Power consumption in voltage regulator

Dear community,

I want to use a voltage regulator(to drop voltage from 9V to 3.3) . However, the one I am using (LDV33 from ST) seems to be consuming a lot of current, about 6mA. I want to power my xbee and my arduino with battery and I want to save as much current as I can. What do you recommend.

thank you for your information,

Luis

The regulator can't "consume" any more current than your circuit does. The nature of dropping 9V to 3.3V means you are wasting most of your battery energy in the regulator, no matter how much current your circuit draws.

I can't seem to find a datasheet for that part number, but if it's a LDO regulator then you could possibly switch to something like a three AA cell holder instead.

EDIT: As others pointed out, there is the quiescent current and it is the current that flows when no load is present. I assume that's the 6mA you were referring to. You really can't do much about that, it's just part of the overhead of using the regulator. Your biggest energy waste is due to the excessive voltage being supplied to the linear regulator. There are modern regulators designed to flow most of the quiescent current thru the actual load (assuming the load is large enough I suppose). This reduces that inefficiency, but again, your biggest waste is the excessive voltage.

How are you measuring the loss? How do you know the regulator is using 6mA?

Regulators have a "quiescent current" which is what the regulator consumes all by itself. Some use more than others. Check your datasheet.

There are two values you should be aware of here.

  1. Quiescent current.
  2. Power dissipation.

The Quiescent current is the current the regulator requires in order to do its work. This will raise your total current consumption by the quiescent current value. For instance, if your circuit draws 150mA, and your regulator has a quiescent current of 5mA, you will need to draw 155mA from your power source.

The power dissipation is the amount of energy that is converted from electricity to heat in order to reduce the voltage to the required level. You can't just magic away the 5.7V you want to "remove" from the supply into nothingness - it has to go somewhere - so the regulator converts it to heat. This is all essentially wasted power.

So, if your system is drawing 150mA through the regulator, and you want 3.3V from a 9V supply, that is (9-3.3=) 5.7V at 150mA that is being lost to the environment as (P=VxR) 5.7x0.15 = 0.855W. Your circuit only uses (3.3x0.15=) 0.495W, so you are wasting almost twice as much energy as you are using.

Kind of silly really...

For voltage drops like this it is much more efficient (though slightly more complex) to use a switching regulator, which basically converts between voltage and current through PWM and low pass filtering. This can make for noisy supplies though, so it is common to use a switching regulator to go most of the way, and then use a small LDO linear regulator to finish the job and give a smoother output.

If your goal is utilize the most available energy possible from the battery (limited by it's mAH rating) in the most efficient manner possible then either:

  1. Run the circuitry directly from the battery voltage if the components in the circuit can handle the battery's voltage variation from max charge to minimum charge.

Or

  1. If the circuitry must have a constant regulated voltage at all times use a DC/DC switching regulator rather then a linear regulator.

Something like this might be useful:

http://www.ebay.com/itm/251066005460?ssPageName=STRK:MEWAX:IT&_trksid=p3984.m1423.l2649

Lefty

You can get micropower voltage regulators that have a quiescent current much lower than 6mA. I use the MCP1702, which has a quiescent current of just 2uA.

afremont:
EDIT: As others pointed out, there is the quiescent current and it is the current that flows when no load is present. I assume that's the 6mA you were referring to. You really can't do much about that, it's just part of the overhead of using the regulator.

You can use a better regulator. Some of them have quiescent currents in the microamp range.

fungus:

afremont:
EDIT: As others pointed out, there is the quiescent current and it is the current that flows when no load is present. I assume that’s the 6mA you were referring to. You really can’t do much about that, it’s just part of the overhead of using the regulator.

You can use a better regulator. Some of them have quiescent currents in the microamp range.

I thought that 6mA seemed kinda high, but I googled up a random 3.3V LDO by ST and it was rated at 700mA. It’s quiescent current was 5mA. If the OP needs to use such a high voltage, then the switching regulator is the way to go. Otherwise he should reconsider his power supply. The quiescent current seems to be the least of the issues.

afremont:
If the OP needs to use such a high voltage, then the switching regulator is the way to go. Otherwise he should reconsider his power supply. The quiescent current seems to be the least of the issues.

Did you mean "If the OP needs to use such a high current" ? If so, I agree with you.

dc42:

afremont:
If the OP needs to use such a high voltage, then the switching regulator is the way to go. Otherwise he should reconsider his power supply. The quiescent current seems to be the least of the issues.

Did you mean "If the OP needs to use such a high current" ? If so, I agree with you.

I guess what I meant to say was "high input voltage of 9V". Seems to me that if efficiency is the primary goal, then linear regulators are out of the question even if you can get one with no quiescent current. It's still going to be dissipating twice the power as is being dissipated by the actual load. Disagree?

The reason for my comment was that it is difficult to find switching regulators that are efficient at low output current. So if the load current is low, then a micropower linear regulator is more appropriate.

This one doesn't look too bad:

Over 90% efficient as long as the load current is at least 1mA.

Even the first switching regulator I found was over 60% efficient at the lowest current rating. Isn't that still more efficient than burning up 5.7V in a linear regulator just to obtain 3.3V regardless of the current?

EDIT: I can't seem to find how much current the OP requires.

I stand corrected, that device looks excellent for low-current applications.

I don't think the OP has told us what the maximum current requirement will be.

dc42:
I stand corrected, that device looks excellent for low-current applications.

I don't think the OP has told us what the maximum current requirement will be.

Even the 750mA device I first found didn't look all that bad, but the minimum current was kind high (10mA I think). This part looks like it might be good for automotive use.

Dear All,

Thank you very much for your comments. They are really appreciated.

Best regards,

Luis

Hi,

I use LE33cz or LE50cz from ST or others in the product family. They have a idle power consumption of 0,5 ma and low noise. Vin max is 20volt so it will protect your low voltage ic like the “atmega 328” whatever power supply you use.

greetz from holland.

I suggest a MAX631, which is a switching regulator, with 80%+ efficiency. Since it is also a step-up regulator, even if the input voltage drops below the output voltage, it can still provide the output.

It requires just 2 external components + a current divider if you want to adjust the output voltage (which is your case, since the MAX631's predefined output is 5V)

I am very found of Maxim's voltage regulators. They have an extremely low dropout voltage.

I often use the MAX604 and MAX603 in my designs.

AlxDroidDev:
I suggest a MAX631, which is a switching regulator, with 80%+ efficiency. Since it is also a step-up regulator, even if the input voltage drops below the output voltage, it can still provide the output.

It requires just 2 external components + a current divider if you want to adjust the output voltage (which is your case, since the MAX631's predefined output is 5V)

I am very found of Maxim's voltage regulators. They have an extremely low dropout voltage.

I often use the MAX604 and MAX603 in my designs.

I think this would be good if the OP was using say 2 or 3AA batteries for the supply, which might be a good idea. Unfortunately the OP is using 9V, this would be extremely bad news for the load connected to the regulator as the output voltage will be 9V - 1 diode drop. The OP needs a buck regulator or to change the supply and use this regulator. I like the idea of 3AA cells though.

I used to use the MAX1724 (2.7, 3.0, 3.3, 5.0 versions available) step-up DC-DC, 1.5uA quiescent current. Worked from ~0.7V input. Only a single inductor and 2 capacitors are required.. :slight_smile: