First, a little background - I'm working on replacing an old (late 1970's) output driver board. The old equipment this came off of has a very dirty and poorly-regulated 5V system. Since I'm using newer pieces and replacing some address decoders with a FPGA, I need to have a more stable 5V on the board.

The 5V coming in can range from 4.75V to 5.5V (or more, if the other boards have bad caps, thermal wear, etc). I was looking at using a 1117 to at least keep the voltage close to 5V when/if possible - to avoid a bad power board destroying the new board.

I haven't used 1117's before (2705's usually in the past), and noticed on the datasheet it says:

Vin: This pin must always be 1.3V higher than Vout in order for the device to regulate properly.

Am I correct in assuming that means that in a case like this, if the inpute voltage was between 4.75V and 6.3V, my FPGA and buffers would get the same as the incoming voltage, and then if say the power board blew up and fed 8V in, it would prevent the 8V from frying everything?

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    \$\begingroup\$ Is there any higher voltage available? What does the old 5 V come from? Would it be too onerous to replace the 5 V supply instead of creating something to try to compensate for the existing poorly performing one? Why do the other boards have such an influence on each other via the power supply? \$\endgroup\$ – Andrew Morton Dec 26 '17 at 16:09
  • \$\begingroup\$ Your quoted line should read "must be at least 1.3 V higher", I think. \$\endgroup\$ – Peter Bennett Dec 26 '17 at 17:16
  • \$\begingroup\$ @PeterBennett - You're right, I just mistyped. Thanks for fixing. \$\endgroup\$ – Coyttl Dec 26 '17 at 19:14
  • \$\begingroup\$ Could you throw some big caps from VCC -> GND to try to smooth out the input power? Are you able to get a scope shot of the 5V line so we can see the spikes? \$\endgroup\$ – Jim Dec 26 '17 at 19:24
  • \$\begingroup\$ @AndrewMorton Unfortunately, that's handled on a completely different board. I can't guarantee anyone plugging my board into the unit would have a newer power-board. The 5V is generated by a LM323 (old sucker!) since there are a few other devices/boards being run off of the 5V. The board that supplies the 5V also handles switching high-impedance loads, so there's definitely a lot of ripple on the feed. I'm more concerned with the voltage range than I am the ripple. \$\endgroup\$ – Coyttl Dec 26 '17 at 19:24

The 1.3 V is the "dropout" voltage for the linear regulator. What it means is that for any Vin < Vout + 1.3 V, the output is not guaranteed to be the specc'ed Vout; 5 V in your case. Vout will likely be approximately Vin - 1.3; depending on output current.

From the TI datasheet (LM1117 Datasheet) : LM1117 Datasheet Annotated

So from a design perspective, you would want the input voltage to be at LEAST 6.3 V, and anywhere up to 15 V (per the datasheet recommended operating conditions) to "guarantee" a clean 5 V out.

Edit 2017-12-26: changed above bold section from 20 to 15 V per recommended operating conditions, instead of absolute maximum's. Thank you @SamGibson for the wise recommendation

Recommended Operating Conditions

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    \$\begingroup\$ Hi - I suggest not to recommend a maximum of 20V input. That is listed under the "Absolute Maximum ratings" section, which means that correct behaviour is not guaranteed at that voltage (see note (1) under that section). It just means that the device should not be damaged by short periods at that voltage. Instead, the maximum operating voltage is 15V (see datasheet section 6.3) - assuming adequate heatsinking, of course. \$\endgroup\$ – SamGibson Dec 26 '17 at 15:54
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    \$\begingroup\$ Good catch. Fixing. \$\endgroup\$ – Jim Dec 26 '17 at 15:56
  • \$\begingroup\$ Thanks, that datasheet gives better explanation that the one I was looking at. (Diodes Incorporated vs your TI). May have to start another question then, and ask for best way to limit input voltage to my FGPAs when it can vary. (FGPAs absolute max input voltage is 5.5V..), since it looks like I'm in a quandry. \$\endgroup\$ – Coyttl Dec 26 '17 at 19:18
  • \$\begingroup\$ Very very very late in getting back to this, but noticed I didn't give your reply as 'Answered'. Thanks again - this did help in the project! \$\endgroup\$ – Coyttl Nov 15 '18 at 18:13

No, the minimum voltage difference between the input and output is called the Drop Out voltage (hence the name LDO, Low Drop-Out regulator). The drop out voltage varies with the current load, but at rated current, Vout will always = Vin - 1.3V as a design minimum. So, at 4.75V in, you can only depend on getting 3.55V out. There are regulators with lower drop out voltages, but for your application you might look into a buck-boost switching converter, possibly followed by an LDO if you need really clean power for the FPGA. If you use a BB followed by an LDO, set the BB output voltage for a bit more than the dropout voltage, many of the regulation characteristics of the LDO improve as you provide more drop out margin in your circuit (but not too much more, the difference is going to be dissipated as heat).


The 1.3V dropout voltage means that the regulator's input must be 1.3V higher than the expected output of 5V, e.g. 6.3V. For this reason most 5V DC power supplies use something like 7V to 10V as the input to meet and exceed the minimum input voltage requirement. The voltage difference between the input and output is dissipated as heat so you don't want too large of an input voltage without an adequate heatsink.

You'll definitely want to use the LM1117 to replace the original 5V supply entirely. It's probably fed by something in the 8V to 12V range. Don't try to clean up the existing regulated 5V output, it's not easily done. Replacing it will likely be less work.

Also for stability purposes check the datasheet and verify the input and output capacitance of the existing system meets the requirements of the LM1117. Chances are whatever is in there is already sufficient, but it's good to be sure.

If the input voltage is lower than the sum of the output voltage plus dropout voltage then the regulator will not regulate properly and the output is unreliable. It could pass the input to the output as you suggested, or it could shut off the output or even oscillate as the load changes. So you want to avoid that at all costs.


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