I wish to control the switching on and off of several LEDs as indicated in the following figure:


simulate this circuit – Schematic created using CircuitLab

The bases of transistors and the ON/OFF pin of the LDO will be controlled from a microcontroller.

My question is:

Can a periodic signal be applied to the ON/OFF pin of the LDO and it works correctly? (Possibly from 1 kHz to 2 Khz)

Note that the schematic represents the first row of a matrix of several LEDs. That is to say that for each row there will be an LDO.


  • \$\begingroup\$ Aside from the question you asked, you should watch out that your schematic shows no way to limit the current through the LEDs, so you're likely to end up with dead LEDs fairly quickly. \$\endgroup\$ – The Photon Oct 5 '18 at 16:44
  • \$\begingroup\$ 1.7 volts? What sort of LEDs are you attempting to drive? \$\endgroup\$ – Andy aka Oct 5 '18 at 17:00
  • \$\begingroup\$ I have not yet searched for the diodes, but I remember that that voltage needed a red diode sometime, the voltage is only referecial, then I must determine exactly what voltage I should use, the current that circulates is approximately 15mA \$\endgroup\$ – Fabián Romo Oct 5 '18 at 17:02

It seems a strange choice to drive the rows using several LDO regulators when there are chips you can buy for the job such as this one: -

enter image description here

Using several regulators will still require a series current limiting resistor for each row or column yet the MAX7219 only needs one to set the whole matrix. The scan rate is also much higher than you could effectively get from a bunch of LDO regulators. MAX7219 has a scan rate of typically 1 kHz but regulators with their output capacitor are likely to be much slower.

The MAX7219 has integrated transistors too. Doesn't seem much of a contest really.

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Possibly. The datasheet gives us some useful info:

First, from page 3, "This pin (out) requires an output capacitor to maintain stability". The output capacitor is required to be at least 22uF.

Let's say your LEDs draw 20 mA each, or 60 mA total. I = C * dV/dT means that if the regulator shuts off instantly, the voltage will only drop by 0.27V in 0.1mS (1/10th of your period). Exactly how long it takes between shutting the regulator off and when the LEDs turn off depends on your exact circuit, and on how much current the LEDs are drawing.

The turn-on transition will have a similar delay, because the LDO needs to have time to charge that capacitor. There aren't any turn-on waveforms in the datasheet to reference, but at 1A it will take about 0.04mS to charge that cap, which is probably OK.

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  • \$\begingroup\$ You might also look at simulating a similar Linear Tech LDO in LTspice - they have really good models of those parts. \$\endgroup\$ – Selvek Oct 5 '18 at 16:31
  • \$\begingroup\$ Thank you very much for answering. In theory, there will be 40 led diodes. Maybe I should look for a better way to manage the rows of the matrix. \$\endgroup\$ – Fabián Romo Oct 5 '18 at 16:37
  • \$\begingroup\$ Yeah, you probably should. This might work, but it's not a normal way to do things. Andy aka's answer is probably a better idea. \$\endgroup\$ – Selvek Oct 5 '18 at 20:42

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