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TLV3201: Output current capability

Mohammed Khan
Intellectual 570 points
Part Number: TLV3201

Hello,

I have connected the output pin of TLV3201 to drive an Opto-Coupler. The idea is to calculate the resistor value to be placed between output pin and Opto (Current range driving opto 5mA ~ 20mA).

1. Should i Refer to figure 14 (Short-Circuit Current vs Temperature) of datasheet to calculate current driving capability or how is it calculated? please guide.

2. Also, guide me on a thumb rule to calculate output current capability for Push-pull / Open-drain

  • 0
    Guru 317260 points

    The short-circuit current specifies what happens when the output is shorted. To find output how much current you can typically use when you want a specific voltage over the load, see figures 15/16.

    In general, you compute the resistor like for any other LED, and subtract the comparator's output impedance from that. (Figure 16 shows a typical output impedance of 50 Ω. Assuming you have a 5 V supply and an IR led with a forward voltage of 1.2 V, and need 10 mA, you'd want 380 Ω total, or a 330 Ω resistor.)

  • 0 in reply to Clemens Ladisch
    Intellectual 570 points

    My bad, I was actually pointing out to figure 16, somehow i got that error and pointed it out to fig 14.

    But nevertheless, I get the point. By looking at the (Figure 16. Output Voltage vs Output Current), the max current drawn can be safely concluded as 35mA between the temp (25 ~ 75).

    Also, what is the point mentioning the output impedance as 50 Ω, does this strictly impacts the resistor value which is chosen ?

  • 0 in reply to Mohammed Khan
    Guru 317260 points

    For reasonably small currents, the comparator output (a MOSFET) behaves like a resistor; the impedance is the slope of the voltage/current graph. So you have this resistor in series with your actual resistor.

  • 0
    TI__Guru 60716 points

    Hello Mohammed,

    The current would be set by the recommended LED current from the opto-isolator manufacturer. As Clemens mentioned, that is the current you should design to.

    We are assuming 10mA for the LED, so looking at the Fig 15/16 graphs, the output voltage drops about 700mV at 10mA, so the output would be about 2V at the worst-case 2.7V supply.

    5V - 0.7V = 4.3V; 4.3V - 1.2V = 3.1V; 3.1V/10mA = 310 ohms, so Clemens 330 ohm LED dropping resistor is just about right.

    10mA is a but much for the newer high-brightness LED's, a few mA should be enough these days. Just re-calculate the above for the recommended opto LED current.

  • 0 in reply to Paul Grohe
    Intellectual 570 points

    Thanks, that was informative.