# Need alternative way to deal with LDOs minimum current requirement

I am designing a circuit that uses an LDO to produce 3.3v from a 10 to 18v input. I sometimes will need 1.2 amps from this regulator and at other times I only need microamps. When I need even less I can shut down the regulator but there will be times when I need regulation and my circuit will only draw possibly .5 or 1 milliamp. My circuit contains a microcontroller.

My datasheet says my LDO requires 5ma minimum load. It's a Micrel/Microchip MIC29150.

So my question is, other than putting a resistor on one of the GPIO pins of my micrcontroller and burning off electricity at around 5ma when I want to, are there any other techniques I can use that are better?

One of my thoughts were to put a zener in series with a resistor so that if the voltage goes over 3.5 or so that it burned off some milliamps but that seems dangerous to me.

• So what's the problem with putting a 680 ohm resistor from the output of the regulator to ground (not one of the GPIO pins) and let it waste 5mA? – JIm Dearden Feb 23 '17 at 19:51
• It's not dangerous but if you are always going to be taking 5 mA you might as well use a standard regulator especially as your supply has tons of headroom and doesn't need an LDOR or do what Jim says. – Andy aka Feb 23 '17 at 19:52
• regulating 1+ Amp over 7~15V difference from input, sounds like you could use a buck converter.. – Wesley Lee Feb 23 '17 at 19:56
• Small LDO + buck converter that can be switched on and off? – pjc50 Feb 23 '17 at 20:11
• @JImDearden: I thought about a buck converter instead. But it's noisier. The reason I need to turn this on and off is because I can disable the LDO and the circuit needs to only pull microamps. It will be living on a supercap charge potentially for days. The uController will wake up every few hours to check a few things and then put itself back into deep sleep. – mark b Feb 23 '17 at 22:49

Per the datasheet for the MIC29150:

The MIC2915x–2975x regulators are specified between finite loads. If the output current is too small, leakage currents dominate and the output voltage rises. The following minimum load current swamps any expected leakage current across the operating temperature range...

So it seems the minimum load requirement is determined by the worst case leakage from IN to OUT through the LDO. 5mA seems like a lot - the leakage is probably far less than this in most cases. Also, leakage is a very strong function of temperature, so if your application will not see high temperatures you could get away with less.

A zener at the LDO output might also be an option, as you stated. This will prevent the output of the LDO from exceeding the zener breakdown voltage in the case that the leakage current is greater than the load. In this case you would want to check that your load can survive the zener voltage, and also make sure that there is no overlap between the regulation voltage of the LDO and the breakdown of the zener - that would cause the LDO to dump current into the zener.

Personally, I would recommend looking for a different LDO that doesn't have such an onerous minimum load requirement. There are many with no minimum load requirement at all.

Most LDOs with minimum current requirements do so for stability. With too little current, the control loop may not have enough gain, or it may lose phase margin and get unstable.

Perhaps try to find a regulator which does not have such a requirement?

Check MIC2940A and MIC2941A

You could also use a switching regulator...

• I chose that regulator because it has automotive load dump and reverse connection features I like. I will look at those others. – mark b Feb 23 '17 at 22:52
• Unfortunately, the MIC294x datasheets say this: The MIC2940A will remain stable and in regulation with load currents ranging from 5mA on up to the full 1.25A rating. – mark b Feb 23 '17 at 22:58
• It also says "The minimum recommended load current of 1µA" for MIC2941 so.... I wonder which one wins! – peufeu Feb 24 '17 at 0:14
• That's weird. I'm going to go ahead and prototype this thing and find out more. – mark b Feb 24 '17 at 4:16
• Those ICs are huuuge. – mark b Feb 24 '17 at 6:47

MCP1703 is specified up to 16V (tolerates 18V), has a $<10\ \mu{A}$ quiescent current and ground current $<0.15\ mA$ under maximum load conditions.

If you need it, you can drop some volts using series diodes at the input.

• Yes. But only 1/4 amp. :( – mark b Feb 24 '17 at 6:47