Fast startup LDO - Page 1
Forum section: electronics / general

author: TC
author_rank: Contributor
timestamp: August 19, 2023, 02:33:56 pm
content: I am trying to find an LDO (1.2V to 1.8V, 10mA to 100mA is fine) with thefastest startup time. I'm hoping some of you might be able to suggest parts from your experience. Unfortunately, many of the LDOs that specifically claim fast startup are specifying turn-on time from an enable pin assuming that the input voltage is already within spec. I'm looking for an LDO with fast startup specified from the time that the input voltage rises from 0 to a valid input range (on the initial power-up).I'd be willing to use a discrete transistor design if it gave me a faster startup.Thanks in advance for any suggestions you may have.

author: ahbushnell
author_rank: Frequent Contributor
timestamp: August 19, 2023, 08:42:18 pm
content: How fast do you need?

author: TC
author_rank: Contributor
timestamp: August 19, 2023, 09:58:01 pm
content: Something under 100uS... but it's one of those things, the faster the better.Vin can be held to under 6V (or even 5V). I don't need a high Vin.Almost all specifications can be compromised in favor of fast startup.

author: Zoli
author_rank: Frequent Contributor
timestamp: August 20, 2023, 12:43:57 am
content: How critical the other parameters are? Because for speed, you can't beat an LED(OK, diodes) in series with a resistor...

author: nctnico
author_rank: Super Contributor
timestamp: August 20, 2023, 01:25:10 am
content: Have you considered a parallel (shunt) regulator?

author: langwadt
author_rank: Super Contributor
timestamp: August 20, 2023, 01:33:31 am
content: have you actually seen it being slow? I'd expect the output of a "dumb" three pin linear regulator to follow the input with very little delay

author: magic
author_rank: Super Contributor
timestamp: August 20, 2023, 08:13:11 am
content: Mind output capacitance.100mA for 100μs is 10μC.For 1.2V output, total load capacitance must be less than 10μF.And I'm not even accounting for load current yet...Quote from: TC on August 19, 2023, 09:58:01 pmWhat's the actual input voltage and do you really need an LDO?Particularly, sinceQuote from: TC on August 19, 2023, 02:33:56 pmis the input (or perhaps other rail) high enough to keep the base of an emitter follower at ~2V?

author: Shonky
author_rank: Frequent Contributor
timestamp: August 20, 2023, 08:20:52 am
content: Why can't you use the enable pin? Does the input supply come up fast enough to meet your requirement?

author: TC
author_rank: Contributor
timestamp: August 20, 2023, 01:30:49 pm
content: Quote from: Zoli on August 20, 2023, 12:43:57 amSpeed is the most important parameter. I can compromise on other parameters.Assuming a ramp from 0V... then low output voltage, low dropout (an LDO), low input and output capacitance, and low load current are all factors since it is a dV/dt problem.

author: TC
author_rank: Contributor
timestamp: August 20, 2023, 01:34:58 pm
content: Quote from: nctnico on August 20, 2023, 01:25:10 amYes. But a shunt regulator generally requires a higher Vin compared to an LDO, doesn't it? So there would be added delay for Vin to reach the higher voltage if this is correct. When I mentioned a discrete transistor design I was thinking about a shunt regulator.

author: TC
author_rank: Contributor
timestamp: August 20, 2023, 01:37:13 pm
content: Quote from: langwadt on August 20, 2023, 01:33:31 amI haven't built and measured anything yet. I started looking at datasheets to try and select a fast startup LDO. That's what prompted my initial post. I was hoping someone might have experience to share.But, yeah... I need to get a few LDOs and do some measurements... but that's right back to which ones?

author: TC
author_rank: Contributor
timestamp: August 20, 2023, 01:44:32 pm
content: Quote from: magic on August 20, 2023, 08:13:11 amAgreed. I mentioned a 10mA to 100mA LDO just to say that I don't need much current. I expect a few mA load current for this fast-start application. You are right re: input and output capacitance, and load current.

author: TC
author_rank: Contributor
timestamp: August 20, 2023, 01:46:37 pm
content: Quote from: Shonky on August 20, 2023, 08:20:52 amThe Vin supply ramps from 0V. The enable pin assumes Vin is already at the required minimum input voltage for the regulator.I simply want fast Vout when Vin crossed the Vmin threshold (during the initial power-on ramp).

author: Zoli
author_rank: Frequent Contributor
timestamp: August 20, 2023, 01:52:30 pm
content: TS3431 has the turn-on characteristics in the data-sheet(attached) on page 7; add a transistor, few resistors and you've got an LDO.

author: TC
author_rank: Contributor
timestamp: August 20, 2023, 02:14:23 pm
content: Quote from: Zoli on August 20, 2023, 01:52:30 pmThis looks like a great suggestion. Thanks. I will give this a try!

author: AnalogTodd
author_rank: Regular Contributor
timestamp: August 20, 2023, 02:14:53 pm
content: Quote from: TC on August 20, 2023, 01:46:37 pmThere are really two things you want to look at to make something as fast as possible:1. Quiescent current of the LDO. Most 10mA to 100mA LDOs are micropower to avoid burning extra power from a limited source. You want something that has several hundred uA of quiescent current. This is needed to charge up all the internal capacitances, including transistor junctions.2. Output capacitor requirement and current limit. The larger output capacitor required and the lower the current limit, the longer it will take to charge the output. You may need to go with a part that is capable of 500mA or more in order to have a good start-up time.Some research online found the MAX1935, a 500mA regulator that runs 210uA quiescent current and a typical 1.4A current limit with a 10uF output cap. Other possibilities are the LT1965 or LT1963A. Check the Typical Performance Characteristics and you'll see the output coming up in 25usec after /SHDN is toggled, which means you should be able to meet 100usec from VIN rising.

author: langwadt
author_rank: Super Contributor
timestamp: August 20, 2023, 02:22:56 pm
content: Quote from: TC on August 20, 2023, 01:37:13 pmprobably ones without enable

author: DavidAlfa
author_rank: Super Contributor
timestamp: August 20, 2023, 03:19:50 pm
content: The fast startup will also depend a lot on the output capacitance.If you have 50 ICs powered by the LDO, each one having their own 1uf/100nF capacitor, forget it!If you are only powering a small load, maybe!Torex has quite a lot to choose from, all their datasheets show the trasient waveforms of the Enable pin.Had a peek at some, some showed about 40us turn-on time:https://www.torex-europe.com/products/voltage-regulators/high-speed/

author: nctnico
author_rank: Super Contributor
timestamp: August 20, 2023, 06:23:14 pm
content: Quote from: TC on August 20, 2023, 01:34:58 pmNo. A shunt (parallel) regulator limits the voltage so it is instant on.

author: nctnico
author_rank: Super Contributor
timestamp: August 20, 2023, 06:24:23 pm
content: Quote from: langwadt on August 20, 2023, 01:33:31 amFor an LDO this is tricky because the control circuitry needs to become active first. Some LDOs will overshoot the output to the input voltage (long enough to do damage) if the input voltage rises too fast.

author: TC
author_rank: Contributor
timestamp: August 20, 2023, 06:45:05 pm
content: Thank you ALLfor the great questions, comments, and suggestions.I've got a clearer idea of how to progress now. I do appreciate it!!!

author: magic
author_rank: Super Contributor
timestamp: August 20, 2023, 06:53:46 pm
content: By they way, does your load include or require supply bypass capacitors?If not, maybe a low voltage R2R opamp could be an alternative to normal LDOs, one which doesn't require (or tolerate) output capacitance, avoiding one source of startup delay.Quote from: Zoli on August 20, 2023, 01:52:30 pmNot sure how that would work?TL431-style chips are inverting, so they need to drive a noninverting power stage, most easily an emitter follower.An emitter follower won't make an LDO, unless there is another (maybe 2.5V or more) rail in the system which can pull its base up for those situations when Vin is barely above Vout.Quote from: AnalogTodd on August 20, 2023, 02:14:53 pmThe Linear parts look like old school bipolar PNP LDOs, they have lower minimum startup voltage than the CMOS regulator from Maxim, which may work to their advantage.Quote from: nctnico on August 20, 2023, 06:23:14 pmYes, but its load may be not so instant on. It's not entirely trivial to make a low dropout 100mA current source.Though on second thought, maybe a stupid PNP current mirror could be good enough to power some Txyz431whatever.

author: Zoli
author_rank: Frequent Contributor
timestamp: August 21, 2023, 12:19:40 am
content: Quote from: magic on August 20, 2023, 06:53:46 pmSorry for not putting the LDO between "quote-marks", since the OP clearly stated that he can go up to 5.5V input voltage, and he was looking for ideas; I hope this will address your concerns.

author: nctnico
author_rank: Super Contributor
timestamp: August 21, 2023, 08:40:07 am
content: Quote from: magic on August 20, 2023, 06:53:46 pmThere is no dropout on a shunt regulator as a shunt regulator is parallel to the load (like a super precise zener diode).

author: AnalogTodd
author_rank: Regular Contributor
timestamp: August 21, 2023, 01:33:23 pm
content: Quote from: magic on August 20, 2023, 06:53:46 pmThey are bipolar and as such will begin to bias up internally at less than 1V. I know, I designed the LT1963A.

author: magic
author_rank: Super Contributor
timestamp: August 21, 2023, 06:00:53 pm
content: Quote from: nctnico on August 21, 2023, 08:40:07 amSorry, I said "load", I meant the current source which supplies the shunt regulator. You have moved the droupout problem from the regulator to the current source.Can't quite use a resistor, because it's either very little current at low input voltage or too much at high input voltage.Not sure about JFETs or CRDs, I think they tend to have too much RDS(on).Bipolar transistor circuits can do it, but you now have to design something.

author: nctnico
author_rank: Super Contributor
timestamp: August 21, 2023, 07:15:59 pm
content: Quote from: magic on August 21, 2023, 06:00:53 pmYou are creating problems that aren't there. OP specifies a supply voltage of 5V. So yes, a resistor is the way to go. In some cases I put a resistor in series with a series regulator as well to provide an extra RC filter and dissipate some heat outside the regulator itself.

author: magic
author_rank: Super Contributor
timestamp: August 21, 2023, 07:47:10 pm
content: Quote from: nctnico on August 21, 2023, 07:15:59 pmIt's possible that OP is inventing problems that aren't there, I still have no clue what's the purpose of this regulator.However, we just go along and try to come up with the fastest starting regulator possible, which means giving 1.2V as soon as the input rail reaches any more than that.This was explicitly stated as one of the goals.

author: TC
author_rank: Contributor
timestamp: August 24, 2023, 01:19:52 am
content: Quote from: magic on August 21, 2023, 07:47:10 pmYes, that's the goal. Fastest startup as the input source ramps up to a maximum of 5V... hopefully way before the source reaches 5V.This is for what is essentially a low-voltage, instant on, sensing circuit. The faster it starts up... the better. Even if this requirement seems strange it is an interesting design challenge.Again, I appreciat all the comments and suggestions.