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I am re-creating a simple low-battery level circuit that turns on a low-power LED to indicate that the battery needs to be changed. The circuit is shown below.

Low Voltage Detection Circuit

There are two things I would like to change on this circuit.

  1. The cutoff voltage is approximately 6.9V. Based on my initial analysis, all I need to do to change this is to swap out the zener diode. The cutoff voltage is the sume of Vz (the zener knee voltage) and the 0.7V drop across the diode, Vd. So, if I want to have the cutoff be sooner during the discharge cycle (ie: 7.5V), my assumption is that I just need to swap out the 6.2V Zener diode with a 6.8V Zener diode. Is this assumption correct? Do any of the resistor values need to change?
  2. I want to remove the green LED from the circuit, and the left-most Bc547 NPN BJT, and just have a "low battery" indicator. How do I reduce the existing circuit to just this?
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  1. You should also change the 10k resistor to limit the maximum current flowing through the zener diode (increase the value if using lower voltage zener). For 6.8 V zener, the current will get smaller so the 10k resistor is ok.
  2. You can replace it with a resistor. 1 k or more should work.
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  1. Your calculations based on the Vz of the zener are a little bit dependant on the zener current. Reverse biased zeners have their specified knee voltage at a specied current, your schematic allows for very small "leakage current", but this also influences the biasing of the zener. This means some experimentation or graph-hunting may be required to find your value.

  2. I would propose a little math to find the best solution to your exact situation with lowest losses:

You want the LED to turn on at about 7.2V. I am going to make the assumption you want it to work in the range of 6V to 7.2V, so we'll use those two as the extremes of our calculation. Further, for ease of the example I am going to assume a hfe of the BC547 of 100. a BC547C type may go up to 500 in the right circumstances.

First off, you mention a low-power LED. This again drives an assumption, as no specifications are made, I will assume the LED current is 2mA and that the red LED adheres to commonalities, needing about 1.9V at that current. Here we go!:

Lowest V+ = 6V, I(led) = 0.002A, V(led) = 1.9V, means: R(led) = (6V - 1.9V) / 0.002A = 2.1k Ohm.

LED current at 7.2V: I(led-7.2) = (7.2V - 1.9V) / 2.1kOhm = 2.5mA

The current into the base of the LED's transistor then becomes 25uA at 7.2V and 20uA at 6V (divided by hfe, which is assumed 100). Which means:

Bias Resistor value, maximum: Rbias = (V+ - Vbase) / I(base) = (7.2V - 0.7V) / 25uA = 260kOhm; or (6V - 0.7V)/20uA = 265kOhm. Meaning 260kOhm is our upper limit. Let's choose 133kOhm to stay well on the safe side:

R(pull-up) = 100kOhm and R(base) = 33kOhm (makes 133kOhm together).

Now, the first transistor is only switched on when the battery is between 9V and 7.2V to keep the LED transistor from turning on. We use that to do the remaining calculations:

I(pull-up) = 9V / 100kOhm = 90uA; or 7.2V / 100kOhm = 72uA. Both values are below 100uA and we want to put at least a couple of uA into the base of the first transistor, so we'll use 1uA as a minimum.

If we want the transistor to turn on at the first sign of a 0.7V difference, we need to assume a differential voltage of 15 to 30mV over the resistor, making:

R(base) = 30mV / 1uA = 30kOhm.

The rest is determined by your choice of Zener and Resistor to determine the initial trip voltage, in this case your transistor set-up will be using about 95uA to 150uA of current when turned off. The rest of the "waste" will be determined by your zener. Once the battery is at the trip level, the zener will start to cut off, so its loss will slowly minimize, but then, of course your LED will be draining 2mA.

For a reliable operation you want Iz not to be too far from its optimum, i.e. not a factor of 100. But there are probably types that can give decent effects with a bias current of 0.5mA.

The schematic becomes:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Is it possible to make it trigger in the entire [0V,7.5] range, rather than just the [6V,7.5V] range? \$\endgroup\$ – Cloud Sep 26 '14 at 22:44
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    \$\begingroup\$ To the extend that the LED will need a voltage to work with, this schematic will leave the LED on the entire time, the range was just taken as a reference for the calculations. Be well aware though, that by increasing the range you are increasing the losses. If you want to do it properly follow my calculations and put in your own lowest voltage, 3V or above in stead of where I used the 6V. But if your device needs a 9V battery, it's very safe to assume at 6V it will not work as designed any more. \$\endgroup\$ – Asmyldof Sep 26 '14 at 23:33
  • \$\begingroup\$ Where I decided for 133kOhm I just divided the maximum resistance by two for a 100% overhead and then split that over a 3:1 between the two resistors. The base-going resistor is quite arbitrary, it's not even absolutely necessary, but provides some safety. If you keep the pull-up as high as possible the losses will be somewhat saved, within the limits of what's possible with your voltage range. \$\endgroup\$ – Asmyldof Sep 26 '14 at 23:39
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  1. Your analysis is approximately correct. If the 6.9V is reasonably accurate, swapping the zener for a similar type of zener with 6.9V Vbr should work. It may not be exact, of course, due to tolerances and other reasons.

  2. Just replace the green LED + 330R with a resistor- something like 10K would be fine. I would also reduce the 47K to 10K.

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