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I have a 12Volts, 2Amperes transformer and I need a lower voltage and lower current power source for my other components. I want to know how to make a step down transformer i want to lower it up to 5Volts and 750mA... Can I make it that low?

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  • \$\begingroup\$ AC or DC and if AC what frequency? \$\endgroup\$ – Andy aka Jun 13 '15 at 13:33
  • \$\begingroup\$ This is likely in connection with your previous question. I would say you're looking for an LM7805 and a small heatsink-kit if you want to start easy. \$\endgroup\$ – Asmyldof Jun 13 '15 at 13:40
  • \$\begingroup\$ Does 7805 work with 2Amperes? \$\endgroup\$ – Alyzha Orejudos Jun 13 '15 at 13:59
  • \$\begingroup\$ I'm kinda worried it wont lower the current... it might destroy my microcontrollers \$\endgroup\$ – Alyzha Orejudos Jun 13 '15 at 14:00
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    \$\begingroup\$ 2A is the maximum current the secondary winding can supply without overheating/damaging the transformer. If your load only draws 100mA at the specified voltage the transformer won't forcefully pump more amps into your load. Therefore @Asmyldof comment still stands. As long you don't draw more than ~1A from your 7805, the transformer will be OK with a large margin and your load won't risk anything, because the 7805 will keep the output voltage constant (and the load will draw whatever current it needs). \$\endgroup\$ – Lorenzo Donati Jun 13 '15 at 14:14
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The below assumes you are using the correct term 'transformer' for a heavy mostly metal device that outputs 12VAC from mains voltage. If you actually have a switching wall wart adapter that outputs 12VDC then you can omit the bridge rectifier and large filter capacitors shown below, but the solutions are otherwise the same.

Your transformer will give you 12VAC (RMS) - actually probably more like 14V with a light load. If you add a bridge rectifier and capacitor you'll get approaching 20VDC with no load and more like 14VDC with a heavier load. One approach is to simply use a linear regulator such as an LM7805 to reduce the voltage to 5V. The main issue with that is that it's going to waste 2/3 of the power and get very hot with a 750mA load (and thus will require a large heatsink to not destroy itself).

Another approach would be to buy a module that uses an LM2596 regulator (or build a circuit using that regulator). There are such modules available cheaply from China (eg. eBay)n, though I believe the 'LM2596' is typically a counterfeit.

The schematic would look something like this:

enter image description here

Though you'd probably want to increase C1 to 2200uF/25V or more.

If you buy a module the parts after the bridge rectifier and C1 are in the module. You can also buy kit power supplies that are missing all but the transformer.

The LM7805 circuit would be more like the below (again with C1 increased to more like 2200 or 3300uF/25V).

enter image description here

The LM7805 will need a heatsink maybe 100mm x 100mm 3mm thick aluminum plate or thereabouts.

As is well explained elsewhere on this site (Olin has written a canonical answer) the voltage rating must match your device, but the current rating needs to be the same or higher than your requirement. At 5V 10A power supply will not force 10A through your device if it is designed to draw 750mA at 5V. A 5V 1A supply will also work, but not a 5V 0.5A supply.

The super-easy solution (and this is not a design thing) is to just buy a switching wall plug adapter that outputs 0.75A or more at 5.0V. For example, an Ethernet router power supply that outputs 5V@2A would work just fine. They're cheap and good enough for many purposes (and generally safety-agency approved so you won't likely get a shock from the mains).

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  • \$\begingroup\$ +1 for LM2596, simple and excellent regulator. \$\endgroup\$ – Bence Kaulics Jun 13 '15 at 14:37
  • \$\begingroup\$ I'm planning to use LM7805 to get a 5V supply voltage for my arduino mega. I just want to clear something... a 2 ampere transformer won't destroy the microcontroller but a higher voltage than 5V can destroy it? \$\endgroup\$ – Alyzha Orejudos Jun 13 '15 at 15:06
  • \$\begingroup\$ Your uC is the load, let's model it with a resistor (very simple model). Now you have a fix voltage (V=5V) and a fix resistance (e.g.: R=100Ω), the current from Ohm's law will be 5/100=0.05A. The load won't change, but if the voltage increases the current will increase as well and to much current can make harm. \$\endgroup\$ – Bence Kaulics Jun 13 '15 at 15:20

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