# designing a switch...

#### sdh314

Joined Jun 2, 2011
18
Hi all,

In a project I'm working on I've a couple of different circuits that run off four different voltages (+9V, -9V, 5V and 3.3V). The device is portable and runs off a bank of 9V batteries, and these feed four regulators to provide the voltages. During operation, I need to kill the power to some circuits and not others, which I plan to do via digital pins from a uprocessor.

My question is, what is the best circuit to look into to achieve this? I need some with very low quiescent currents because power consumption is the biggest constraint. I've noticed that some regulators have enable pins, but the ones I plan to use dont so I need to design up a switch instead.

Any suggestions much appreciated!

#### SgtWookie

Joined Jul 17, 2007
22,221
Logic level MOSFETs. They don't require any power to maintain a state, ON or OFF.

You might want to run the whole circuit by the forum. Sounds like you are using expensive 9v batteries that don't have much power, with regulators. Many older linear regulators consume a good bit of power even when idle, and have a considerable dropout voltage that wastes even more power.

#### thatoneguy

Joined Feb 19, 2009
6,359
#1: Do not design circuits for 9V batteries. Only test ideas with them, or try to use 6V instead (4AAs).

#2: ±9V can be gotten with 12AA batteries, 6 on the positive side in series, 6 on the negative side, which is a lot of batteries. This is very bulky.

Before going any further, there are several ways what you want done can be accomplished, but some more information is needed:

1) how much current will be drawn at each voltage?
2) Does any component use both the + and - rails at the same time, such as an op-amp (they'd need to be switched at the same time)
3) How long do you need the power supply (batteries) to last?

A 9V battery has only 200-400mAH capacity @ 100mA draw 1 AA Alkaline battery has between 2,000mAH and 3,000mAH@100mA Draw capacity.

If your project draws very little current (< 50mA) it could be run on a 9V battery, if the project runs on < 10mA, it can be run on 3 CR2032 coin cells stacked together for a 240mAH capacity and very small size.

That being said...

If your circuit really does require all those different voltages, I'd suggest a 4AA battery pack (6V), and a few DC-DC Boost/Buck Converters to get the other voltages you require.

#### thatoneguy

Joined Feb 19, 2009
6,359
With a 6V pack, you can get:
3.3V, 5V, and 9V with 2A total capacity

Some external stuff is required, but it'd work.

Now it is a question of how badly you need the ±9V. Will ±12V Work? If the negative is powering an op-amp, can a single rail to rail positive voltage op amp take it's place?

Will give you ±12V @ 41mA

There are undoubtedly many other DC-DC Switching Supplies, those just came up on the list quick in a search. Linear supplies that burn the excess power off as heat are extremely bad for efficiency, usually 50% to 60%. The Switching Regulators above are 85+ percent, with 90+ percent in some cases like full battery and medium current draw.

With those 2 ICs, and 4 AA batteries in a holder, and some external components (resistors, capacitors, inductors), ,but not too many, you will have a +6V source from the AA Batteries good for 1A draw. 5.3V (Logic Power) at 1A from the 6V batteries going through a diode drop. Nearly all logic will run at 5.3V. In addition you will have 3.3V, 5V, and 9V with a total capacity of 2A split between outputs. Finally, you'll have a ±12V supply rated for 41mA, which is enough for about 1 or two op amps.

#### sdh314

Joined Jun 2, 2011
18
Thanks for the awesome replies. I'm learning heaps. Didn't realise how rubbish the 9V batteries are.

This is a portable device which takes a quick measurement a few times an hour, and for the rest of the time, everything is powered down and the uprocessor goes into a sleep mode. So the whole thing is only 'on' for perhaps a minute of every hour, so the the quiescent draw of the uprocessor while in sleep mode (10s of uAs) and the draw of the 5V regulator which powers it are the real bottlenecks of the system. Ideally want the thing to run for a few months without having to change/recharge the batteries. There is also a RTC clock which needs to be powered at 3.3V and stay on the whole time (which draws 400uA).

When the device is in full operation, it needs to drive a LED (~40mA), a stepper motor (300mA), run a bunch of op-amps, most of which need dual supplies (can take +/- 9V, but would run better at +/-12V), and a couple of other devices at are only expected to draw a few mA.

Once the measurement is complete, I was planning to use the uprocessor to shut down everything and then go into sleep mode. Ideally it would be good to find a regulator with an enable/disable pin such that the regulator switches off completely. Or - design a mosfet switch which would open circuit the connection between the battery and the input to the regulator. Would this be the best method?

It seems as a general rule the quiescent current is proportional to the difference between the supply voltage and the regulated voltage. In which case, would it be best to use a 6V supply to feed the 5V and 3.3V regs, while using a boost reg to get the dual +/- 12V (or +/- 9V). I'm worried that over time if the voltage drops below the minimum necessary to feed the 5V regulator and the thing stops working. So - do I go with a larger voltage battery, (ie load up a bunch of AAs in series), or just go with buck regs?

Thanks again for all your help. You're awesome!

#### thatoneguy

Joined Feb 19, 2009
6,359
Both supplies that I posted have enable pins.

For sleep mode, and basically everything else, just power the PIC from the AA battery pack with 1 or 2 silicon diodes in series with the 6V to drop down the voltage a bit.

The PIC can then enable the other power supplies as needed, and test for power good on the + voltages (using a divider for the +12V), this puts the quiescent current at nano-Amps, roughly the same as the self discharge rate of batteries. PICs are extremely good at power saving, even the 5V versions. Voltage regulators are where the waste occurs, especially linear in standby.

You mentioned how much current each device took, but not what voltage it ran at.

What I'm getting at, is that there may be an easier solution to your problem, so if we know the overall problem or task, we can help with a cleaner solution, rather than trying to come up with a way to make your solution work.

Usually, a project shouldn't require more than 2 power supplies.

The op amps, for example, could those be Rail to Rail single supply instead? There are quite a few new op amps with parameters that would make something like an LM741 look as old as it is (40 years).

#### sdh314

Joined Jun 2, 2011
18
I noticed that TI -/+ 12V component is classed as a _unregulated_ supply. I'm more after a regulated supply because I'll be switching LEDs on and off at a reasonable rate (kHz) and I need a stable supply for the op-amps and other circuitry running off +/- 12V. Do you know of any other options that provide +/- 12V off a 12V battery pack, that include an enable/shutdown pin that would cut down the current to a few uA or less?

I'm using Arduino, not PIC, but it runs off 5V. The RTC runs of 3.3V
The motor and a few other sensors run off 5V. And the op-amps and some other ICs (AD630) need dual supplies to run, and I'll run them off either +/- 9 or 12V.

#### thatoneguy

Joined Feb 19, 2009
6,359
I'd suggest using Single Rail op amps to eliminate the negative rail requirement.

A 12VLiPo battery with purpose built LiPo charger is all you'd need for a source, unless the motors are a high current draw, then use a separate supply for motors/H-Bridge than you use for logic.

Then use switching supplies, such as Simple Switchers or similar, there are thousands out there, to get the other voltages you need. For the current drawn by a RTC, you can make an "LED Regulator" to power it when on, and from a coin cell when off. LED regulator is just a 3V Blue LED with a current limiting resistor to indicate power, then take the 3V from the LED anode and power the RTC with that, since it is a very low current draw.

Search around for "ready to go" switching regulators, most have an enable line as well, so one switch controls them all.

#### SgtWookie

Joined Jul 17, 2007
22,221
That RTC seems like a power hog. 400uA = 0.4mA. Maxim has low-power RTC's with SPI interfaces that draw only 250nA, or 0.00025mA.

When you're on batteries, anything you do to reduce current use is a big plus.

You mentioned:
drive an LED (~40mA),
Do you really need to use that much current to drive an LED? Unless you need to use it for lighting, or as in an optoisolator, that might be reduced or eliminated.

a stepper motor (300mA)
You might find yourself ahead if you used a lower power motor that was geared for greater mechanical advantage. Heavy loads cause batteries to dissipate power internally, and the problem gets much worse as the batteries become exhausted.
Capacitors across the batteries can help reduce power losses when a load is being driven has a rapidly changing current requirement, such as a stepper motor. At the same time, if the capacitors have a relatively high leakage rate (as in aluminum electrolytic capacitors), they can cause more drain over time than without.

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