8 D-cells wired in series held in a used plastic peanut butter jar (yeah, really!). Thanks!

 

[waltr]

If you only pull 15 Amps then the Voltage drop across the D-S is low with 6V gate. Actually looks like from the graph that a higher G-S Voltage would not decrease the D-S drop.

All this is why many MOSFET drive circuits have a 'boost' - this increases the Voltage used on the Gate-Source.

 

[brockrwood]

I tried simulating the MOSFET-based circuit in Falstad. See pic. It occurred to me that simulating the Estes igniters as resistors is incorrect. What they really are are FUSES. I mean, they are a piece of wire that heats up and, at about 1 amp or so set off a rocket engine. In my experience, using the 12V, SLA battery (9aH), the wires get so hot they break in half. Hence, they are fuses. They get hot, set off the rocket motor, and then break. Plink.

When I tried simulating the Estes igniters as fuses, however, I could never get them to last the typical 1 to 2 seconds I know it takes the real igniters to heat up and break. They would break in milliseconds - 10ms to 20ms. All the while lots of amps were flowing through the circuit. Hmm. I may just need to prototype the MOSFET circuit, hook it up to an Estes igniter, and see what happens.



The ignition push buttons and safety key switch are simplified down to just a single, SPST switch in the simulation. You can see the “continuity test” switch and LED hanging off the right side of the schematic.

 

[sghioto]

Might try the simulation using slo- blow fuses for the igniter if available in the library.
I highly recommend protyping the circuit as well.

 

[waltr]

Yes, do prototype the circuit. It is the only way to know for sure.
Sims are great but they tend to be "Ideal" components.
As for the sim:
Try adding a little R in series with the fuze. 1 Ohm may be enough. May also try adding a high R across the fuze.
This should slow down how fast the fuze blows.

 

[brockrwood]

Here is the most recent version of he circuit as simulated on Falstad.com. This is getting close to the final version I am going to prototype onto perfboard. Actually, if the prototype works, I will just mount it into the project boxes and make it the final, production version.

 

[FrugalGuy]

I just joined this forum tonight, so I know nothing, understand nothing, and don't wish to create confusion, but instead I wish to learn what I can.

I have a question: OP was concerned about having 2 parallel resistor/LED circuit paths powered at the same time, apparently thinking that would increase the current passing through the igniters. Please correct me if I'm wrong, but that isn't correct, is it? This is about post #40.

The continuity circuit passes through the igniter, but the power LED circuit does not.
The current passing through the igniters faces the continuity resistance plus the igniter resistance. That affects the current flowing through those resistances.
The current passing through the power LED (and resistor) has zero effect on the current passing through other portions of the circuit, I believe.

R2 is the 1K resistor for the continuity LED plus the igniter resistance, which we'll say is about 1 Ohm (I think .7 Ohm was quoted earlier).
R1 is the power LED circuit resistance. The actual values don't matter. What matters is that the current through each is calculated independently of the other.

So, I think all the concern and increasing the resistor sizes and extra circuitry to avoid sending extra current through the igniters is unnecessary? There won't be any, as far as I can tell. I've got to be missing something. What is it I'm missing?

 

[FrugalGuy]

As for battery cut off switches, I think one of these might work nicely:

or just use a pair of XT60 connectors, and short one together as a kind of link you plug in when you're ready to go. They're rated for 30 Amps:

 

[OzHybrid]

Here is the most recent version of he circuit as simulated on Falstad.com. This is getting close to the final version I am going to prototype onto perfboard. Actually, if the prototype works, I will just mount it into the project boxes and make it the final, production version.

View attachment 615408

The REASON you should wire in series and NOT parallel is that if a single igniter fails, you get an off-centre thrust from the remaining motors firing. This can be mitigated by having the line of action through the cg of the rocket. Also, a significant number of ignitors go short-circuit during firing and sometimes after. This can prevent the remaining ignitors from firing. So the parallel connection method INCREASES THE CONSEQUENCE FROM A FAILURE AND INCREASES THE CHANCE OF FAILURE. However convenient it might seem, because you don't have to calculate the voltage required for series firing.
So this particular theoretical model of a circuit above does not match the real-world behavior of an igniter in parallel.

 

[sghioto]

I've seen series igniter fail as well. Never had an issue with parallel igniters.