Why not simply like this?
Raspberry Pi switches between 0 and 3V3, more than enough to saturate Q1, which takes over the "heavy" work: switching the +5V relay on/off. Depending on the relays you're using, small modifications for D1 and Q1 might apply.
If you want to drive a single low-voltage relay, using a 7-channel driver chip will be an overkill. You'll be perfectly fine with a single NPN transistor and a flyback diode:
simulate this circuit – Schematic created using CircuitLab
Note: R1 can be replaced by a LED and a 200 - 500 Ohm resistor in series if you want a visual clue about the state of ...
Your wiring diagram is correct, as (per the Sainsmart.com website you linked) the specs of the device are:
Input control signal voltage:
0V - 0.5V Low stage (SSR is OFF),
0.5V – 2.5V (unknown state).
2.5V - 20V High state (SSR is ON).
The Raspberry Pi uses 3V3 signals on its GPIO pins; a voltage level which is high enough to trigger the High State in ...
You should not directly drive a relay from the Raspberry Pi. An individual GPIO can only safely provide about 16mA at 3V3 which is unlikely to be enough to energise the coils of a standard electromagnetic relay. Even if you could the back EMF caused by the collapsing magnetic field in the relay coils when it was switched off could destroy a GPIO and the Pi....
The black housings are available for more than just one contact, as one row or two row version. The contacts are crimped to the wires , which means you either have a crimp tool or somehow solder it. Ideally, you can use your cable, remove contacts from the housing, and just push them into the bigger housing.
Edit (by Milliways) To add to the answer I have ...
# Introduction #
The OP would like to use Rpi to safely control a bank of 5 Sparkfun's Beefcake relay module. He had a problem because Rpi GPIO logic level is 3.3V, but his relay uses 5V logic control. He wants to know how to modify Rpi to get around the logical level disparity problem. His choices including the following: using the transistor BC5468 to ...
I had the same problem as you, but I finally solved it.
As you are powering it with 5V, both 0V or 3.3V coming from the GPIO pin are considered as "low level", so it won't actually switch.
You need to power it with 3.3V (it seems it was designed to work also with lower voltages), this way it can correctly distinguish between 0V (low level) and 3.3V (high ...
Relays are pretty safe to use without any special isolation like optocouplers. The reason people recommend optocouplers in DIY projects... well because its DIY and an optocoupler is safer in case of something going wrong.
The input pins power a coil and push or pull a lever using some kind of snap hinge to eliminate bounce and sparks. So you are already ...
I'm not sure from your answer whether you're using a relay module or a bare relay. I assumed the latter in writing the following.
You'll need a flyback diode to protect whatever output pin you're using. When you switch off an inductive load (i.e. a coil), which includes relays and motors, a large voltage spike can be generated as the coil resists ...
On many electronics boards, GPIO pins are set to a "floating voltage" (not on, not off, just somewhere in the middle) until the pin actually gets initialized or used for the first time.
Remember, when you turn the Pi on, it has to load a bunch of kernel modules and drivers, including the ones to use the GPIO pins. The pins are part of an electrical circuit, ...
Most of the pins are configured as inputs. This is normal, and usually the safest option.
All of these inputs are put into a defined state with either a pullup or pulldown. The normal values are shown in http://elinux.org/RPi_BCM2835_GPIOs These are quite high impedance ~50kΩ.
The only pins set as output are those specifically configured as such e.g. TxD. ...
The problem is that your relay board is designed to run at 5V, but your Pi's GPIO is 3.3V. The relay is a 5V relay, so the power supply needs to be 5V, but you can alter the design of the driver board slightly to make it work at 3.3V input. Looking at this image, we can clearly see the transistor is marked 2TY which means it's an S8550 PNP ...
There is nothing wrong with the GPIO, it just can't supply enough current for your coil.
If the coil has just 24Ω, at a 3,3V the GPIO would need to supply 137mA.
This source at raspberrypi.org says that the maximum current it 51mA for all GPIOs together and 16mA per GPIO.
You can connect the GPIO to a transistor that can supply the necessary 137mA.
#! /usr/bin/env python3
import RPi.GPIO as GPIO
# 5 * * * * sudo python /home/pi/fan.py
# A crontab will run every hour and check the temp. If the temp is > 49 the script will start the fan
# until the temperature goes down to 28. When it does, the script will end, shutting down the fan as ...
It will be best to power the Raspberry Pi using a good 2Amp power supply.
You will still need a transistor to drive the relay.
But really relays are for high power, 24volts up to 220volts and heavy loads that use dedicated power supplies not connected to Micro controllers.
You can get rid of the relay and just use a 2N2222 / TIP127 transistor to drive the ...
The programmable pins (General Purpose Input/Output) can be used in a variety of ways - but not all at the same time. Before the point where power is applied to the RPi's Systen-on-Chip (Soc) AND it has performed a reset the state of those pins is entirely dependent on the design of the circuity inside the SoC - it is likely (I haven't checked the ...
I am interpreting the picture of the printed circuit board of the relay module with accompanying notes as follows:
the board includes a transistor and a diode - and while it is impossible to tell from the image without looking at the wiring and/or schematics - it is reasonable to believe this is the freewheel diode you're refering to. Conclusion here: no ...
The Relay module in your picture is what so called Relay shield specific designed for directly interfacing with micro controller such as Raspberry Pi or Arduino, the board already consists of the protection diode and switching transistor and active/disabled LED indicator.
You can connect 5v, GND, and GPIO directly to VCC, GND, and IN at the shield.
I found OP's "GPIO pin Low not Low enough" question very interesting.
I agree with Jaromanda X in chat saying:
... relay triggers on LOW not on HIGH ... when ... LOW, it should be in it's triggered state ... so that may be why you think GPIO21 being LOW is not doing what you want!
In other words, OP seems to have asked the wrong question. ...
GPIO4 is the default for the 1-wire interface. If that is activated in /boot/config.txt it may interfere with other hardware connected to that pin.
Check config.txt for a line like this:
To deactivate it you can remove this line manually or use one of the Pi configuration utilities (raspi-config or the one in the GUI under 'Preferences')....
It sounds like you are wanting to wire the wall switch and the relay as a three-way switch arrangement. This way, if the relay has the light turned on, flipping the wall switch will turn it off, and vice versa. You could wire a single-pole double-throw relay to work as a 3-way switch, and install a 3-way switch in the wall (if it isn't already a 3-way switch)...
The Pi's gpios are 3V3 not 5V.
I couldn't make out from the datasheet the needed current to operate the relay contacts.
Assuming it is less than, say, 250 mA, I'd use a ULN2003A or similar to operate the contacts. It has in-built fly-back protection and can be operated by the Pi's gpios.
I'm a software type though. There are probably better sites for ...
You should never connect a electromechanical relay directly to a Pi gpio. You will kill the gpio and possibly the Pi. When the electromagnetic field collapses a large voltage will be fed into the gpios. A gpio probably wouldn't be able to supply the needed current for the relay contacts either.
The Pi needs protection, either use a relay board which ...
I had that same problem for my Pool Timer. Schematic, parts and reasoning are here: http://upon2020.com/blog/2012/12/my-raspberry-pi-pool-timer-electronic-assembly/
It's been working in production for almost 2 years now.
I actually have this same relay, but in single relay package, not the 4x package.
Anyway, from the testing I've done, I've concluded:
The relay's 5V source cannot be powered by the 3V3 rail (not surprising)
When powering the relay's source from the 5V rail:
Signal driven by 5V rail, current draw is 14.5 mA and relay operates
Signal driven by 3V3 rail, ...
OK, I found this diagram. It appears that the relay(s) can be opto-isolated or not (the default jumper position) so that the power for the coil can be separate (supply on pin JD-VCC) or use the same power as the input (reference?) voltage. I got the impression I should be using 3.3v on VCC since my signal is 3.3v gpio and 5v on JD-VCC.
Added diagram from ...
Use readymade remote-controlled sockets. They should be tested AND work already.
The hard part is replacing the supplied remote with a cheap 433 Mhz transciever , which might need reverse-engineering , see here :
The specifications for the board state a maximum of 250 volts AC, which covers your 220v reference. The current draw that the board can handle is rated for AC voltage at 15 amperes. Considering the poorer application of 110vac, a 500 watt lamp will draw slightly more than 5 amperes. With 220vac available, if the lamp will so operate, you'd be drawing less ...
I think you have misinterpreted a few things. First, keep in mind the product page notes "low level suction close, high level release". This means the contacts are closed and the circuit will be engaged when the input level is low.
Next, digital logic deals with thresholds. 0.05V is certainly low if the logic level is 3.3 or 5 volts. It does not need to ...