voltage-regulator
I want to use TPS799 for an LED driver circuit.
Here is the datasheet: http://www.ti.com/lit/ds/sbvs056k/sbvs056k.pdf
I am assuming that input will be 5V through power jack on the board, and I will be driving an LED , after regulating the voltage to 3 V.
By what I understand, I don't need a capacitor at either, EN or NR terminals.
I am I understanding its usage correctly? And I am not sure how to have the output to 3 V. I am very new to idea of using a regulator so any help is appreciated.
The 100 nF capacitor on the output is an important component for the stability of the regulator's control loop. It's not there to catch fast load changes; for that its value is too low.
A voltage regulator needs a short time to respond to load changes. This may seem like an imperfection but it's a requirement actually; a regulator with zero delay would not be stable.
Switching the reader on and off causes fast load changes. It won't be anything like 1 A/µs (which is damn fast), but fast enough to let the 5 V output sag for a moment. Russell calculated a value of 40000 µF to allow a 100 ms holdup, but you won't need that, the LM7805 will respond much faster than that (I would love to see more dynamic performance graphs in datasheets, especially step response!). For 200 mA steps a value of 100 µF should be fine. If you want to add a much larger value, like the 40000 µF it should be at the regulator's input, not the output.
edit
The guys at Diodes do include dynamic performance data in their AP1117 datasheet:
This shows an output voltage spike of a mere 15 mV for a 700 mA load change. The LM7805 is a much older design, and the figures won't be that good, but they give you an idea.
(end of edit)
I second Russell's suggestion for a series resistor to take most of the dissipation away from the regulator. At 200 mA a 56 Ω resistor will still give you 8 V input with a battery voltage of 20 V. The resistor will dissipate 2.25 W, so take a 5 W part for that. At 24 V in the regulator will have to handle 1.6 W, which it can do with a moderate heatsink. (Russell get a much lower dissipation, but he doesn't have any headroom in case the input voltage will sag.)
You mention that you want fast switching. None of the devices will be able to do that. Any the feedback circuits internally are designed to be slow so that they don't to much incorporate noise. And in fact some of the feedback is thermal feedback. Put another way, the dominant pole is low, 100's of KHz is typically the the fastest response time. This is true whether you are using a disable pin or connecting/opening the load to the source.
The best way to fix this is that you build an external circuit that current steers between two limbs. On one limb will be your LED chain and on the other a dummy load that draws the same amount of current (and hopefully also has similar load characteristics). An example of such a circuit is a differential pair. You should be able to get 10's of ns switching time then.
The challenge with the two limbs will be that the load characteristic differences may disturb the feedback of the LM317 etc. which then reacts slowly. If you can't match the two limbs then I'd suggest building a current mirror that decouples the current output the LM317 etc. so that it only sees a constant load the mirror transistor sees the variability.
You should be able to simulate all of this in LTSPice or similar before building.
The other factors you mention above not show stoppers, so pick what you want.
Best Answer
Are you just driving an LED? If so you don't need a regulator. Also you shouldn't be driving an LED with a constant voltage, you need to drive it with a reasonably well defined current. You can do that with just a resistor.
simulate this circuit – Schematic created using CircuitLab
EDIT:
If the input voltage only changes over a moderate range (2:1) then you don't need a regulator. Just choose an appropriate resistor value to give the LED current you need.
If for example if you have a 5V supply and you want 20mA through the LED which has a 3V drop (common for a white LED). Then the resistor you will need is:
R = (Vinput - Vled) * 1000/LED_Current.
For the example values this gives (5 - 3) * 1000/20 = 100 Ohm.
The LED voltage will vary with the color. A Green LED is probably about 2.1V and a Red LED about 1.7V, Infra red ~1.2V.
If the input voltage varies the current will as well but you probably won't even be able to see a 30% change in brightness. This would occur if the voltage changes between 4.4V and 5.6V.