Does this Schematic look right? (Firing Electronic Matches)

[NickS]

Sure there are ways... I am just trying to think of low impact options.

The problem isn't the MOSFET its the green LED. You would find if you break the line going to the anode of the green LED that the red does not come on(because there is no place for the pnp base current to flow). One thing you definately don't want to do is put a diode inline with your high current path.

let me think some more.

 

[funman1]

Another thought....
Is there any way to create some sort of non-inverting amplifier?
IE: Tap off the +3.3 volts from Arduino output and amplify that to +12V?

I'm looking at like opamps but they are smoking my brain right now, I don't have anough background in this to quite get them...

Worst case I use a NPN driven by the Arduino, and then use that output to drive the I have in the circut now PNP(2N3906)?
So instead of the PNP(2N3906) negin driven off the mosfet, just add a NPN (unk number) and drive from the microcontroller?

I wish I had some better reading materail on this type of stuff
I guess I need a class of some sort....

 

[funman1]

After giving myself a crash course in op amps....
I think this will work??

The op amp should infinitly multiply the voltage all the way up to the supply (12vdc)
So the 3.3 volts becomes 12.... Atleast if I'm understading them correctly?

 

[NickS]

I am pretty sure that will not satisfy your criteria. First off the green LED will be on all the time except when firing. This includes when there is no ematch in place. So this is actually no improvement over what you had before.

I think I have something but I have not fully tested the idea yet. The theory is essentially the same as original with one exception. When the ematch is out of circuit the current that has been going through the red LED is re-routed by Q3 to bypass the red LED. This did require getting creative with the resistor scheme but it is the lowest impact fix I could come up with that satisfies your criteria. The net part list change is one extra transistor and two extra resistors.

****EDIT
Sorry I forgot that your LEDs are tied together at the cathode. So in my schematic simply reverse the position of LED2 and R2.
btw- LED2 is green and LED1 is red in my schematic

 

[funman1]

Sorry it was late and I didn't look, and was very tired.
Here is what I meant hehheh

 

[NickS]

Hmm it was late for me too because I can't quite understand how I came to my previous conclusion. It would actually be that the Green LED would be off all the time except when firing. But that is in the past. So looking at your new schematic it is ok on logic now but the way you are using the opamp could give you bad results.

When an op amp is used to achieve some reasonable gain there is always feedback involved to reign in the gain and increase bandwidth. When you run an opamp with no feedback like you are then the gain is out of control huge meaning a very slight input would amplify all the way to a rail(on the order of 100V gain per mV input). Now that is what you want to make your 3.3V signal boost up to 12V. But on the flip side when you try to pull it low can you ensure that the voltage on pin 4 never climbs a mV above ground? The answer is no. So with this configuration you cannot gaurentee the op amp output stays low.

Not to worry though if you still want to go the op amp route we just have to add a resistor divider. You then tap off of the divider to set a non zero voltage on pin 3(1V would be fine). I should preface by saying that opamps are differential thus whith massive gain it is just a game of pulling one input lower or higher than the other. Back to the fix. So if pin three is fixed at 1V and pin 4 is held to 10mV(nearly zero) then pin4 is lower than pin3 and the amp with slam into the Vss rail (in your case ground). Then if you were to bring pin4 up to 3.3V now pin 4 is larger than pin3 and the output slams into the Vcc rail(in your case 12V). That is the behavior that you want.

Now for the details. This kind of use is better suited to a comparator. A lot of work and focus goes into making an op amp preform well in controlled gain situations. That is NRE that you don't want or need. So a comparator is basically an op amp that has been modified to only support this sort of rail-to-rail switching you are after.

Other considerations include that the output drive on many of these devices are not very strong so you need to make sure your amp/comparator can handle the current you need for the LED. And you want to make sure your pick also offers rail-to-rail output. Some devices will not output all the way to the rail voltages.

 

[funman1]

Well I just snagged a quad package op amp chip, and bread board tests look VERY good!
It's working exactly how I had expected / wanted. Do you think I should add pull down resistor of like 10K to gnd for each op amp input just to make sure?
It's working without them but just as good measue?

Yes it's not pulling all the way up to the rail, only about 11.3 volts instead of 12v But it's just for the LED and it looks great so?
The quad package says it can sink 20ma per channel, and with the led being current limited to about 10 ma via the 1K resistor this should be ok, right?

I'm so close now hahaha...


~Steve~

 

[NickS]

I think you missed my point. You can add a pull down if you want but if you are going to do that you really should just add one more resistor and properly bias all the neg inputs.

The only way to guarantee stable operation is to actually pull the voltage on pin4 below the the voltage on pin3. That can't be done it they are tied to the same voltage(gnd). The only reason it looks to be working is because you must be driving pin4 in your test with a pretty clean 0V. That 0V is still being amplified by around 10k but 0V*10k is still zero. You may get away with it working sometimes but if you ever get a little noise on that input it will start to flicker. Perhaps you don't care about a little flicker. But if you re-spin the board to accommodate the op amps then you really should add a non zero voltage reference to pin3's.

Congrats on nearing the end of your project.

 

[funman1]

Hahah You mentioned a flicker, and I looked REALLY close and it does!
When ever the unit sends a command out the XBee it does have a tiny flicker.
So then should I add like a 100K Resistor between pin 3 and gnd?

 

[funman1]

It appears that got rid of the flicker...
But is the solution correct?

Well far from the end of the project!
This is just the reciver, once this is 100% done the I begin work on the Sending unit! and possibly more recivers....

 

[NickS]

Actually I was thinking a voltage divider so the connection would be 12V to Ra to Rb to Ground you tap off between Ra and Rb for your reference voltage.

A 10 to 1 ratio(Ra:Rb) on the resistors should put the level in the right range(just over 1V) So you could use 20k and 2k or 100k and 10k. I am not sure what opamp you are using but you just have to pick values low enough to satisfy the input current spec of the op amp.

 

[funman1]

Do you have a pic / simple schematic of that?? I think that just spun my head hahah

 

[NickS]

Vout = Vin * R4 / (R3 + R4).
Idraw = V / Rtotal, Rtotal = R3 + R4

So keep the ratio of R3:R4 =10. Thus if You pick R3 = 100k then R4 = R3/10 = 10k.
You will see that as long as the ratio stays the same the voltage will stay the same. But using smaller value resistors will source more current. see below

R3 = 10k, R4 = 1k
Vout = 12V * 1,000 / (10,000 + 1,000) = 1.09V
Idraw = 12V/11k = 1mA

R3 = 100k, R4 = 10k
Vout = 12V * 10,000 / (100,000 + 10,000) = 1.09V
Idraw = 12V/110k = 0.1mA

 

[NickS]

This might help you visualize what I am saying. The circuit below is just an opamp like you have. The inverting input(-) is tied to a voltage divider and for simulation sake the AC voltage is taking the place of your uC drive signal. So now looking at the plot
Green = the output
Red = (+ input) uC drive signal(I know its not really a sine wave)
Blue = (- input) Voltage divider

If you look carefully you will see that the output transitions from high-low or low-high whenever the +input crosses the voltage level of the -input.

Does that make sense?

 

[funman1]

Well I think I will stick to the single resistor since it's working correctly.
I'm just straight out of room on the board.

I updated all the pics of the board with the new configuration.
Let me know what you think.

 

[NickS]

Nice work! The rework you did on the power FET layout is excellent. And the rest of the layout looks nice too.

Sorry to beat a dead horse but from a component space standpoint I see you have 2 unused op amps right? You could use one of those and 1 more resistor to drive the -input level for all the rest of the op amps. I understand its not a big deal and you are probably ready to move on but if your interested here is how you would do it.

You should already recognize R1 and R2 but this time the level is fed into one of your unused amps that is configured as a unity gain(0 gain) buffer. Then the output does the work of driving the other 10 -inputs to the desired 1v level.

 

[funman1]

Now that's slick if it works
OK respun it again and updated pics.

Now I only have 2 resistors instead of the 10 surface mount (SMT) I was going to have to use.

 

[NickS]

Can you give me the op-amp part number?

 

[funman1]

Op Amp LM324

 

[funman1]

Doing last min checks on all the traces and such before I pay to have the PCBs made up.

Quick question on battery monitoring....
Right now I'm using 2 resistors a 3.3K tied to gnd and a 6.8K tired to 12V positive and I take my reading from the junction between the two resistors and feed that into the Arduino.
Looking at this chart
http://www.scubaengineer.com/documents/lead_acid_battery_charging_graphs.pdf

Looks like I should program my system to read low battery at say 11.8 volts?
To me that does not seem very discharged?