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?
Step down transformer
transformer
Related Solutions
You are correct. The transformer will only reduce the voltage (and increase the available current), so you need to add additional circuitry to rectify, smooth and regulate your 12\$V_{AC}\$ transformer output to 5\$V_{DC}\$.
This is the type of circuit you should be looking to build:
- The transformer reduces the voltage from mains to 12\$V_{AC}\$ (RMS).
- The Diode Bridge (known as a bridge rectifier) will convert 12\$V_{AC}\$ to 15\$V_{DC}\$. The voltage is \$\sqrt{2}\$ times minus two diode voltage drops higher than the input voltage because the rectifier output is the peak AC voltage, not the RMS AC voltage.
- The first capacitor will smooth out the ripples that come from the output of the AC to DC bridge rectifier.
- The LM7805 regulator will maintain a constant voltage as the load varies. For example if you are switching a light bulb on and off, the current will go up and down, and if you didn't have a regulator then the voltage would drop as the bulb is switched on. The regulator keeps it at the 5\$V_{DC}\$ your microcontroller needs.
- The final small capacitor filters out any noise or interference on the regulated side of the circuit.
1. Second question first:
If power does not change, then how can an 11W LED light put out 50w of power based on power equation and transformers performance?
It is often claimed that an eg LED light will out as much light as much light as a 50W incandescent lightbulb. This relates to the electrical to light power (lumens) conversion efficiency - related mainly but not solely to the absolute power levels.
LED lights are more efficient at producing visible light than incandescent bulbs - they produce more visible light energy out for a given amount of electrical input power. The key word here is visible. An incandescent bulb converts almost 100% of its input energy to electromagnetic energy BUT most of this is in the infrared spectrum where the human eye sees little or none of it. This is the main factor in the difference in effective light outputs.
Perceived brightness depends on the human eye's sensitivity and many LED lights make light which is closer to the most sensitive wavelength of the human eye than incandescent lights do. This is a smaller but significant part of the difference.
LED lights can have their beam shape more closely tailored to their application so may be able to place more of their light on the target surface,Incandescent bulbs radiate over a wide part of a 360 degree cross section and reflectors are often used to attempt to utilise the otherwise wasted light. Much of the light is lost. As LEd lights can have a larger % of output usefully used this can also be used in statements about "effective output power".
Perceived light power by a user is expressed in lumens and is a measure both of the actual optical power (Watts) and the reponse of a typical human eye. The human eye is most sensitive to a yi=ucky green yellow colour - which is why SOME fire engines are pained this colour like many of these ones or this:
Eye response curves:
Dotted line is "low light" sensitivity.
Black bordered solid colour area is eye response as per 1931 standard.
In 1978 the red bordered area at left was added to the curve. NOTE that the left hand "reddish" coloured area is in fact blue-violet!
Power In / Power Out
- In our universe, as far as is known by main stream Science, so far, energy is "conserved". ie energy can neither be created or destroyed in a closed system.
Power is the rate at which energy is provided.
So power into a system at a given moment must equal power out.
The form the power out takes can vary.
In a transformer the main power outputs are the desired form (electrical energy), and the incidental byproducts of the process. Mainly:
Heat - usually 1% to 10% range. (Mostly winding "I^2R" losses, core losses)
Sound - Only a small amount. Lamination "buzzing".
Light - very little indeed except when things go very wrong.
Electromagnetic radiation - minimal and minimised by design.
Electrical power out and heat are the usual major energy outputs.
Related Topic
- Using a 5KVA step-down toroid transformer as a step-up transformer
- How to identify a step up or step down transformer
- Electrical – convert step down to step up transformer
- Electronic – How does the secondary coil of a transformer affect the primary coil? How is it in case of Step-up and step-down transformer? Does it obey Lenz’s law
- Electrical – what determines the lower limit on the number of turns in a secondary coil in a step down transformer
Best Answer
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:
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).
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).