May 2015
What would be the most efficient way to reduce the voltage from a nine volt battery to 5V? I could use a 7805 but it seems to bleed off a lot of power as heat. Is there a more efficient circuit or part?
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Take a look into DC-DC buck converters. Many of the IC manufactures offer low part count devices, and depending on your load requirements, may offer a completely integrated solution.
Things to consider when selecting your buck converter:
1. Internal switch/es to the IC — ease of implementation
2. Synchronous design — higher efficency
3. Load capabilities — This is one of the most important things to consider, especially with an integrated design. You want to select an IC that meets your load requirements, ie output current, but not something that blows them out of the water.
In other words, If your output requirements are 5V @ 100mA (0.5W), don’t choose a device capable of 10W, or 2A at 5V. The reason being is, these devices are more efficient when operated at the loads they were designed for.
One last thing, review the EVKIT the manufacturer has available for the device. These are really good starting points for layout and design with the device. I hope this helps.
A very economical solution are the LM25XX switching regulators on a board, which sell cheap from MPJA or the internet. I also use the MC34063 which is often found in the “12 volt to 5 volt” cell phone adapters at your local thrift store, but easier to work with purchased new.
For low current applications, my favorites are the encapsulated 1W or 3W switching DC-DC regulators from CUI, Murata, etc. via Digi-Key.
The UA78s40 is an old switching regulator that is obsolete, but parts are available on Ebay. The datasheet has a schematic for 26 volts to 10 volts but you only need to change two resistors for 5 volts output: R1 = 30K, R2 = 10K.
A more modern solution is the LM2576 buck regulator. It only uses two capacitors, a diode and an inductor for external components. The datasheet has a schematic that you can use directly.
The package is TO-220-5, Mouser part number is: 926-LM2576T-5.0/nopb, cost: $2.82. The IC is good for 3 amps but if your load is less than 20 mA, it would not be a good choice.
Yes, the 7800 and 7900 series seem to just dissipate heat to drop the voltage. Try the LM2940-5 - I found them at Jameco for $1.39 each......
You might try using an LDO regulator like the LM2596S chip. I bought a 1.5V-37V DC/DC buck convertor from MPJA for $1.95 to reduce 12V to 9V and it runs very cool.
