# How do you calculate regulator lost current?

We are using on the board an az1117 regulator to some kind of toy. The toy works with 3.7v, hence we use this regulator.

The question- when the processor sleep, the output current will be 5uA, but I guess a 9V battery as an input with this regulator will be a bottle neck in terms of quiescent current.

1. How do you calculate from the data sheet, what would be the quiescent current for 9V or 5V inputs? it says on the data sheet that its 5mA for VIN=VOUT+1.25V. What about VIN=VOUT+4.25V?
2. Is there anything you can do to eliminate this bottle neck? the only option is to disable it but then there will be no voltage to the sleeping processor.

https://www.diodes.com/assets/Datasheets/AZ1117.pdf

• I don't understand what you mean with "bottleneck". A Bottleneck with respect to what? – Marcus Müller Mar 5 '17 at 8:07
• Presently, your question is unclear. Which current is 5uA? Is that 3.7V output current, or 9V input current? Generally, input current is a bit more than output current. Even if output current is zero, there will be some input current. This is called the "quiescent current." Have you measured the input and output current? Are you trying to minimize the input current so that the battery will last for a long time when processor is asleep? – mkeith Mar 5 '17 at 8:09
• Also note that when going from 9 V to 3.7 V, it's very likely that a LDO is not the linear regulator of your choice. In fact, if you care about battery life, linear regulators wasting down ~60% of your voltage seem to be the wrong pick. What's the current you draw on the output? – Marcus Müller Mar 5 '17 at 8:10
• OK, two things. Minimum output current is 5mA, and quiescent current is also a few mA. This will not be a good choice for your application. You should find another LDO regulator rather than try to make this part work. – mkeith Mar 5 '17 at 8:39
• what @mkeith said, this is certainly not the IC of choice in a low-power scenario. It's really not hard to find a less wasteful linear regulator. You should also ask yourself: why are you using 9V if you need 3.7V? wouldn't 3x AA batteries -> 4.5 V be better? – Marcus Müller Mar 5 '17 at 9:10

An XY problem?

A good alternative to the 1117 (quiescent current) for low-power applications would be the low-quiescent current LDO MCP1702, which has a 5 uA maximum quiescent current over its operating temperature and input voltage range.

Of course, a linear regulator itself wastes a lot of of power, but this should only be a concern when there is an appreciable load current during extended periods of time. In that case, a switching regulator could be a better option, at the expense of higher consumption when the toy is sleeping.

But if your toy will spend most of its time sleeping, then your overall power budget will be better off with a low quiescent linear regulator instead of a switching one.

As a final note: reducing battery voltage (from 9V to 4.5V = 3 x 1.5V AA), as Marcus Müller said, will always be less wasteful and less thermally-stressful for the regulator.

• Thanks a lot ! I used to think we will use a 3.7V Lipo, but there are many problems shipping them. 9v is just smaller then 3 AA .So if the one you gave me has much lower quiescent current, whats his downsides ? is it has less LDO ? is it more expensive? – Curnelious Mar 5 '17 at 14:09
• It's only rated up to 250 mA, whereas the 1117 is rated up to 1A. – Enric Blanco Mar 5 '17 at 14:11
• Oh I see. it also has lower max input voltage of 12V . So usually speaking, thats the downsides of a low quiescent current? I am trying to think about it, and everyone wants lower current spent.. so there must be some downside. I guess its the output current. – Curnelious Mar 5 '17 at 14:17
• Indeed, an usual downside is less current handling for the same chip area. Worse regulation performance can be another downside. All these downsides are related to the need of using a MOSFET as the pass element for LDO and low quiescent current. I won't go deeper, as this will be a subject for an entirely new question. – Enric Blanco Mar 5 '17 at 14:56
• The 1117 regulator is very very old. It was extremely popular long ago, before CMOS technology, and there are lots of old designs using it. Basically, you should forget about the 1117. In modern LDO's, you pay for low quiescent current, low noise, and high input voltage tolerance. But it is very competitive. There are many manufacturers, and if you are shipping in high volume, the price may be US\$ 0.07 or so for a SOT23 regulator. Maybe a little more if you need high input voltage. – mkeith Mar 5 '17 at 16:34