Author Topic: LT3083 current limiting: how does it work?  (Read 7355 times)

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Offline exscape

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LT3083 current limiting: how does it work?
« on: March 08, 2012, 02:43:48 pm »
Hey, everyone! I've been following the blog for a while, and have decided to make my own power supply. However, I'm not following along Dave's design exactly, but rather making things a bit more simple. I'm using regular analog control via pots, though will use digital readouts of voltage/current. Anyhow.

I HAVE watched the PSU videos, but at least in simulations I couldn't get the single-LT308x with the transistor limiter to work properly (one, it didn't go to 0, so a shorted output causes >2 amps to flow! - and two, it tended to oscillate a bit now and then). I've almost decided to go a simpler but more expensive route and simply use two LT3083s.

I've simulated the "lab supply" schematic from the LT3083 datasheet, with some slight modifications:


Unfortunately, I don't fully understand the current limiting part, and the datasheet doesn't really discuss it more than mentioning the circuit diagram.

I attached the LTSpice circuit, though without the MAX4080T current sense amp, since that component is in a separate file on my computer. It didn't affect these questions whatsoever, though.

Also, note that "Iset" and "Vset" are potentiometers wired as rheostats; I model them as resistors in LTSpice since there's no built-in pot component. In each case, they're controlled by a 50 µA current from the LT3083.
Vout = 50 µA * Rset for the right regulator, but I don't understand the left one.
Rlimit limits the current to 1 V / R, and I've chosen 0.5 since I want the maximum to be at 2 amps.

What I'd like to do is to use Rlimit (0.5 ohms) as the current sense resistor (with a 5x gain amp instead, of course), to eliminate the extra 250 mV drop over Rsense at the output. The reason I "can't" right now is that there is a lot of ripple between the regulators when the current-limit mode isn't active; the left LT3083 simply outputs as high a voltage as it can, which lets the input ripple through unchanged:


Here's a closeup of the current ripple through Rlimit:


So, with all that ripple, the measurement isn't going to be that great unless I average a ton of values etc.
I figure that a solution might be to limit the output voltage to just below the "ripple voltage" (i.e. make sure that the maximum output voltage is below where the input ripple makes it through), but I don't quite understand how the current limit works!
The set resistance is 10 ohm per mA (so with 2k it lets 200 mA through to the load), but I'm not sure "why" that is...

Would it even be possible to add this? To be clear, I'd like (if possible) to have the adjustable limit of 0 - 2 A, and to have the left regulator output at most ~16.8 V, which should keep it ripple free at all times. Needless to say the max final output voltage will be slightly lower than now, but that's OK. (I can stand to lose at least 0.75 V. Also I should gain up to 0.25 V at load from dropping the 125 mOhm shunt.)

Sorry for all the text; I figured that details make it so that there's less work to be done in order to help. :)
 

Offline Ajahn Lambda

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Re: LT3083 current limiting: how does it work?
« Reply #1 on: March 10, 2012, 04:53:17 pm »
If you're after analog control, take a look at the attachment, an excerpt from the LT3080 data sheet.  I don't see any reason you couldn't replace LT3080s with LT3083s (they are essentially the same except for their current output ratings).  This way, you avoid using any more external parts, op amps, transistors, etc.


Don't always trust simulation software, either.  Things like step time, initial conditions, etc. can have an effect upon the results, and they may not always be right.  Most of the time, they're fairly close, though.
I am not a smatr poney.
 

Offline exscape

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Re: LT3083 current limiting: how does it work?
« Reply #2 on: March 10, 2012, 04:56:34 pm »
If you're after analog control, take a look at the attachment, an excerpt from the LT3080 data sheet.  I don't see any reason you couldn't replace LT3080s with LT3083s (they are essentially the same except for their current output ratings).  This way, you avoid using any more external parts, op amps, transistors, etc.


Don't always trust simulation software, either.  Things like step time, initial conditions, etc. can have an effect upon the results, and they may not always be right.  Most of the time, they're fairly close, though.
That actually looks very similar, except it parallels two LT3080s instead. The design I use is almost identical to the "Lab supply" schematic in the LT3083 datasheet, though, which ends up a lot cheaper (even then, 2 rails with 2 each is on the order of $50 for just the regulators).

I'd have to wonder, though, if the same problem doesn't apply to that setup! I would almost assume it does.
 

Offline Miroslav

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Re: LT3083 current limiting: how does it work?
« Reply #3 on: March 06, 2015, 08:07:52 pm »
It's been almost three years since original post, but I hope this still might be useful for future readers :)

You have posted several questions, so let's go through one at a time.

1) The cost of regulators.
Might be that prices now (2015) and then (2012) are not comparable - currently on digikey LT3080 is under $4.5, and LT3083 under $8.5, so quite less then stated $50 in one of configurations mentioned below

2) Input voltage +/- ripple vs. output voltage
No doubts here - your projected output voltage HAS to be less then MINIMUM value of the input voltage (Vin-Vripple) but also don't forget to subtract voltage drop on all components in line between the Vin and Vout (in your case those are required dropout voltages on two LDO regulators (in your case with Vcntrl and IN pins connected, it's up to 1.6V; if you "separate" Vcntrl and IN pins, that can go down to up to 0.5V!), and also voltage drops on two current sense resistors). In your case that would be:
18.5V (Vin) - 0.3V (Vripple) - 1.6V (first regulator) - 1V (current sense) - 1.6V (second regulator) - 0.25V (the last current sense) => Vout should be up to 13.75V. You have it set to 12V, so that should be OK

3) "Magic" behind the current limit
LT3080/3083 internally assures that voltage supplied to the pin SET is identical to the voltage on the pin OUT (it's called error amplifier, if OUT pin goes higher, that causes internal circuits to "increase resistance" of the regulating transistor, and in turn the OUT pin's voltage goes down. In the case of the Voltage Regulation circuit, potential (voltage) of the SET pin is defined (for LT3083) as 50uA (provided by the IC) times resistor you call Vset. Note that one side of that resistor is connected to the GND!

In the case of the current limiter, that "same" resistor (now you call it Iset) is connected "after" the current sense resistor (Rlimit).  At this point in diagram, potential is OUT-Rlimit*Ilimit. To have OUT and SET at the same potential, voltage on Iset has to be 50uA*Iset, or simply said Rlimit*Ilimit = 50uA*Iset => I=50uA*Iset/Rlimit (note: Iset is actually resistor, but that's how it's marked in your diagram  :) ).

Unless it reaches Ilimit, voltage drop on Rlimit is lower then votage drop on Iset resistor, so error amplifier sees that OUT pin is not "high" enough, and internal circuitry opens the transistor that regulates the output to the max, effectively not impacting the next stage. If the current gets to Ilimit value and attempts to go higher, error amplifier will see that OUT has higher potential then SET, and will take actions to "reduce" the Vout as much as needed to get current under control. That will in turn lower the supply voltage of the next block (voltage Regulator) and in turn lower the output to your load

4) Getting 0V output
If you check the data sheet, you will see that to function, this LDO regulator requires certain minimal output current (1mA). When output is close to zero, output current goes under that value. You either need to accept that limit, or to have a bit more complicated schematic that ensures that requirement to be met at all times (for example, you will need a simple constant current circuit drawing at least 1mA out of the OUT pin, and for that you will need additional negative supply rail)

5) Occasional oscillations
Two major reasons are:
   a) incorrect selection of capacitors - read the "Stability and Output Capacitance" section in the Data Sheet! Especially check where it mentions "low ESR requirement". (You need high quality electrolytic or tantalum/ceramic). 

   b) incorrect "wiring" (or PCB design). See design recommendations in the Data Sheet. The main issue is that LT308x uses fairly small current for the SAT pin (10uA or 50uA) and any stray capacitance and noise can result in instability (oscillations).

You also might want to add additional capacitors between IN and OUT pins of the current limit block (see Data Sheet where it describes current limiter).
 


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