how do switch mode regulators save you power...? 2

[groundhog1]

If you use a linear regulator at say 10 volts input power and 5 volts out drawing 1 amp, you are burning 10 watts.

If you now insert a step down regulator and step the voltage down to 5 volts at your device, doesn't your device still draw 1 amp? And isn't that 1 amp still comming from the original 10 volt supply? Or does the current draw now become less at the 10 volt supply?

Other stupid questions:
What's the difference between a buck and a step down converter?

My task in a nutshell:
I have this pager sized gadget that will run on batteries. One part of the circuit needs 3.3 volts. I can only fit a 9v battery or maybe 1-2 AAA batteries in the enclosure. At first, I was thinking of stepping down the 9v to about 3.5 and then do a LDO (low dropout linear reg) from 3.5 to 3.3v. I can use the 3.5 for the digital circuits and the 3.3 for the RF circuits.

Then.....
I thought of just using one AAA cell and stepping it up! (BRILLIANT!! - funny if you've seen the TV commercial.) Is that a good idea? Which approach is better?

The 9v is rated at 400 mAh to .8v per cell or 4.8 volts. So this circuit would just keep going as the battery voltage drops.

The 1.5v AAA is rated at 540 mAh to .8v. So if you upconvert, the upconverter has to work harder (more gain?) as the battery discharges to .8v? .8v X 4.375 = 3.5v

Thanks in advance for any comments..
groundhog1

 

[felixc]

Power efficiency is described as the ratio of the power used by the target device, on the power entering the regulator.

In your case, a linear regulator will have an efficiency of 50%. Ten watts come in, and five watts are used. The rest is dissipated as heat; wasted power for a battery.

With a switcher, using the same example, using say a National LM1572. This part has an efficiency of 85%, thus 5.88 watts are entering the circuit to generate 5 watts of power. Less than one watt of wasted power, and less than 600 mA of current at the input. Quite different from a linear eh! This is the general idea, and it applies for both step up or step down switchers.

Felix

 

[groundhog1]

Thanks Felixc! I think i see. So it seems you do draw less current at the 10v supply by inserting the switcher. You mention less than 600mA, is that 10/5.88 = 588mA to be exact?

What do you think of me upconverting an AAA battery up to about 3.5v for LCD and digital, then linear Vreg down to 3.3v for the RF circuits? Do you see any land mines? I haven't done this before.

groundhog1

 

[mrkenneth]

groundhog1, how will you be upconverting the AAA battery? Using an IC or diodes? You might need a voltage tripler to get from 1.5V to 4.5V and still have some room left for the voltage from the battery to drop. I am not sure whether voltage triplers are availabe in an IC package.

The only problem I see is that the voltage may not be high enough when the battery voltage is low (e.g. when a rechargable NiCD or NiMH AAA is used). A LDO regulator would be a must. Or you can use a voltage quadrupler instead and have a higher input voltage to the voltage regulator.

 

[Mstrvb19]

Groundhog1-there is another side to this. How much power does your circuit consume? As you decide between the 9V battery at 400mAh and the AAA Battery at 540mAh, keep in mind that the 9V battery will be supplying more energy (significantly) than the 1.5V battery, even though the 1.5V has a higher mAh rating. When you are using power converters, it is more important to consider what kind of input of input power you will be needing. Additionally, you had asked what the difference is between a buck and a step down converter: nothing. Two names for the same device. Also, another advantage of a power converter versus a linear voltage dropping device (other than the power losses)is that the power converter can produce the same ouptut for several different input voltages, while the linear device always has an output proportional to the input. This is very important in electronics, particularly electronics that is battery supplied, becuase all batteries lose terminal voltage as they discharge. An electronic power conditioner can keep your electronics operating off of the same battery successfully at much lower battery voltages than a linear device (and thus longer). I would suggest using the 9V battery if you can get it to fit, as it will last much longer. Good luck

 

[groundhog1]

Well, that makes it much more clear. Thanks Mrkenneth and Mstrvb19.

So from your instruction, I am calculating that I would get...

For a 9v battery and linear Vreg down to 3.5 volts. I would get 26.7 hours continuous operation at 15mA draw.

For a 9v battery and stepdown Vreg to 3.5 volts. I would get 69 hours operation at the lower current draw of 5.8mA from the battery (the 15mA still being drawn at the circuit at 3.5 volts).

For a 1.5volt battery and step up to 4.5 volts. I would only get 12 hours due to the increased current draw at the battery of 45 mA (15ma still at the circuit).

This seems consistant with what you were saying.

Plus the difficulty of multiplying the 1.5 volts up so many times.

 

[ScottyUK]

Hi Groundhog,

A PP3 9V battery is one of the least cost-effective primary cells available. If you have space for such a battery, you could almost certainly accomodate a 1.5V 'C' cell which has a greater energy capacity than the AAA cell. You could then use a boost converter as described by some of the other useful posts above.

BTW, according to Duracell's datasheets, a 9V PP3 has an energy storage of approx. 5 watt-hours, and a 1.5V 'C' cell has an energy storage of approx. 10 watt-hours, i.e. roughly double that of the PP3. A 'C cell is also half the cost of a PP3, so the cost of operation from a 'C' cell is approximately one quarter of operation on a PP3.






----------------------------------

If we learn from our mistakes,
I'm getting a great education!

 

[logbook]

ScottyUK

I would be interested to see switching devices that work efficiently at 1.5V. The 9V solution is easy in terms of switching voltages, but 1.5V sounds hard to me. Although the energy storage in the battery is apparently higher, the efficiency of the boost converter will certainly be lower.

 

[IRstuff]

http://pdfserv.maxim-ic.com/en/ds/MAX1722-MAX1724.pdf

Would be 65% efficient with 1V input.

Many other choices:

http://para.maxim-ic.com/compare.asp?

Fam=DCDC_All&Tree=PowerSupplies&HP=PowerSupplies.cfm&ln=&SOR
D=612&FT_612=7102&ITEMLIST=115680,115681,115682,115683,11568
6,115687,115705,115706,115717,115718,115719,115721,115730,11
5731,115734,115735,115736,115737,115738,115739,115762,115765
,115766,115771,115773,115774,115775,115783,115784,115785,115
786,115809,115810,115811,115816,115817,115818,115826,115827,
115832,115848,115849,115850,115851,115865,115866,115870,1158
71,115874,115875,115883,115884,115885,115886,115912,115913,1
15918,115919,115920,115921,115927,115928,115942,115943,11594
8,115949,115950,115951,115952,115953,115954,115961,115962,11
5963,115964,115965,117304,119823,119825,119828,120357,120863
,120864,120865,128298,128299,133976,145533


TTFN

 

[felixc]

So you see that different solutions exist. Finding the right one depends on the project requirements. Your job is to achieve the best tradeoff between, the weight and size of your device, its cost (including or not the battery), the current draw (or targeted battery life), and the ease of design. Noise and emissions regulations may also be to consider.
If you can afford its size, weight, and capacity, the AA battery is a very cheap battery.
Using a step-up from a single cell, I would go directly to 3.3 volts and use filtering to isolate between the digital and rf sections.
Depending if your circuit can handle the voltage range, using two AAA batteries might mean no regulating device at all.

 

[Skogsgurra]

groundhog,

Your question: "What's the difference between a buck and a step down converter?" deserves an answer.

There are two main types: Buck and Boost converters. Then there are also combinations. No surprise that they are called buck/boost converters.

A buck converter always outputs a lower-than-input voltage and boost converter outputs a higher-than-input voltage.

Both can be isolated or not. The isolated version sometimes is referred to as a two-port converter and since the transformer can have any turns ratio they can easily be both up and down converters in the same beast. The terms buck and boost converter therefore mostly apply to non-isolated converters where the fly-back topology is common in boost converters while the forward topology is common in buck converters.

There is a lot more to it, but this is a rather complete first introduction. I think.

 

[ScottyUK]

Hi logbook,

It's (quite) a while since I've been in circuit design, and micropower stuff was never my field of expertise, but since the O.P. indicated that he was planning to use a 1.5 AAA cell it seemed reasonable to assume that he had a circuit. Certainly the efficiency will be lower with a low voltage input, but even at the 65% efficiency of the device kindly posted by IRStuff, the cost per watt from the battery or cell is cheaper with the C-cell than the PP3.

Something else I should have thought of before, but which the efficiency discussion has triggered, would be to use two AA cells in series giving 3V output. Physical size is slightly smaller than a PP3, converter efficiency is higher than for a single-cell design, weight is lower than for a C-cell, and cost per watt is lower than PP3.




----------------------------------

If we learn from our mistakes,
I'm getting a great education!

 

[IRstuff]

Apologies to all for mucking up the display. The new site still has some idiosyncrasies.

The commercial consensus, based on the stuff I own suggests that the commercial marketplace has gone the route of 2 AAA or 2 AA vs. 9V. The price difference is quite substantial and the 1.5V cells are much easier to find around the house in any case.


TTFN

 

[logbook]

Thanks to IRStuff for the data sheet links.

Those devices are remarkably efficient at low input voltages. All objections to the lower input voltage solution are removed