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Timer MOSFETs - inside or outside the timer PCB

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Should I use a single PCB or seperate boards for the timer and the high-power switch?

  • Use a single PCB.

  • Use two PCBs, and power the high-power switch directly from the baterry

  • Use two PCBs, and power the high-power switch through the timer PCB


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DexterLB

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I've decided, while waiting for the pressure sensor for the altimeter, to make a homebrew timer based on PIC12F629. It must handle drogue and main 'chute deployment (e.g. 2 outputs). That means I must put two giant MOSFETs and a super-sized 1mF (that equals to 1000uF:eek:) capacitor. I think it'll be better if the MOSFETs and the capacitor are on a "high-power switch" board, seperate from the timer PCB, but there comes another problem - where will the seperate high-power switch get its power from - directly from the battery or from a cable between the timer and the high-power switch? Or should I put everything on a single PCB?
 

FROB

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Dexter,
I see no reason why you'd want to keep them separate-
When it doubt, use the KISS principle:
2 boards plus the extra wires and connections adds unnecessary complexity and potential points of failure. 1 board is simpler.
What battery do you plan to use? that will tell you if a large cap is really necessary.
Also, you don't need a "giant" mosfet, so-8 or smaller dual mosfets that can handle 10 or more amps are quite common.
 

DMcCauley

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I've decided, while waiting for the pressure sensor for the altimeter, to make a homebrew timer based on PIC12F629. It must handle drogue and main 'chute deployment (e.g. 2 outputs). That means I must put two giant MOSFETs and a super-sized 1mF (that equals to 1000uF:eek:) capacitor. I think it'll be better if the MOSFETs and the capacitor are on a "high-power switch" board, seperate from the timer PCB, but there comes another problem - where will the seperate high-power switch get its power from - directly from the battery or from a cable between the timer and the high-power switch? Or should I put everything on a single PCB?
A few things.
Firstly, keep them all on a single PCB board.
For MOSFETs, they can be very small. An 8-SOIC package is pretty small.

Better yet, don't even use a MOSFET. Use a MOSFET Gate driver IC. You can get small ones (dual packages) in an 8-SOIC which are good for 8-9A or so.

A 1mF capacitor at say 12V is not very big at all.

Think smaller.
 

jderimig

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A 1mF capacitor at say 12V is not very big at all.
A lot bigger than the mosfet. Capacitors are not needed. Fresh 9v batteries will deliver plenty of current (5A) for several seconds.
 

DMcCauley

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A lot bigger than the mosfet. Capacitors are not needed. Fresh 9v batteries will deliver plenty of current (5A) for several seconds.
It depends on the implementation.

If you run your uProc and pyros on the same battery, then the resulting ESR drop of the battery during pyro discharge may cause the voltage on your uProc to drop to low levels where it could be reset or shutdown.

By using capacitors, you can effectively isolate your high current path from your uProc.

Also, capacitors will ensure you get the necessary current to your pyros, even if the battery is low.
 

RCBrust

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Capacitors do not deliver the energy that many people think they do. For example, a 1000uF capacitor charged to 9 volts only stores about 0.04 joules of energy. Another way to look at it is, it will take less than 0.01 seconds to completely discharge that capacitor if you draw 1 amp of current out of it.

Capacitors are better used on the electronic side of things where power draw is very small. Of course, isolation is needed between the high current and low current supplies for this to be of any advantage.

Randy
 
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DexterLB

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It depends on the implementation.

If you run your uProc and pyros on the same battery, then the resulting ESR drop of the battery during pyro discharge may cause the voltage on your uProc to drop to low levels where it could be reset or shutdown.

By using capacitors, you can effectively isolate your high current path from your uProc.

Also, capacitors will ensure you get the necessary current to your pyros, even if the battery is low.
That's why I do the thing with the capacitor - i charge it through a 510ohm resistor, and discharge it directly through the mosfet into the e-match. The resistor does not let any current from the battery. But I'll put the pads for the capacitor & resistor, and i could put a 0ohm resistor and short the capacitor, and try it that way. If it resets the MCU - i'll put the capacitor.

As jderimig said - i may not need a capacitor.

As for the MOSFETs - I found a dual N-channel 5.2A MOSFET (in SO-8) in a drawer, and I think it'll do the job. 5A are enough, right?
 

RCBrust

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Depending on the size of your capacitor, this may only work with the most sensitive e-matches.

Another option is to let your high current supply be connected directly to the battery, then diode couple to your low current supply which will include a large capactitor. That way the full current capability of the battery can be used on your outputs and if the battery voltage sags under load, the low current supply won't be pulled down with it.

Randy
 

FROB

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Like John said, if you use a 9V battery, no capacitor needed. your MCU regulator @ 5v or 3.3v allows for plenty of headroom so a liittle sag wont be seen by the mcu. try to use a low-dropout regulator here, it helps tremendously in this regard. if your not sure which, PM me for some suggestions.
On the other hand, if your planning to use much smaller batteries like a couple of coin or button cells, then your approach with the capacitor will work, but best to use a bigger low-ESR supercap, like 1F or more. Physically, they are quite small.
If your battery is in between, say like a small 3.6V lipo cell that can directly fire an ematch but might sag too much for the MCU, then what RCBrust suggested is the best approach- if your MCU only draws 1mA for example, a 1000uF cap will only drop 1V/second while the ematch is firing, which should only take a couple 100's milliseconds at most- so you can probably tweak it to use a smaller cap in that case, again it depends on actual MCU current draw. It's not too hard to put most micros in a uA low power mode for a short time and have them wake up by a timer so a really tiny cap there might be sufficient. I'd use a small shottkey diode in parallel with a high ohm resistor to feed the cap, to help on power up and any high current peaks.
 

DMcCauley

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Just keep in mind, your design depends on what your load requirements are.

Sure, a simple low current ematch may not require much energy, but if your application demands paralleled ematches or even a high current igniter, than your energy storage requirements may differ.

However, good pulse power practice is to always utilize a cap when instantaneous power is required to a pulsed load, i.e. ematch. In this case, you would size the capacitor to hold all the necessary energy to fire your load, and the battery would only be required during recharge.

Of course, with a single discharge event, even thats a bit overkill and the 9V battery would suffice.
 

jderimig

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However, good pulse power practice is to always utilize a cap when instantaneous power is required to a pulsed load, i.e. ematch.
No, good practice is to design a power supply that can deliver the required current to the load continuously.

Its not that hard to accomplish. Most ematches have all-fire currents of 1.25A or less. Increasing the continuous current much above the all-fire current does not statistically increase your reliability.

A 9v battery will comfortably fire an ematch or 2 in parallel. The risk as Randy has stated in a previous post is that the voltage drop may reset the mpu which can be prevented by capacitance and isolation on the logic power supply as FROB has explained.

Putting extra capacitance on a poor pyro power supply is putting lipstick on a pig.

--john
 

DMcCauley

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No, good practice is to design a power supply that can deliver the required current to the load continuously.

Its not that hard to accomplish. Most ematches have all-fire currents of 1.25A or less. Increasing the continuous current much above the all-fire current does not statistically increase your reliability.

A 9v battery will comfortably fire an ematch or 2 in parallel. The risk as Randy has stated in a previous post is that the voltage drop may reset the mpu which can be prevented by capacitance and isolation on the logic power supply as FROB has explained.

Putting extra capacitance on a poor pyro power supply is putting lipstick on a pig.

--john
Sorry, thats incorrect. Sure, it will work for the given application (i.e. rocketry altimeter with ematches), but for pulse power systems in general, you NEVER design the power supply to handle the full load continuously. That defeats the purpose.

In pulsed power design, you have a power supply and energy storage. The energy storage (i.e. capacitors) deliver the actual energy to the load when the pulse is switched on, and during the OFF time, the power supply will recharge that energy storage. This is true of both single-shot type pulsed systems (i.e. camera flash, Marx Generators, Sandia Z-pinch, etc...) and of repetitive pulsed systems (i.e. pulsed radar, lasers, etc...)

The power supply in these applications is effectively isolated from the load (usually with a high inductance to "smooth" the load so the supply never "sees" a pulsed load, but rather an average current)

You can equally apply the same principles here. However, since ematches are basically one-shot events, you really don't need to worry about pulse repetition and a 9V battery is more than enough to supply a single ematch.

So again, your initial statement is incorrect as well as impractical, especially in higher power systems.

Putting extra capacitance on a poor pyro power supply is putting lipstick on a pig.
Sure it is if you don't understand the principles at hand.

That said, for some high performance small rockets, i'm actually designing an altimeter that uses a single 1.5V watch battery. It provides more than enough energy to provide the necessary current to 2 ematches. An onboard simple booster converter steps up the voltage which charge two capacitors to about 50V, and then stores it until the altimeter is ready to fire. The result is an altimeter and battery that fits to the size of a quarter. Initial testing is 100% successful.
 
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jderimig

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.........Sorry, thats incorrect......

........So again, your initial statement is incorrect as well as impractical, especially in higher power systems.

.......Sure it is if you don't understand the principles at hand.

.....i'm actually designing an altimeter that uses a single 1.5V watch battery. The result is an altimeter and battery that fits to the size of a quarter. Initial testing is 100% successful.
Congratulations!!!!!:p

You are rude and unnecessarily argumentive.

Last time I checked this thread is about firing ematches which is not high power pulsed application. (Its not radar, camera flashes, or Hadron collider stuff) For such applications a simple power supply that can supply sufficient continuous current is simpler and thus a superior topology.

Other than tiny size, which is of limited value in most, not all, rockets that use electronics, what advantage does this more complex topology offer other than a bells and whistles factor?

By the way like you I have designed and marketed altimeters also. It really isn't that hard or does it require unusual above average skills. How about coming down a few notches from your superiority complex and realize there are valid points in every argument (even yours have some, acknowledge that I may have some).

(Sorry FROB for hijacking your thread)
 

FROB

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No worries! ;):p
 
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DMcCauley

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You are rude and unnecessarily argumentive.
John,

I'm merely correcting two statements you made which are not correct. No need to take it personal. If you post blatantly incorrect statements in a discussion forum, expect to be corrected. Plus, its a discussion, so no need to take it personally. I'm sure if we have this discussion on the field, it would be a most civil discussion. Ask anyone here that knows me.

No, good practice is to design a power supply that can deliver the required current to the load continuously.
This statement as I said before is not a true statement. The poor choice of words on your part here is "continuously." Thats the problem i have with your statement. You are basically saying that if my ematch has a fire current of 1.25A, i need to design a power supply that can continously deliver 1.25A which is not true. It only needs to pulse 1.25A of current for a very short duration of time. The power supply (in this case the battery), does not need to be sized to deliver 1.25A. Thats the point i'm trying to make. If you use a capacitor bank, you can use a battery that only puts out say 100mA maximum. Its a trade off which needs to be made. Or of course, you can go brute force and just put a large 9V battery in there.

The second statement i have a problem with is the following:

Putting extra capacitance on a poor pyro power supply is putting lipstick on a pig.
Firstly, and you can correct me if i'm wrong, i take a "poor pyro power supply" as one not capable of continuously supplying the necessary current to fire an ematch. Now by saying putting capacitance on a low output current power supply (changed "poor pyro power" to low current power supply) is putting lipstick on a pig is ABSOLUTELY false. You can absolutely do this and with great advantages too, such as reduced size, weight, and cost (battery)

Since this thread started out as a DIY altimeter thread, these points i'm making are absolutely relevant.

I've hijacked this thread enough. This is my last post on the subject.
 
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jderimig

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If you post blatantly incorrect statements in a discussion forum, expect to be corrected.
Likewise. Peace.

PS. If you come to LDRS28 we can share design philosophy over a cool one (soft drink of course).

--jd
 

DexterLB

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Right.
I tested the "5.2A" MOSFET and it went off at 2amps. My fault - didn't notice the "MADE IN CHINA" title. So i'm staying with the two 75307D MOSFETs.

As I said, I am going to make the PCB without the capacitor, but with possibility to add one.
One question remains: where can I get a good 9v battery from?
 

DMcCauley

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Likewise. Peace.

PS. If you come to LDRS28 we can share design philosophy over a cool one (soft drink of course).

--jd
And my apologies for sounding like an argumentative ass. I do tend to get outspoken at times.

I'll be at LDRS, so save me a cool one!
 

DexterLB

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Another option is to let your high current supply be connected directly to the battery, then diode couple to your low current supply which will include a large capactitor. That way the full current capability of the battery can be used on your outputs and if the battery voltage sags under load, the low current supply won't be pulled down with it.
I simulated this in Proteus, and I found out that it's a better solution. A 100uF relatively small capacitor will keep my MCU alive for 1/4 segond if the power drops - that's more than enough. And, the best of all - this way the routing is simpler and the board is smaller.

Again, thanks for all the replies! You can't imagine how helpful this thread is!
 
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bobkrech

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Just a quick review of pyrocircuits used in altimeters.

There are two types of pyrocircuits currently used in hobby rocketry altimeters: 1.) Switched DC pyro circuits. (conventional); and 2.) CD (capacitor discharge) circuits.

The power developed by a pyrocircuit is the product of the supply voltage and the current drawn by the igniter circuit.

P = I*V = Vbattery^2/Rtotal = Rtotal*I^2 where I is current, Vbattery is battery voltage, and Rtotal is the total series resistance of Rigniter, Rtransistor, Rwiring and Rbattery.

The power delivered to the igniter is P = Rigniter*I^2.

With a switched DC pyrocircuit, current sourced from a battery is switched electronically to activate a pyrotechnic device. At a minimum, the battery, switch and wiring must be able to source, at a minimum, the rated all-fire current of the e-match or pyrotechnic device to be activated (This is only known if if it is measured or supplied by the manufacturer.). Ideally you should be able to source 2-3 times this value to insure that if the pyro device is an e-match of igniter, it activates it's pyrogen in only a few milliseconds.

There are several reasons to use this type of circuit. If you use an unboosted (pyrogen free) resistance wire to activate a Pyrodex(R) ejection charge (or BP charge for that matter), you may need to heat the wire of several tenths of a second to ignite the charge. Or if you are activating a high current device that requires a lot of energy (E = P*t where E is the energy in Joules and P is the power in watts and t is time in seconds.).

To deliver high currents for long periods of time (tenths of seconds) your pyrocircuit requires a battery with a low internal resistance (high current rating), a low resistance electronic switch (transistor, FET or MOSFET), and low resistance wiring. The electronic switch must be rated for the short circuit current of the pyrocircuit, otherwise the maximum current that can pass through the switch must be limited with a series resistor.

The disadvantage of this pyrocircuit is that you need a large battery to supply the current. You also need to protect the microprocessor from a brownout when the pyrocircuit is activated. This can be achieved by using separate batteries for the microprocessor and the pyrocircuit, or by buffering the microprocessor from voltage dropouts with a current limiting resistor feeding an energy storage buffer capacitor.

The CD pyrocircuit was developed (and is used in many other applications) to delivery high current (high power) pulsed power for "short" time periods from devices employing small batteries. The PerfectFlite MAWD altimeter is an example of this applied to hobby rocketry. http://www.perfectflite.com/Downloads/MAWDManual.pdf

In the PF MAWD, a pyrocircuit consisting of 4700 uF capacitor and a low on-resistance MOSFET can deliver up to 27 amps of current for a short time. The capacitor stores 0.19 Joules which is more than enough to activate a properly selected commercial e-matches. For example the on-resistance of the MAWD pyrocircuit is 0.33 ohms. (R = V/I = 9/27 = 0.33), and a common e-match resistance is 1.6 ohms. The old DaveyFire N28F e-match had a 1.6 ohm resistance, a 1 amp all-fire current, and a 3x all-fire activation time of ~2 milliseconds. This is an all-energy of only E t*R*I^2 = 0.002 * 1.6 * 3^2 = 0.029 Joules. The Rtotal of the MAWD pyrocircuit is ~2 ohms, so the igniter will draw an initial 4.5 amps from a capacitor charged to 9 volts.

The discharge current of a capacitor follows an exponential decay equation. http://www.antonine-education.co.uk/physics_a2/module_4/Topic_7/Topic_7.htm

I = Io*e(-t/RC) = 4.5 * e(-t/0.0075) where RC = 1.6*0.0047 = 0.0075 seconds. In 0.002 seconds, the current will have dropped to ~3.5 amps, so in 0.002 seconds, the CD circuit has delivered an average of 4 amps or ~0.05 Joules into the igniter which is more than required to activate the pyrogen on the e-match. Since the energy is stored and delivered by the capacitor and not directly from the battery, the current capability of the battery can be much lower. The microprocessor in the MAWD draws only about 0.008 amps and can be powered with only 5 or 6 type SR-44/357 Silver Oxide cells in series. These tiny button cells can supply 0.1 amps on a short, but are rated for ~0.020 amps of continuous draw. 6 AgO batteries in series supply 9 volts. If the charge current is limited to 0.020 amps, the 0.19 Joules of energy stored in the capacitor can be recharged in 1 second! 6 silver oxide cells weigh a fraction of the weight of a 9 volt alkaline battery, but will provide the equivalent function if a capacitive discharge circuit is employed.

Hope this helps.

Bob
 

DMcCauley

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Well said Bob.

The microprocessor in the MAWD draws only about 0.008 amps and can be powered with only 5 or 6 type SR-44/357 Silver Oxide cells in series.
This is actually how my micro-altimeter works, but it uses only a single watch battery cell. The capacitor discharge circuit is sized to fire (2) parallel daveyfire type electrical matches and features both a drogue and main firing circuit.
 

DexterLB

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Bob, thank you for that great physics lesson! Anyway, I figured another circuit out: I put a diode, which does not allow the high-current part to draw voltage from the low-current part, and a small (100uF) capacitor which can hold the MCU running for 1/5 of a second, and I'll be able to put a second big capacitor on the power line (e.g. paralelly with the battery), which will actually power the e-match. Or i can alternately put a larger battery. In both cases the diode and the 100uF capacitor will keep the CPU from resetting :)

tomorrow I'll post a circuit.
 

DMcCauley

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Bob, thank you for that great physics lesson! Anyway, I figured another circuit out: I put a diode, which does not allow the high-current part to draw voltage from the low-current part, and a small (100uF) capacitor which can hold the MCU running for 1/5 of a second, and I'll be able to put a second big capacitor on the power line (e.g. paralelly with the battery), which will actually power the e-match. Or i can alternately put a larger battery. In both cases the diode and the 100uF capacitor will keep the CPU from resetting :)

tomorrow I'll post a circuit.
Just remember where you put the diode as there is a voltage drop.

In this arrangement i would put the diode in series between the 9V battery and then the 5V regulator and not directly on the Vcc of the uProc. This way, when power is disconnected, you'll droop on the input side of the 5V regulator, and the regulator will continue to maintain regulation even during the droop until you reach the minimum voltage level for that particular regulator (2V for 78xx, or 1V or LDO types etc...)
 

DexterLB

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Just remember where you put the diode as there is a voltage drop.

In this arrangement i would put the diode in series between the 9V battery and then the 5V regulator and not directly on the Vcc of the uProc. This way, when power is disconnected, you'll droop on the input side of the 5V regulator, and the regulator will continue to maintain regulation even during the droop until you reach the minimum voltage level for that particular regulator (2V for 78xx, or 1V or LDO types etc...)
That's what i did - once that voltage drop turned on my head, so now I was prepared :D
I'm planning to put 78L05 as a voltage regulator, which is not an LDO, but if the CPU resets during the ground-test, I'll replace the 78L05 with an LDO-7805, which is bigger (TO-220).

Here's the circuit:


And the PCB:


View attachment timer_proteus_pcb_and_sch.zip
 

DMcCauley

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You need some resistors on the gates of the MOSFETs. They should be sized so you don't violate the maximum current ratings of the PIC as the MOSFET gate will "look" like a short circuit (capacitor) when you first turn on although the gate charge characteristics of those particular MOSFETs are quite low anyways. Probably something on the order of 1k - 10k. Also, with a gate resistor, you may get excessive ringing on the gate (no critical damping) which could sometimes cause your MOSFET to oscillate a bit, however, this is usually critically damped with a low value resistor (such as 5 - 10 ohm) - then again, for single shot, it really doesn't matter anyways.

Just add some 1k-10ks and it should be golden.
 

FROB

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You need some resistors on the gates of the MOSFETs. They should be sized so you don't violate the maximum current ratings of the PIC as the MOSFET gate will "look" like a short circuit (capacitor) when you first turn on although the gate charge characteristics of those particular MOSFETs are quite low anyways. Probably something on the order of 1k - 10k. Also, with a gate resistor, you may get excessive ringing on the gate (no critical damping) which could sometimes cause your MOSFET to oscillate a bit, however, this is usually critically damped with a low value resistor (such as 5 - 10 ohm) - then again, for single shot, it really doesn't matter anyways.

Just add some 1k-10ks and it should be golden.
Actually, a series resistor isnt really needed, the very brief current pulse wont hurt the pic, but i agree its a good idea from a ESD protection standpoint (on all i/o's). Much more important though, is a pulldown resistor, because the gate can float momentarily into conduction on powerup or brownout when the pic is in reset, and before it has had a chance to switch its i/o's from high-impedance inputs into outputs and pull them low. something in the 100k range is usually used.
 

ben_ullman

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Likewise. Peace.

PS. If you come to LDRS28 we can share design philosophy over a cool one (soft drink of course).

--jd
Hey Ill referee that match no problem. Set a time Ill sell tickets and cut you guys in ;)

I like the timer layout. Its got some SOIC but I would feel comfortable soldering that myself. I like the design and I actually am (hoping) to get someone to write some software for my all in one altimeter/GPS/camera/downlinking/feedback/beer serving setup :)

Ben
 

DexterLB

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I put two resistors for protection and two pulldown resistors, printed the pcb and put it into acid. Now i'm waiting.
 

DexterLB

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Hey Ill referee that match no problem. Set a time Ill sell tickets and cut you guys in ;)

I like the timer layout. Its got some SOIC but I would feel comfortable soldering that myself. I like the design and I actually am (hoping) to get someone to write some software for my all in one altimeter/GPS/camera/downlinking/feedback/beer serving setup :)

Ben
LoL!
Well, that ultimate embedded systems software designer won't be me! I just know the basics - it took me a month to make the timer software. (with a linux&windoze version of the pc part) :D

As for the soldering - it's my favourite part! I really enjoy it
 

DMcCauley

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LoL!
Well, that ultimate embedded systems software designer won't be me! I just know the basics - it took me a month to make the timer software. (with a linux&windoze version of the pc part) :D

As for the soldering - it's my favourite part! I really enjoy it
Then you should definitely check out the following kits!
Will give you your fair share of soldering fun!
http://www.easternvoltageresearch.com
(Sorry, shameless self-promotion!)

Now, i'm just trying to figure out how to utilize a Tesla coil inside a high power rocket. Will definitely require some thinking on my part! ;)
 
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