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I would like to get some input for my very first electronics project.

I would like to split the 5V power Voltage to multiple outputs. Furthermore the individual power lines are supposed to be switched by a microcontroller.

schematic

simulate this circuit – Schematic created using CircuitLab

I know that the MC cont controll the "push botton switch" but that is supposed to be representive of a transistor or such.

the 3 output voltages power small loads ( recharge them)

any ideas to bring this idea to life?

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  • \$\begingroup\$ Does your microcontroller have GPIOs that output 5V? \$\endgroup\$ Sep 9, 2017 at 13:44
  • \$\begingroup\$ yes it has 5v GPIO pins \$\endgroup\$
    – beacon1993
    Sep 9, 2017 at 14:07
  • \$\begingroup\$ Depending on the load you are driving (how many amperes your load will take), you can drive directly from the GPIO pins. Generally, GPIOs can drive loads up to 10 ~ 20 mA, but make sure to check the datasheet for you MCU. If you need to drive more current to the load, the simplest options are to use either a MOSFET or a BJT transistor. The other you connect them will matter for the type of transistor you choose. For the configuration you drew, you may be looking for a PMOS (P-type MOSFET). \$\endgroup\$
    – Lucas
    Sep 9, 2017 at 15:45
  • \$\begingroup\$ specification changed ... the MCU now has 3,3 Voltage GPIO \$\endgroup\$
    – beacon1993
    Sep 14, 2017 at 14:52

3 Answers 3

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Do you need to switch the positive voltage (high-side switch, more complicated) or can you switch the ground side of the load low-side switch, simpler).

If your load is, say a LED then low-side will be simpler (you only need a BJT or MOSFET).

If your load requires a constant connection to ground, then you need to switch high-side.

Aside from next-hack's answer, you can also consider integrated load switches. These will usually offer interesting features like current limit or short circuit detection, also it's one component instead of several, but they're specialized parts so less easy to get unless you already have enough stuff in your mouser/digikey basket...

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  • \$\begingroup\$ This sounds like a perfect IC. So my load is basicly a beacon that gets charged. the charger is supposed to be switched by the MC. So I want to have a steady 5V DV going into the circut ( because the MC will be powered with that as well) and charge the beacon with 5v. so I would get 5 IC like you mentioned and get them contolled by the MC right? \$\endgroup\$
    – beacon1993
    Sep 15, 2017 at 8:18
  • \$\begingroup\$ Yeah. If there are connectors and cables, the short-circuit protection feature of these load switches is good to have, if someone plugs in a damaged cable which is shorted, for example. They usually have an ERROR pin which will tell your micro that a short has occured. Make sure you pick one which is rated for the current, and pay attention to its internal RdsON and the associated voltage drop. Overdesigning the part won't hurt... \$\endgroup\$
    – bobflux
    Sep 15, 2017 at 14:09
  • \$\begingroup\$ Thank you so much for the input. It sounds like a good idea. Do you have a suggestion for a IC? \$\endgroup\$
    – beacon1993
    Sep 15, 2017 at 14:15
  • \$\begingroup\$ Not really, there are so many, pick your favorite shop, filter and order by price as usual ;) also decide if you want an ERROR output or not, etc. \$\endgroup\$
    – bobflux
    Sep 15, 2017 at 14:17
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You might consider this circuit. I draw this only for one channel, but of course you can replicate for as many channels as you want.

A little explanation: V1 is your 5-V power supply. V2 is your GPIO of your microcontroller. R1 keeps the pMOSFET in the OFF state. This is very important, as many microcontrollers have floating GPIOs during reset, and you do not want the gate of that MOSFET floating around, picking all the noise. "LOAD_1" is your load. I just put 10 Ohm for the simulation. Of course you must put your load instead of that 10-Ohm resistor :).

M1 MUST BE A 5V LOGIC LEVEL MOSFET!

R2 - C1 ensure a soft turn on, limiting inrush currents. This is mandatory if your are switching capacitive loads, otherwise you'll have a huge drop on your 5V line, possibly causing the RESET of your microcontroller. Adjust the time constant as per your requirements.

EDITS:

I have added also a possible suggestion if you have 3.3V GPIOs. Note that in this case the function is inverted, because of the presence of the BJT Q1, which is put as a level shifter.

The function of R5 is to make sure that the BJT is OFF when the GPIO is not initialized (e.g. during reset). In other words, it has the same function of R1, i.e. to prevent that during reset the OUTPUT1 powered.

D1 can be omitted. I just placed it, because in this way, the turn OFF time is not determined by R1+R2, but just by R1 (this is useful if you want a faster turn OFF than a turn-on).

The resistor values are randomly placed. You should adjust them as per your requirements! (Still, with these values, it will work).

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Hey Thank you so much for the answer, what if I only have 3,3 V GPIO Pins \$\endgroup\$
    – beacon1993
    Sep 14, 2017 at 14:51
  • \$\begingroup\$ Do you still want to switch 5V loads, but using a 3.3V gpio ? \$\endgroup\$
    – next-hack
    Sep 14, 2017 at 15:52
  • \$\begingroup\$ @beacon1993 I added the schematics for 3.3V gpio switching 5V loads. \$\endgroup\$
    – next-hack
    Sep 14, 2017 at 16:08
  • \$\begingroup\$ R2 and C1 in first schematic and D1, R4, C2 in second schematic overcomplicates things. You generally want fast switching to keep power dissipation low on the MOSFET. Also, if LOAD is very heavy (e.g. tons of decoupling caps), then the MOSFET will hit saturation and limit the current to the load anyway. (Better at controlling inrush than a mechanical switch that way.) \$\endgroup\$ Sep 14, 2017 at 17:46
  • \$\begingroup\$ @VincePatron I do not agree. 1) The OP asked for a load switch. He didnt' specify he need to PWM it. Therefore the condition "high I&V" is "rarely" reached (and for few ms). 2) The Rds(on) of power MOSFET can be so low (even on SOT23), that a capacitive load will act like a short. To overcome this, you could place a large decoupling capacitance, to provide the charge for the capacitive load. But at this point it's worth adding some 0603 components such R2-C1 instead. I wrote that D1 is optional and the values of other components can be adjusted according to requirements. 0 is a value. \$\endgroup\$
    – next-hack
    Sep 14, 2017 at 18:07
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You might want to look into load switches. TI makes a variety of them and without knowing more about your specific application it's hard to identify a specific component. Here's a place to get started:

http://www.ti.com/power-management/integrated-load-switches/overview.html

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