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I am trying to determine how much current this circuit will draw under rest conditions when the Ign- is at roughly 12.5V. Would current run through the blue path? Sourcing from Ign- passing through 330 and 1k ohm resistors and optocoupler to ground?

If this is true it seems that the circuit would draw 12.5/(1K+330) + 12.5/(10k+12K) = 9.4mA+1mA = 10.4mA. Is this a correct assumption? optocoupler If so is there a way to limit this? Part number for zener is BZY55B5V1 RBG zener

tach circuit

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    \$\begingroup\$ Consider the D204 Zener voltage - the Zener will likely draw some current from the R213/R214 junction. \$\endgroup\$
    – Peter Bennett
    Commented Mar 30, 2022 at 1:57
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    \$\begingroup\$ What is the part number for D204? \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Mar 30, 2022 at 1:58
  • \$\begingroup\$ @SpehroPefhany I will add this to the question real quick \$\endgroup\$
    – Feynman137
    Commented Mar 30, 2022 at 2:02

1 Answer 1

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It will approximately be the greater of:

12.2/12K + 11/1.33K (D204 not conducting)

and

12.2/12K + (12.2-Vz)/1K

Assumptions: Vf for Schottky 0.3V, Vf for IR LED 1.2V, actual zener voltage Vz.

If the zener type is less than about 5.1V, it may be necessary to do a bit more work to find the actual zener reverse voltage (and that current will account for a majority of the total).

Edit:

Given the correct part number for the zener, the first equation is correct, so:

Iq = 1mA + 8.3mA = 9.3mA. The current through the LED will be ~8.3mA (D204 does not conduct significantly, the voltage across it is 3.9V).

The L1 bin optocoupler is guaranteed to work with 1.6mA, but you should allow perhaps 3:1 more for reliable operation over a wide temperature range and for aging, so using that safety factor you could reduce the current to 5mA-ish.

It's not clear what benefit the optocoupler is providing since the grounds are common and you are feeding a divided signal directly into the microcontroller on another pin.

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  • \$\begingroup\$ This has been quite insightful and too much power drawn at rest condition for my application. What if I added a PNP transistor between D203 and R213? Such that I could allow the high current portion to operate only when necessary? \$\endgroup\$
    – Feynman137
    Commented Mar 30, 2022 at 2:22
  • \$\begingroup\$ Also won't the isolated 3.3V circuit be consuming 3.3V/330 = 1mA \$\endgroup\$
    – Feynman137
    Commented Mar 30, 2022 at 2:24
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    \$\begingroup\$ You could add a MOS SSR after D203. Anything much cheaper requires more knowledge. Yes, that is just for the IGN current draw. The other side of the circuit (if powered) will draw current from the 3.3V bus. Depending on the regulator design that may be some smaller current from the 12V (presumably) source, or a bit higher. The H11 itself draws as much as 5mA (1.6mA typical) and the output would be another 10mA (3.3V/0.33K), so as much as 15mA, but typically closer to 12mA. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Mar 30, 2022 at 2:29

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