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How could this "Battery OK" indicator be improved, without adding too much complexity (or maybe even simplifying)?

Battery Status Circuit

The intended function is to have the LED illuminated at relatively-constant brightness when the supply is above a threshold voltage. As the supply voltage drops below that threshold, the LED should transition as rapidly as possible to completely off. Idle current should be as low as possible.

With the circuit as shown, this is a plot the current through the LED for supply voltage up to 10V:

Voltage Sweep

The R2/R3 voltage divider is set so that when the supply is at the threshold voltage, the voltage across R3 is one VBE drop.

R1 sets the LED current above the threshold. (Q3 serves as a current regulator.) This way there's not a dimming of the LED as voltage decreases from maximum until the threshold is reached.

The Sziklai pair is used so that the transition occurs over a narrower voltage range, and so that Q1 base current doesn't load the divider. A Darlington pair could be used here, but the divider would have to be adjusted so that the lower voltage is 2 × VBE.

Edited to clarify the specifics of the sample circuit:

  1. The 9V source was just chosen because a 9V battery is readily available for testing. I would like something readily adaptable to other voltages.
  2. The 2.5 mA current through the LED is also somewhat arbitrary. Having the brightness be relatively constant is more important than the actual value.
  3. I could replace the JFET and its source resistor with a packaged current regulator diode and it would be essentially equivalent. (I didn't have a model for one of those.)
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  • \$\begingroup\$ Replace the circuit with a humble network of three elements in series - a resistor, Zener diode and LED... You can try it without resistor but be careful... \$\endgroup\$ – Circuit fantasist Feb 9 at 20:54
  • \$\begingroup\$ Do you want a 2.5mA cutoff range of 0.5V or less instead of 1.5V at some threshold like 8.5V or 8.0V? Then specify in Question for a better answer. (BTW programmable Zener is simpler with more gain.) What about temperature sensitivity spec? \$\endgroup\$ – Tony Stewart EE75 Feb 11 at 14:38
  • \$\begingroup\$ Just to clarify: does a component like a tl431 meet you simplicity requirement? It is certainly more "readily adaptable", with a simple trimpot, precise, stable e require less components. \$\endgroup\$ – vangelo Feb 11 at 15:39
  • \$\begingroup\$ @vangelo: The TL431 wouldn't be a solution on its own, but it could replace the fixed Zener diode in something like the circuit in the answer by Michal Podmanický. \$\endgroup\$ – Theodore Feb 11 at 16:00
  • \$\begingroup\$ @Theodore Well, the TL431 could also be used with positive feedback for a sharper voltage transition. Please see below the modification to my answer using a voltage follower with a zener. The component count is increasing... ;) \$\endgroup\$ – vangelo Feb 11 at 16:17
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The simple way is to use zener diode, for 9V battery is 6v8 or 7v5 good choice. When the voltage across R2 is higher than 0.7V, i.e Vbatt is higher than Vzen + 0.7 ,the led is gloving.

enter image description here

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  • \$\begingroup\$ This is simpler, but the LED will dim more gradually. \$\endgroup\$ – Theodore Feb 9 at 22:52
  • \$\begingroup\$ Than you go to opamp comparator or 2 transistor differential amplifier, you will get a "sharp" knee. \$\endgroup\$ – Michal Podmanický Feb 9 at 22:57
  • \$\begingroup\$ in what way is this "better" than the original? \$\endgroup\$ – Jasen Feb 11 at 10:45
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Add a resistor south of Q2 (in series with the collector) connect the south end of R3 to Q2 collector instead of ground.

That will give hysterisis boosting the switching speed of the led. try 22 ohms to start.

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  • \$\begingroup\$ I don't think this works since gain is < 1 for Rc/Re of Q2 \$\endgroup\$ – Tony Stewart EE75 Feb 10 at 3:58
  • \$\begingroup\$ I don't think so, it's a Sziklai pair, so the effective collector has a current source and the effective emitter is grounded, (or has the 22 ohms - except it doesn't because Q1 emitter current bypasses it) does that make the gain infinity divided by zero? Talking about voltage gain from Q2 is counterproductive, there is no feedback setting the gain of Q1 or Q2. I just want 50mV of feedback to give this "V_be comparator" some hysterisis. \$\endgroup\$ – Jasen Feb 11 at 10:30
  • \$\begingroup\$ You are describing adding a bootstrap circuit to a unity gain emitter that will not add hysteresis voltage gain to the Battery threshold. It will make it worse as the voltage gain is reduced Rbe effectively being increased by bootstrap feedback. thus reducing voltage gain slightly and not adding hysteresis. The gain however will be increased by drawing more Ic to reduce Rbe, but not a solution \$\endgroup\$ – Tony Stewart EE75 Feb 11 at 12:59
  • \$\begingroup\$ Q2 collector is decoupled from Q1 emitter. it's not a follower because Q1 emitter is not raised above ground \$\endgroup\$ – Jasen Feb 11 at 22:17
  • \$\begingroup\$ Agree with Jasen, this is the most simple mod. Q1 is acting as a comparator (think of its Vbe as a reference diode), not a linear amplifier. The resistor added to the Q2 collector is part of the voltage divider that sets the trip point, and the Q2 signal polarity is in phase with the Q1 base. That is positive feedback. To eliminate one transistor, I would go with the current source in another answer; but you still get more "snappy" switching when you add this technique. \$\endgroup\$ – AnalogKid Mar 4 at 15:50
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I agree with Jasen. The net change to the circuit is the addition of one resistor, so this is the most simple mod. It is a variation of a standard 2-transistor Schmitt Trigger circuit. Q1 is acting as a comparator (think of its Vbe as a reference diode), not a linear amplifier. The resistor added to the Q2 collector is part of the voltage divider that sets the trip point, and the Q2 signal polarity is in phase with the Q1 base. That is positive feedback, and gives a more "snappy" response to the changing input.

I'm not great at Spice, but here are before and after sims that show a much more rapidly changing LED current. Note that both trip points move when hysteresis is added.

Before: R4 and R5 are not in the circuit. enter image description here

After: R2 is moved and adjusted to become R4, and R5 is added. enter image description here

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Could a circuit with positive feedback be an initial idea?

enter image description here

It borrows your idea of using the 68k resistor to polarize the BE junction, but the feedback causes an abrupt increase in the current:

enter image description here

It is not perfect, since the brightness will be reduced gradually before the LED abruptly turns off. If you adjust the hysteresis and the max. LED current maybe it fits your needs. Limiting the LED current (more bjts or mosfets) or voltage (zener and resistor) would also reduce the current variation.

Edit: I'm keeping the above circuit for now and uploading one of the modifications I proposed:

enter image description here enter image description here

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  • \$\begingroup\$ one way to explain the problem with this proposal is the Darlington gain drops rapidly as diode conducts towards 10 Ohms making the gain 10/47 instead of high gain as a comparator should be so the sensitivity about 0.3mA per V of Vbat. Then the emitter R conducts more current than LED when Vbat is OK so inverted logic for indicator \$\endgroup\$ – Tony Stewart EE75 Feb 11 at 14:01
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Try this, a 2-transistor constant-current sink that is driven through a zener diode as the detection threshold. Here is an innergoogle schematic grab that is close:

http://lednique.com/power-supplies/simple-constant-current-driver/

enter image description here

To this circuit, add a zener diode in series with R1, and connect its anode to Vbb. Not counting the LED, 5 components instead of 6, 2 transistors instead of 3.

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Answering my own question with a specific-purpose IC solution that's not particularly clever, and a little more costly:

Comparator with reference IC, MAX931

The comparator output can provide enough current for an LED indicator (with resistor).

Adding one more resistor at Pin 5 will add some hysteresis so that the light doesn't blink due to slight load fluctuations when the battery is near the threshold.

Editing my answer to note that @MathKeepsMeBusy gave an answer to Undervoltage lockout that could also do the job.

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  • \$\begingroup\$ It does look simpler than the tl431 for use with positive feedback. \$\endgroup\$ – vangelo Mar 4 at 18:02

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