Author Topic: Raptor 40a enable circuit (Read 10797 times)

snap crackle · « **on:** November 16, 2014, 10:23:47 PM »

I am having an impossible time understanding the enable circuit on the 40a Raptor variant....

It says the maximum input V for high logic on is 5.5v on page 4....
http://old.lineagepower.com/BinaryGet.aspx?ID=cef8d303-9566-4f16-ae31-de97e2a0e763
I am using an 11.1v lipo and I'm not sure if I understand the enable feature correctly...

I was originally going to use + through a 9.1 zener to a momentary switch and to the enable pin
with a 4k7 resistor from the enable pin to ground...

Can someone please help me?!?!

wa9w00d · « **Reply #1 on:** November 17, 2014, 07:08:40 AM »

Not an EE, but I would try just a voltage divider. I would start with 100k over 10k, see what the voltage at the enable pin is like.

Wayne

snap crackle · « **Reply #2 on:** November 17, 2014, 01:39:14 PM »

I feel like that would allow a lot of parasitic current drain.... even a divider circuit of say 330k 33k is still going to have a substantial amount of parasitic
draw....
Someone please correct me if I'm wrong.

wa9w00d · « **Reply #3 on:** November 17, 2014, 01:55:52 PM »

Well, even if the bottom half was shorted, 100K ohm will only pass 111uA at 11.1 volts. The data sheet that you linked to gave an impedance range for the enable line of anywhere from 300 ohms to 11K ohms, if I remember correctly. According to what the actual behavior of that input is, you might be able to go higher than 100K. If you needed to a lot lower, then yeah, you would need a more involved solution, but as long as I had a master disconnect, I would be satisfied with 100uA loss. YMMV.

Peace,
Wayne

Visus · « **Reply #4 on:** November 17, 2014, 02:12:31 PM »

A good power on switch will eliminate that issue if I was building that unit and not using enable I wouldn't personally IMO....

Problem is there are not many 40a NO switches but in my use I have had 0 issues with a 4A c&k switch at 8.4v 70W --11v is substantially higher voltage for that switch and IDK..

Heres a vol div calculator. low/off is 200ua @-0.3 thru .4v high 1v- 5v
200ua will take a long time to draw your batts down but it will..
http://www.calculatoredge.com/electronics/voltage%20divide.htm

snap crackle · « **Reply #5 on:** November 17, 2014, 02:30:47 PM »

The enable circuit won't draw anywhere near 40a and this is why I wanted to use that instead of a mosfet on the Vin circuit.
If I use the voltage divider circuit after the momentary switch then it wont have any parasitic draw correct?

Visus · « **Reply #6 on:** November 17, 2014, 03:00:12 PM »

Makes it that much easier to build lol.

You will have a draw if using it it needs to pulled low in order to be off if using the enable function. There's no getting around that part to my knowledge maybe there is.. lol

Totally forgot about fets lol
boom there goes the snap crackle -- git er dun lol

wa9w00d · « **Reply #7 on:** November 17, 2014, 03:01:18 PM »

You could use a momentary, heavy switch, but I would recommend something like a slide or rocker, and use the enable to fire it. Any time the chip is powered, it's gonna use some of your battery, even if you have it disabled. You shouldn't be drawing 40 amps on the input side. That's the output max. Since you're using 11.1 volts in, current there might be approaching 23 amps, but you would need to run the numbers to get a better idea.

The good thing about using a disconnect switch and then the enable line to fire is that your disconnect switch "switches" under low amperage conditions. Most of the damage that happens to a switch comes from making or breaking a high current circuit.

Wayne

snap crackle · « **Reply #8 on:** November 17, 2014, 04:16:28 PM »

The module is OFF when enable pin is logic low. Module is ON logic high or pin open....
Am I reading that right?
That would mean that there wouldn't be any continuous loss unless the enable pin was indeed "enabled"
I would like to minimize any additional components if possible.

Could I just use a normally closed switch to ground for the enable circuit?

wa9w00d · « **Reply #9 on:** November 17, 2014, 04:45:17 PM »

It's a good thing you're asking questions, 'cause it make me check myself

. If I'm reading that data sheet right, you shouldn't even need a pull up circuit. The enable pin should float high, and the board will output. You SHOULD be able to just put a N.C. momentary switch with a pull down resistor on the enable pin. When you open that switch, the pin should float high, output comes on.

But you will have some current draw even when output is not enabled. Like anything that is "remote controlled", off ain't really off. If I read correctly, 200uA. That's why you want a master disconnect, or disconnect the batt. when it's gonna sit for a long time. If you put it on a charger every night, you're never gonna miss those 200uA.

Wayne

ETA - If I did my math right, it would take over 7 months for 200uA to drain a 1100mha batt.

snap crackle · « **Reply #10 on:** November 17, 2014, 05:13:12 PM »

I wish someone could confirm the normally closed switch operation on this unit....
It would simplify everything so much...
If I use a normally open switch I need to use a pull down resistor and a voltage divider to make this unit function properly....
Then I'm left with the zener issue and voltage loss on my Vin circuit.... possibly need to run two 9.1v zener parallel

snap crackle · « **Reply #11 on:** November 17, 2014, 06:29:41 PM »

I am by no means an EE either but if I've used my noggin correctly this should double as a
pull down resistor and a voltage divider with minimal parts correct?