Powerblock also controlling power of LCD Screen
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@golaat What kind of switch do you have connected to the PowerBlock? If it is an SPST switch, replace it with a DPST type. The DPST acts as two switches in one. One side of the switch will be connected to the PowerBlock and the other for the LCD power. When the switch is turned on both PowerBlock and LCD will receive their respective power and when off, both turns off.
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Alternatively, you could use the 5V out pins to control a MOSFET or relay, which in turn feeds the LCD screen.
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@ortsac It is a SPST; I thought about replacing it, but now that I have mounted it, it would be a bit of work to get it out. I am interested in the relay approach that @petrockblog mentioned. I think it could be a learning experience too.
I have one of these sitting around:
https://www.sainsmart.com/sainsmart-2-channel-5v-relay-module-for-arduino-raspberry-pi.htmlI am assuming I would take the jumper off and then have one 5volt power feed the left jumper pad, the other 5 volt feed the right pad (from the PowerBlock?), of a separate 5V rail. What about the in1 and in2 pins?
Why are there three posts on the output terminals for each relay?
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@golaat This would work well since an opto-isolator is a low current device and would not take away much from the +5V of the Pi/PowerBlock. The board you are interested is two-channel relay port device. You would only need one of the circuit unless you want to use the other for something else.
I do not know which jumper you are referring to.
Using the schematic provided you can wire the circuit as follows:
VCC: connect to the +5V controlled by the PowerBlock
IN1 or IN2: connect to Pi/ControlBlock GND (ground)
Relay side connector: The relay has three post terminal for Normally Open (NO), Common (C) and Normally Close (NC). You would wire the Common (C) from the power source adapter and Normally Open (NO) to the LCD.This is what would happen:
When the switch on the PowerBlock is turned ON, the Pi will get its power and that will also trigger the opto-isolator on the relay board to close the relay and supply power to the LCD(assuming that the power adapter is already turn on).
When the switch on the PowerBlock is turned OFF, the Pi will shutdown and also cut the power on the relay board opto-isolator and thereby opening the relay and cut power from the LCD. -
@ortsac My LCD is powered by a separate power rail. I can't use the 5V coming off of the Pi because the draw is too much for the Pi to handle at its max of 2amps. When I power the LCD from the Pi's 5V, it turns off and on and the display also whines. The LCD draws about 900ma.
The jumper I was referring to is on the 3 pin connector on the relay. I don't believe the schematic details it. If you follow the download link from that page there is a word document that says "VCC for the system power supply, JD_VCC for the relay power supply. Plug in the jumper cap". By default, there is a jumper on the 3 pin connector between JD_VCC and VCC. I am assuming, that with this jumper installed, the relay is powered by the 5V coming in on the 4 pin pad, however, with the jumper removed , could it be possible to power the relay separately by the VCC on the 3 pin pad? Could I use the PowerBlock to power the relay mechanism, and then use the other 5V rail to feed the relay? This way, the LCD would get a dedicated 5V and full 2amps
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Just a side note from me: You should be able to power the LCD display with that relay and the PowerBlock. @ortsac pretty much points into the direction that I also would have given.
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@golaat Now I understand. Remove the jumper. Connect VCC to PowerBlock controlled 5V which will only trigger the opto-isolator and then connect JD_VCC to the LCD power rail. This way 5V power on the Pi is isolated from the LCD 5V power rail and the relay.
What is the voltage/current rating of the 5V power on the Pi side? On my set up, I added a barrel connector on the PowerBlock and use a 5V, 4A power supply. This provides ample power to the Pi and whatever USB device I connect to the Pi that might require power. I don't experience the power shortage problem.
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@ortsac I have a 5V, 2.1A power adapter on the Pi side. Once I have my keyboard controller and momentary touch sensor plugged into the Pi, there is not enough headroom for it to power the LCD, thus why I have a second 2.1A power rail. It is my understanding that the Pi can't handle a draw faster than 2A.
Can the PowerBlock be used to power the Pi if you give it 5V via the barrel connector?
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@golaat My setup is exactly that. I installed (soldered) a barrel connector on the PowerBlock and use a barrel-type power supply (5V, 4A) which provides power to the Pi via the GPIO pins. I do not have to use the micro-USB connector to power the Pi. The Pi will draw whatever it needs from the power supply and whatever USB device that is connected to it that requires power. I do not experience any power shortage at all. The Pi does not overheat either and I do not have to use a fan, but if it does I have enough power for the fan as well.
If you use the same type of setup and power supply unit as mine, you will not need the extra power supply and relay board to power the LCD. The PowerBlock has solder holes for 5V_OUT to connect to external devices such as in your case a 5V LCD. The 5V_OUT is also controlled by the switch on the PowerBlock.
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@ortsac Thanks so much for this information. I believe this is the way I will also go. I never realized that the Pi can be powered via the GPIO pin connection from the power block!
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@golaat Glad to be of help. Let us know how your project is going.
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