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I want to buy an external power supply for my breadboard (this one). It says that the input should be between 6.5V and 9V, which is easy enough to understand. To power the supply, I planed on buying a wall plug with a barrel jack as an output. There are plenty of wall plugs which supply 9V. However, I found that they differ in the amount of current they output (i.e. ranging from 1A to 2A while both supplying 9V). My question now is, which one is suitable to power the breadboard, and what the benefit of one other the other is.

Cheers!

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  • \$\begingroup\$ It depends on what you want to build on the bread board. The board itself has no need for high current. The MB102 PS can only handle 700mA of output. So a 1A power supply would be sufficient by far. But, the circuit you build on the breadboard must not use more than 700 mA = 0.7 A. \$\endgroup\$ Feb 14 '21 at 11:51
  • \$\begingroup\$ And the PS must also handle the heat that comes up with lowering the voltage. The module uses two AM1117 (5V and 3.3V) to regulate the voltage to the respective value. If you power it with 9V and you only need 3.3V on the bread board with let's say 0.5 A of current. The PS wastes a lot of energy that is converted to heat. (9 - 3.3)*0.5=3.35 watts. That a lot for such small parts like the AM1117. Just to have mentioned it. ;-) \$\endgroup\$ Feb 14 '21 at 11:57
  • \$\begingroup\$ Thank you very much for your answer! I planned to use it to power a set of relays (this one) and a small water pump which also uses 5V. So I think going with the 1A supply would be better in my case if I am not mistaken. \$\endgroup\$
    – nickhir
    Feb 14 '21 at 14:27
  • \$\begingroup\$ You should read your product specifications more carefully. Eingangsspannung: 6.5 - 12V DC. \$\endgroup\$
    – Seamus
    Feb 15 '21 at 17:54
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I suggest you buy a 5V regulated adapter for your ~1A requirements. A standard adapter (or power bank if you want battery power) and use the USB socket rather than the barrel jack.

Similar (but not interchangeable) boards with the same part number are available with a mini or micro USB socket to match standard power supplies as used with Raspberry Pi, cell phones etc. Make sure the two are compatible, there are a number of combinations. The USB power connections are connected directly to the +5/0V and nothing else.

This will avoid excessive dissipation on the board, and help prevent oopsies that can destroy your parts.

You can use the power supply module you linked to create a 3.3V bus if you need it. The 5V from the USB socket is passed along directly when switched on. See my answer here for the schematic of a similar product sold with the same part number.

Note that the one you've linked does not include a resettable fuse, which I would consider at least an advantage if not a 'must'.

As another detail, it would be nice if the polarity of output matched the color of markings on your breadboard (red = +, blue = GND), but most seem to be reversed.

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  • \$\begingroup\$ Thanks for your answer! Can you maybe elaborate why you recommend using the USB socket instead of the barrel jack? I thought that it doesn't really matter if I use the USB or the barrel jack. \$\endgroup\$
    – nickhir
    Feb 14 '21 at 16:42
  • \$\begingroup\$ The USB socket is connected DIRECTLY to the +5V output pin. So no regulator is in the middle. For 5V it only provides the connection between USB socket and breadboard (maybe with a switch or fuse). \$\endgroup\$ Feb 14 '21 at 16:44
  • \$\begingroup\$ Does this also mean that I can only use the 5V output pin on the MB102 (the one which is kinda in the middle of the board next to the 4 GND)? Or can I still use all output pins and determine the output voltage with the jumper (e.g. 5V or 3.3V) \$\endgroup\$
    – nickhir
    Feb 14 '21 at 17:05
  • \$\begingroup\$ If it matches the schematic I linked to (modulo the switch connections as someone commented aboot), you have the choice between 3.3V from the regulator and +5V from the USB connector (via jumpers). The 5V AMS1117 does nothing. \$\endgroup\$ Feb 14 '21 at 17:06
  • \$\begingroup\$ Thank you very much for all these insights. I think I have one last question. You said that I should use a power adapter which provides 5V (and ~1A), which makes a lot of sense considering the "heat problem". However, the manufacturer specifically requests an input current of 6.5 - 12V DC. Do you think using the 5V input will be an issue then? Furthermore, why do they even recommand something like a 12V DC input to power the power supply? \$\endgroup\$
    – nickhir
    Feb 14 '21 at 18:23
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The current capability of your input power supply doesn't matter as long as it's enough to run the circuit. This PSU that you link to is a linear power supply, which means that its input current will be equal to its output current (plus a negligible amount for powering the power supply itself). That said, with a 9 V input, a linear power supply outputting 5 volts will dissipate \$4\ \mathrm V · I_{out}\$ in heat, which can be quite a lot, so you do need to consider this. And 3.3 V output will be even worse, at \$5.7\ \mathrm V · I_{out}\$.

This type of power supply is not good for providing high current, and generally shouldn't be used for anything involving motors or solenoids--simply put, don't use this PSU for your relays or pump. It's fine to power electronics, but the high current draw of a pump or several low-coil-voltage relays will be a problem for it. It will be less of a problem if you power it with a lower-voltage input, since the power dissipation (the limiting factor) is proportional to the difference between input and output, but I would still recommend looking for a switching power supply module, such as this one or this one.

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