Im planning a new & 1st project for electric imp & April board.
Im going to use power supply and not a battery, so i was thinking of just taking one of my mobile phones chargers, which usually are 5v @ 0.7-1A and plug them to the board. will it work fine? Of course, with voltage regulator to 3.3v.
Will it do the trick?
another issue i wonder is, if i have a 5v power supply, but not usb only two wires, so i can plug it to P+/P-
If i find a mini usb, will it be fine just to plug it directly to the board?
and if i do, can i draw voltage for other electronics from the board, since the power supply is plugged directly to the board?
If so, what is the output voltage i can draw? Is it regulated?
Another question I have, what is the output voltage of the pins?
https://www.sparkfun.com/products/11400?
http://www.ti.com/lit/ds/symlink/tps62172.pdf
what I understand is, that you can power the imp with 3 to 17 V or use USB, it has a regulator on board
Yep, USB plug is the easiest way (set the 3-way jumper to USB). If you have a 5v power supply with 2 wires, connect GND to the GND pad and 5v to VIN pad.
Great. so the vin /usb actually support up to 17v.
but what about the output pins, what voltage does it output?
because i have a relayto control (needs 5v signals) and serial lcd display (needs 3.3v)
if i attach the power supply directly to the board (either USB or VIN), what voltage can i draw from it and how?
the imp works on 3.3 V, so I suppose the output is max 3.3 V. But I don’t think you can drive a relay directly with the output of the imp (maybe with a solid-state relay)
if its only 3.3v and not 5v, so im in trouble.
what can i do then?
could you suggest a method of controlling a relay with the possible 3.3v i can output?
The way i see it i look at few options
- nmos transistor? plug in parallel the 5v source (does it matters regulated or not?) to the drain, source to relay, and the gate is the signals i read from the board?
- test the relay and see if 3.3v can turn it on? (connect the multimeter to both edges and see if it beeps while 3.3v)
- any other relay that works with only 3.3v?
- some kind of voltage amplifier?
http://devwiki.electricimp.com/doku.php?id=exampleswitch
don’t forget the diode otherwise you WILL destroy your transistor
other than seeing the light controller is my main wish, i couldnt see any details about voltage output or relay used in the process
look at the second schematic (with the diode, transistor, resistor and relay)
Im not sure about the schematic.
My relay needs to be 1 leg to the home electricity, the second to a 2.5kW device (the water heater), which needs about 12A.
The schematic shows the relay being connected to VIN (which is going to be either 3.3 or 5v supply. so its not my intention.
other than that, what is the diode feedback for?
when you switch of the relay, the coil generates a large current, which will destroy the transistor. The diode will prevent this. If the relay is capable to switch 12A, there is no problem, if not, you can use this relay to drive a second heavy-duty relay.
in the schematic, you will have to remove the connection from VIN to the switch connection of the relay and replace it with the home electricity.
the switching of the relay generates large currents? But the intended current to flow through the relay is supposed to be high, 12A, so it will fry anyway.
i dont understand this configuration much, also that the input of the relay is vin and not my wall socket, which is really what i need.
no, not the switch generates the high voltage, but the coil of the relay.
http://in.answers.yahoo.com/question/index?qid=20070913060736AAihyES
second problem, i explained.
Are you sure you want to play with high-voltage switching??
Maybe i wasnt getting the right impression about relay.
I, at the moment have a regular switch to turn on/off, manually, the 2.5kw device.
instead, i wish to controll the switch by a simple pulse from the imp.
I understood that the relay is just for that, switching high voltage with low voltage circuit, so that if the relay has vin/vout and v_coil,
vin is my house electricity, vout is my 2.5kw device and v_coil is the imp telling the relay to close/open.
according to you, or atleast, sorry if i dont understand correctly, you say i need vin to be the board’s vin (=5v), so theres no way my 2.5kw will work
does this make sense?
More sense, yes
i thought at first the wiring is using vin/vout.
if i connect 5v to vin to power the board, i also have 5v to v_coil
i dont understand the part with the diode,bjt and wiring. i uploaded a photo.
i understand from it that GND of the relay is wired with diode to vcoil. what for?
and the transistor sends the signal from the board through the diode.
if pin1 is 0v, so the bjt is closed, therefore the diode is closed and no current flows in that branch, but the coil is still connected to vin, so its still on.
if pin1 is 3.3v,bjt is open, and what happens then? I have -5v and its emitter, and the collector is wired to GND and with diode to vcoil?
if pin1 =0 nothing happens
if pin1 =1 (3.3V) relay is activated trough transistor. if vin = 5V you’ll need a 5V relay. If vin = 12V you’ll need a 12V relay (capable of switching 12A)
triangle (?) means ground (GND)
in college triangle is a minus voltage, but if in the sketch they meant ground, so ok.
the transistor is indeed connected to both ground and vcoil?
what i dont get is how i can have vin connected to vcoil and still get there voltage only when diode is closed. they are connected in parallel, so it should matter if diode is open or closed
have you read the link about protective diodes above and did you understand it?
again:
http://in.answers.yahoo.com/question/index?qid=20070913060736AAihyES
diodes do not open or close, just look at the polarity, and it’s only to protect the transistor. it has no affect on the functional working. you can leave it out but then you will destroy the transistor
OK I got the idea behind the diode, its only for protecting the transistor.
but its still connected in parallel to v_coil with vin, so even when pin0 = 0, i still have voltage there, enough to power the relay. so what am i missing?
and the second thing thats still missing is why does the transistor and diode are connected to relay’s GND too? I would plug the transistor & diode directly to vcoil together with vin (assuming i got it already, from my first question) and both relay’s GND & transistor second leg to the ground