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I'm planning to use this LDO namely LT1085CT-5 and couldn't figure out the minimum input voltage.

The datasheet says "Operates Down to 1V Dropout" and the LT1085CT-5 means 5V fixed output. So does that mean a 5.9V output will not work or it just means to obtain 5V the minimum input should be 6V?

My aim is to obtain a regulated output(anything between 4.5V to 5.5V) from a 5.9V input.

The detailed datasheet given here. I was planning to use a 5.9V adapter as input voltage to this regulator and obtain a regulated voltage anything between 4.5V to 5.5V.

Would 5.9V have enough headroom for this regulator to obtain a regulated voltage(anything between(4.5V to 5.5V) or should I use an adjustable version?

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    \$\begingroup\$ The dropout voltage is dependent on current. How much current you need? \$\endgroup\$
    – Justme
    Jan 29, 2023 at 19:29
  • \$\begingroup\$ If you're getting something other than 5V from a 5V fixed regulator, how do you call it "regulated"? \$\endgroup\$
    – hobbs
    Jan 29, 2023 at 19:30
  • \$\begingroup\$ Definitely less than 1A. Around 600mA \$\endgroup\$
    – user1245
    Jan 29, 2023 at 19:30
  • \$\begingroup\$ @user1245 The LT1085 requires more than 0.4 V overhead, no matter what the current (or temperature) is. See upper left corner of page 7 of the LT1085 datasheet. That is if you must have 5.5 V. If less, then you may be okay. Why select something with so much current compliance? \$\endgroup\$ Jan 29, 2023 at 19:31
  • \$\begingroup\$ @hobbs It will act as a post filter \$\endgroup\$
    – user1245
    Jan 29, 2023 at 19:31

3 Answers 3

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So does that mean a 5.9V output will not work or it just means to obtain 5V the minimum input should be 6V?

If the dropout voltage is 1V, then 5.9V in will not produce 5.0V out, but will allow 4.9V out. But you will need an adjustable regulator. The 5.9V is the lowest value of the input including ripple.

Edit for comment: If the dropout voltage is 1V, operation at 0.9 V dropout will not work. The input voltage must be greater than the desired output plus the dropout voltage.

Looking at the data sheet shows that the dropout voltage for the 1085 could be as high as 1.5V depending on temperature and current.

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  • \$\begingroup\$ Many didnt get my question. Im ok with any output down to 4.5V as long is it is well regulated. I dont need exact 5.0V anything down to 4.5V is fine. What Im asking is not about precision but will the regulation work for 5.9V input? By regulation I dont mean to obtain 5.0V output but obtaining any voltage which is stable with current changes. So still headroom is problem? \$\endgroup\$
    – user1245
    Jan 29, 2023 at 20:26
  • \$\begingroup\$ @user1245:Head room is the problem. See my edit and read the data sheet. Don’t trust the third line down. Look at the chart on page 7 \$\endgroup\$
    – RussellH
    Jan 29, 2023 at 20:42
  • \$\begingroup\$ Alright that was I was asking. \$\endgroup\$
    – user1245
    Jan 29, 2023 at 20:44
  • \$\begingroup\$ "The input voltage must be greater than the desired output plus the dropout voltage." What happens otherwise? Would the proper regulation fail? \$\endgroup\$
    – user1245
    Jan 29, 2023 at 20:46
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    \$\begingroup\$ Yes, line regulation will fail. Probably the load regulation will be much worse also (but I'm not familiar with this particular part) \$\endgroup\$
    – The Photon
    Jan 29, 2023 at 20:59
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Look at the chart labeled "Dropout Voltage vs Output Current" -- the least amount of dropout voltage is 0.95V or so. Then look at the "Output voltage" section of the "Electrical Specifications" table -- the part will regulate to anywhere between 4.9V to 5.1V (and hold steady -- for that particular part).

"Dropout voltage" is the voltage where the regulator stops regulating, and just generates a constant drop from input to output. So the most voltage you'll be able to get with that regulator and a 5.9V input will be 4.95V, and probably less -- and unless you have a part that's on the low end of the output voltage range, it won't be regulating.

Put another way: if your input power really only ever drops to 5.9V and never less, and you really want to slap in a regulator that may drop out of regulation at 5.1V, then you need a dropout voltage of 0.8V or less.

So, no, you don't have enough headroom.

I'd go to a distributor that has parts that were introduced within the last 30 years, and get a real low dropout regulator. You can easily find them with dropout voltages of around half a volt or so, there's some that go down to 0.2V. Mind the recommendations for the output filter cap, and enjoy.

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  • \$\begingroup\$ Many didnt get my question. Im ok with any output down to 4.5V as long is it is well regulated. I dont need exact 5.0V anything down to 4.5V is fine. What Im asking is not about precision but will the regulation work for 5.9V input? By regulation I dont mean to obtain 5.0V output but obtaining any voltage which is stable with current changes. So still headroom is problem? \$\endgroup\$
    – user1245
    Jan 29, 2023 at 20:26
  • \$\begingroup\$ Or perhaps we're reluctant to give you answers which, if you can't work them out for yourself, you may not be able to use effectively. So -- what's 5.9V - 4.5V? Is that bigger than your chip's dropout voltage, or less? If it's bigger, do you have a problem? If it's less, do you have a problem? \$\endgroup\$
    – TimWescott
    Jan 29, 2023 at 20:31
  • \$\begingroup\$ 5.9V is SMPS output. I want to try an LDO to postprocess 5.9V. In my circuit anything will work between from 4.5V to 5.5V. But I cannot find any LDO with dropout less than 1.2V. I also need to use through holé and a specific vendors which limits me. So I will either use the fixed 5V one in question or buy adjustable version and set the output to say 4.6V(to be sure not violating dropout). I was asking whether first option is fine. \$\endgroup\$
    – user1245
    Jan 29, 2023 at 20:37
  • \$\begingroup\$ If you're limited to RS components -- my sympathy. That's all obsolete hobbyist part numbers. Through-hole, true LDO linear regulators are a thing, but it looks like an adjustable, higher-dropout regulator is what you'll need to use. \$\endgroup\$
    – TimWescott
    Jan 29, 2023 at 22:16
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"Operates Down to 1V" is the third line of the data sheet you supplied via link (thanks). Dropout is the overhead required for the regulator to work. Dropout is the difference between the input and output voltage where the input voltage is the higher. The output voltage + dropout is the minimum voltage required for proper regulation. If there is an AC component on the power input consider the lowest point the voltage that needs to be your output voltage plus 1. Also there is a curve posted showing that the dropout voltage changes with current.

There is a tolerance associated with the part. You need to figure that into the output voltage. Multiply the output voltage rating by the tolerance and add or subtract that from the stated output voltage and that will be the range the output should be.

What this is saying if you do not have enough headroom voltage the regulator cannot work properly, that is why it is on the data sheet. If there is ripple on the input and it goes below the threshold voltage even for part of a cycle it will also show up in the ripple. You need to read the rest of the data sheet to see what effects this overhead. per the data sheet linked it will not work properly with 5.9V

These parts have a guard band and it may work in some cases but there is no guarantee if it is out of data sheet specification. Different manufacturers may specifications may not be the same even if the same part number.

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  • \$\begingroup\$ So what's your conclusion being explicit? 5.9V is not high enough input for this part? \$\endgroup\$
    – user1245
    Jan 29, 2023 at 20:02
  • \$\begingroup\$ I still dont get an answer whether 0.9V headroom is problematic. \$\endgroup\$
    – user1245
    Jan 29, 2023 at 20:14

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