A Look at Switching DC-DC Converters

switcher module - smallTo use my USB charger boards in the car, I need to convert the nominal 12 volt supply to 5 volts at up to 2 amps (a 10 watt output).  The first choice in voltage regulation is often the ubiquitous 78xx (where xx is the desired output voltage).  For low power loads, this is an effective, inexpensive choice, but for higher power applications, these regulators don't work so well.

The input current to a linear regulator will be equal to the output current at the lower voltage plus a negiliable amount to operate the regulator itself.  In this case, the output is 2 amps at 5 volts, equaling a 10 watt output.  The input is therefore 2 amps at 14 volts (when the car is running), which equals 28 watts of input power.  28 watts input for 10 watts output???  Where does the extra 18 watts go?  HEAT.  A nightlight bulb is 4 watts for comparison and that's more than enough to burn your fingers, so 18 watts is a huge waste!  It would take a large heat sink to dissipate this much heat to keep a linear regulator at a safe temperature.

There are two important equations to keep in mind when using a linear voltage regulator.  The first consideration is the power dissipated by the regulator in the form of heat.

eqn 1

Two factors determine the power dissipation: the output current and the voltage drop across the regulator.  Consider a 5 volt regulator supplying 500 mA from a 12 volt source.

eqn 2

Thinking of that 4 watt nightlight bulb, that's a lot of heat to dissipate!  If we could reduce the supply voltage, say to 7 volts, what would the result be?

eqn 3

That's a nice reduction in wasted power.  Many dev boards use a 5 volt regulator with a higher voltage wall wart.  Even though a 7805 regulator can supply 1.5 amps, it's still limited by the power dissipation.  Using a supply closer to the desired voltage can prevent overheating problems.

 The second equation to note is the efficiency of a linear regulator.  It is almost entirely controlled by the voltage drop across the regulator; the quiesent current of the regulator reduces this slightly but it's largely insignificent with any large load.

 eqn 4

So for the first example,

eqn 5

Only 42% of the power is going to our circuit with 58% going to heat!  Reducing the input voltage from 12 volts to 7 really improves the situation.

eqn 6

Getting back to the iPad/iPhone car charger, the supply voltage will range from 12 - 14 volts or more and I need to draw 2 amps of current.  Clearly, a linear regulator is not a good choice for this application.

Switch Mode Power Supplies

Switch mode power supplies are fundamentally different than linear regulators.  Instead of dissipating energy as heat to control voltage, a switch mode power supply converts the low voltage DC into a high frequency AC voltage and converts it back to DC at the desired voltage.  There are many different topologies for switch mode power supplies and a detailed discussion is beyond the scope of this article.

For the car charger, I'm using a switch mode module that accepts 12 - 24 volts input and outputs 5 volts at up to 3 amps.  There are a variety of these available from eBay for under $10.


So how does this $10 module compare to 7805 linear regulator?  This module claims to take up to 24 volts, and provide a 3 amp output at 5 volts.  From the above equations, we can calculate that a linear regultor would be dissipating 57 watts!  Since there's no massive heatsink, we can guess this module is better than a linear regulator.

To evaluate this module's performance, I connected a 12 volt supply andmy variable load through my power monitor and measured input voltage and current and output voltage and power.  From these measurents, I calculated input and output power and efficiency.

eqn 7

The efficiency ε is calculated by

eqn 8

The results are shown in this graph.  The efficiency is greater than 80% over most of the operating range.  Very little power is wasted as heat.  With no load connected, the module draws about 10 mA, an important parameter for operating in the car.

 switcher performance

I elected to use pre-built modules in this case but the design and layout of a switching power supply is simple with modern components.  Many vendors offer on-line calculators to aid in circuit layout and component selection.  Depending on output parameters, modern switching power supplies are nearly as simple as inefficient linear regulators.

One additional difference between linear and switching circuits is worth noting.  The input power to a linear regulator increases with supply voltage, resulting in increased power dissipation.  With a switching circuit, the input power remains essentially constant with supply voltage.  This is generally a good thing but bear in mind, as the input voltage is decreased, the current must increase to maintain the power.  If the voltage is too low, the required current can overload the components of the regulator circuit.  I managed to burn up an switching LED driver module by starting at a low supply voltage (outside the normal operating range).  Because I was being cautionous and starting slowly, the excessive current draw blew up the module!

Posted: 4 years 3 months ago by Jon Chandler #13376
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I hope you find this comparison interesting.
Posted: 4 years 3 months ago by Baldor #13374
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Thanks for the cristal clear explanation.

I'm planing to use one of these for an small project: http://www.ebay.com/itm/400358408922

9V(batery) to 5V, less than 300mA needed, but is the cheapest one I found in ebay.

I think this regulator also fits your needs.

(I was planing to build it myself in the circuit board, but is more expensive, and also the complete circuit exceeds the 50x50mm size for the cheapest ones)
Posted: 4 years 3 months ago by Jon Chandler #13375
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Those look like great modules. I like the adjustable output voltage.

It's amazing what ou can find on eBay sometimes. To buy the parts would cost more than these modules.
Posted: 4 years 3 months ago by Baldor #13384
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One question: For batery operated circuits, what wold be beter, from an eficiency point of view? Buck, or boost? For example, 2AA bateries (3V alcalines, 2.4V NiMH) and boost to 5V, or a 9V batery and buck to 5V. ¿How I will squeeze more energy, asuming the same energy stored in the bateries?
Posted: 4 years 3 months ago by Jon Chandler #13385
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I'm probably not the right guy to ask, but I'll take a stab at this. Look at the efficiency of various designs. I think they're pretty close between step-up and step-down.

So the next consideration is the energy available in the batteries available. An AA alkaline battery capacity is shown below.

At a current drain of 100 mA, a single AA cell will supply about 2500 mA-Hr. Two cells, with about the same volume as a 9-volt battery, will supply 5000 mA-Hr. Remember, with a switching supply, the power remains about the same over the working range, so supplying twice the voltage cuts the current in half.

Compare this to an alkaline 9-volt battery.

At the same draw of 100 mA, this battery only provides about 500 mA-Hr, about 10x less than a pair of AA batteries. Another consideration is the cost of a 9-volt battery – usually a lot more than AA batteries.

Looks like AA batteries with a step-up would be the best option.
Posted: 4 years 3 months ago by Baldor #13386
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Seems like what i was thinkiing. Now, the final decision is only constrained by the space available. A 2AA batery holder is biger than a 9V batery with the conector (The 9V is about 50mm long with the conector, ideal for a case apropiate for a 50x50mm board). If the aplication is not space constrained, go with the 2AA and boost.

Also, the 9V batery will not fully discharge to the 4.8V shown in the table with a buck converter, it will stop working somwere a litle over 5V, while the AAs with a boost will fully discharge to 0.8V.

So... I need to look furter for a case suitable for my next project. (IR barrier camera trigger, PIC12F508 for the receiver and a PIC10 for the transmiter, batery operated, and long batery life needed)
Posted: 4 years 3 months ago by Jon Chandler #13387
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True, AA batteries are slighly longer Han 9-volt batteries but if you add a snap connector they are pretty close to the same length. Many of the enclosures I've seen with battery compartments say they fit either.

A AAA cell is smaller than a 9-volt and a pair would provide about 4x the capacity of a 9-volt.
Posted: 4 years 3 months ago by Baldor #13388
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The problem is not the size of the batery itself, but the batery holder, and AA batery holders are longer than 9V with the conector.

I've been brosing enclosures with batery holder at farnell, and nothing fits my needs, (Usually, too expensive) so I decided for a simple box that could fit the 50x50mm board, the dc-dc converter, and the bateries. Time to recheck to make room for a 2AA batery holder.
Posted: 4 years 3 months ago by Baldor #13389
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What about this: http://www.ebay.es/itm/DC-DC-Boost-Conv ... eff&_uhb=1 ?

Seems idealy suited for operating with 2AA bateries. Other boost converters I have seen in ebay operates from 3V. This, operates from 1 to 5V, and the output is fixed at 5V. (500mA max, enough for my aplication)
Posted: 4 years 3 months ago by Jon Chandler #36
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Looks like that would do the job and be able to suck everything out of two or three AA batteries. You sure couldn't build it yourself for that price!

There's not much information in the eBay ad to guess what the efficiency might be but at the price, looks like it's worth trying it out.

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