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I am trying to create a project based on TI chip BQ78412 (Pb-Acid Battery State-of-Charge Indicator With Run-Time Display). There is an evaluation board whose schematic is published in this pdf on page 18.

I am struggling to understand why part of this schematic works (see below the one in question).

My main problem is that the VCC is 3.3V (and all signals) but the LEDs are switched on/off by the transistor through 12 V from a separate battery that has a common ground with the rest of the schematic. I assume that when I have logical 0 on the output pins of the IC the LED will be on as it will be connected to ground. But if I have logical 1, it would be still 3.3V so I will have effective voltage of 8,7V that would be sufficient for the LEDs to be on - the current would be about 66 mA! So I guess I am missing something very basic and important...

I also don't get why the current limiting resistors for the LEDs are 100 ohm - the current should be 20 mA (limited by the IC) and the forward voltage should be 2,1V if I am not wrong (see datasheet for SML-LX0603YW-TR) so according to Mr. Ohm it should be around 495 Ohm.

enter image description here

Regards,

Evgeni

EDIT: In my question I have made some false assumptions due to lack of knowledge and experience:

  1. I didn't recognize that the transistor is acting as emitter follower which was a very important fact.
  2. I have also assumed that the driving current of the serial shift register is max 20 mA rather than 5 mA - I am still not sure how to interpret the data sheet - I'l ask a separate question on this.
  3. I have assumed that 1-3 mA would be quite insufficient for the LED to emit light as it's forward current according to the data sheet is 20 mA.
  4. I have also underestimated the HC164 Vlow assuming that it is always 0 V which apparently is not true either.
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2 Answers 2

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What you may be missing is that Q1 is acting like a voltage regulator. It's actually an emitter-follower and the emitter voltage is approximately whatever voltage is on DS1EN less about 0.7V. If DS1EN is a 5V signal, the emitter is sitting at about 4.3V. If DS1EN is a 3.3V signal, the emitter is sitting at about 2.6V.

You don't say what color the LEDs are. Red LEDs drop about 1.7V, old-school green & yellow LEDs drop about 2.1V, modern green, blue, white LEDs drop about 3.2V. I don't remember what the forward voltage of a modern yellow LED is - sorry.

You can also assume that the output LO voltage from the 74hc164 is anywhere from 0.1V through about 0.5V if the LED current is less than 10 mA.

So: the LED current is going to be about: (voltage @ Q1e) - (Vled) - (hc164 Vlow) all divided by 100 Ohms. You need to fill in the missing information.

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  • \$\begingroup\$ From the data sheets, the typical voltage drop for all 3 LED colors is in the range 2.0 (for red) to 2.2 (for yellow). \$\endgroup\$
    – WhatRoughBeast
    Commented Mar 24, 2015 at 23:52
  • \$\begingroup\$ You are right, I missed the point that Q1 is emitter-follower. Are you sure about the LED current calculation - if I assume Voltage@Q1e = 2,6V, Vled=2,1V and HC164Vlow=0,5V then the formula results 0 mA... \$\endgroup\$
    – Evgeni Nikolov
    Commented Mar 25, 2015 at 0:15
  • \$\begingroup\$ The output stage of the hc164 looks like a low-value resistor to whatever rail it is supposed to be (gnd if o/p is LO). In other words, the output pin of the hc164 starts off at zero volts but then rises above ground as the current increases. \$\endgroup\$
    – Dwayne Reid
    Commented Mar 25, 2015 at 1:03
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When the display is enabled, DISEN is held at about 3.3 volts. Q1 then acts as a regulator with the emitter at ~2.6 volts. With the logic outputs held at ~0.2 volts for an active low, and the LEDs dropping ~ 2.1 volts, this leaves about 0.3 volts across the resistors. 0.3 / 100 ohms is 3 mA.

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  • \$\begingroup\$ the datasheet says "steady current 30 mA". Would then 3 mA be sufficient? \$\endgroup\$
    – Evgeni Nikolov
    Commented Mar 25, 2015 at 0:19
  • \$\begingroup\$ The Maximum Recommended LED current may be 30 mA, but LEDs will emit useful light at much lower currrent - I'm using some green LEDs at a bit less than 1 mA, and they are still very visible. \$\endgroup\$
    – Peter Bennett
    Commented Mar 25, 2015 at 0:37
  • \$\begingroup\$ @EvgeniNikolov - As Peter Bennett says, you do get useful output at much lower current. And particularly on a board like this, where you're not trying to be seen at long distance, it doesn't take much light to show what's going on. \$\endgroup\$
    – WhatRoughBeast
    Commented Mar 25, 2015 at 4:22
  • \$\begingroup\$ @ WhatRoughBeast - Thank you for the information! My problem is that the module that I would like to design should be used outside under sunny conditions. It seems I would need maybe to experiment a little bit and in worst case think of driving LEDs else to get better visibility. \$\endgroup\$
    – Evgeni Nikolov
    Commented Mar 25, 2015 at 13:40

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