Zener behaving strangely

Question

I have recently put together a trivial low-voltage interlock circuit to prevent over-discharge of a 9v battery pack. I use a Potential divider of 1600 / 5600 ohm to produce 4.7v when Vcc is 7.2. I have a reverse biased 4.7V Zener going from the middle of the Potential Divider to the base of an NPN transistor (2n222). The emitter is connected to gnd and the collector is the gnd for a load. Vcc to collector gives me the + and - for my load.

(R3/R5 is the potential divider. R4 is the nominal load. U1 is 1N4732 Zener)

The problem is that emulation (on TinkerCAD) says that Vcc below 7.2v reduces current to the load down to about 1mA. But breadboarding it produces real world behaviour where the load is able to draw 10+mA down to Vcc=4.5

Somehow this 4.7v Zener is not blocking current even when the voltage it is opposing is only 3v.

Presumably I am missing some fundamental point - so if someone can explain to me why a reverse biased Zener triggering an NPN transistor is not a viable low voltage cutoff I would be most grateful. (I realise that it will only kill the majority of the load and the Potential Divider itself will continue to discharge the battery - but at a tiny fraction of the rate.

Answer 1

Your design is easier to read when drawn like this:

^{simulate this circuit – Schematic created using CircuitLab}

There are a few problems. R4 is so small, that collector current will be huge if this transistor were switched on. To switch it on would require tens or hundreds of milliamps of base current, which must come via R3, and R3 is too large to accommodate that.

R5 is only required to switch the transistor off, and can be significantly larger than 5.6kΩ. Additionally, there is a current path via R3 and R5, directly across the battery. That path is responsible for a milliamp or so of battery current, regardless of the conduction state of D1. To remove that path, it is sufficient to move the diode:

^{simulate this circuit}

Now current only flows via R3 when the diode begins to conduct, which happens when battery voltage approaches 6.8V. Until then it's only microamps, due to D1 leakage.

This is battery current, as voltage rises:

That design will work up to 500mA or so, but your load (10mΩ) will try to draw many amperes of current, and base current will have to be a significant portion of that, perhaps one-hundredth, depending on transistor \$\beta\$. That will require a prohibitively low value for R3. To overcome this, it might be better to employ a second transistor:

^{simulate this circuit}

I've reduced load resistance R4, which would draw 4A with 8V across it, requiring a beefier Q1. These two cascaded transistor stages have a combined current gain in the thousands, requiring much less base current for Q2, and R3 can be kilohms.

Battery current still drops to near zero, and transition of Q1 from off to on is quicker now: