how do switch mode regulators save you power...?
how do switch mode regulators save you power...?
(OP)
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
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





RE: how do switch mode regulators save you power...?
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
RE: how do switch mode regulators save you power...?
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
RE: how do switch mode regulators save you power...?
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.
RE: how do switch mode regulators save you power...?
RE: how do switch mode regulators save you power...?
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.
RE: how do switch mode regulators save you power...?
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!
RE: how do switch mode regulators save you power...?
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.
RE: how do switch mode regulators save you power...?
Would be 65% efficient with 1V input.
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TTFN
RE: how do switch mode regulators save you power...?
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.
RE: how do switch mode regulators save you power...?
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.
RE: how do switch mode regulators save you power...?
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!
RE: how do switch mode regulators save you power...?
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
RE: how do switch mode regulators save you power...?
Those devices are remarkably efficient at low input voltages. All objections to the lower input voltage solution are removed