Electronic – How to calculate ripple voltage in this bridge-rectified zener-regulated circuit

A zener is a shunt regulator, which means that it is a load parallel to your actual load.

The zener voltage is 6 V, so you'll need a bit more to allow a voltage drop across the series resistor R. If the input voltage is constant, say 8 V, R will see a constant current if also the 6 V is constant. That's Ohm's Law: V = R \$\times\$ I.

Your load may vary between 500 Ω and 20 kΩ , which is a wide range for the zener, as we'll see. The current through the load will then vary between 300 µA and 12 mA. So we'll have to calculate R for at least that 12 mA plus some for the zener, let's say 10 mA. That's 22 mA, and if we assume the 8 V input then our R will be 90 Ω.

The 10 mA for the zener is needed for proper regulation. If the current is too low variations may cause rather large voltage changes. As you can see at the maximum load aof 500 Ω the regulator is only about 50 % efficient; you have the same current through the zener as through you load. At the minimum load it's really bad: the zener will have taken the 12 mA from the load and have 22 mA total, while the load only has 300 µA. Efficiency is then only 1.3 %! That's a reason why zeners are not recommended for loads varying that much.

Why would the zener take the 12 mA from the load? If it didn't R would only get 10 mA instead of 22 mA and the voltage drop would be 900 mV, for a 7.1 V output. It's the zener voltage which forces the total current to be 22 mA.

So the zener current will change with varying load, but also with varying input voltage, because if the 8 V would become 9 V then the total current would become (9 V - 6 V)/90 Ω = 33 mA. If the load takes 12 mA of that then there goes 21 mA through the zener.

Your zener will keep the voltage around 6 V, but with these changing currents you can't expect wonders: the 6 V won't be constant. A zener is not a good voltage regulator.

A better solution would be a TL431 (Russell! ;-)). This is already much more stable with much better regulation, and costs hardly more than a zener.

But the real solution is of course a series regulator. The 78L06 is a 6 V/ 100 mA part, which is easy to use: an input pin, an output, and a common ground. Add capacitors to both input and output and you're set.

edit after adding requirements to the question
If you can only use a zener you'll have to relax your requirements. 10 mV accuracy is absolutely impossible. That's less than 0.2 %, while the 1N5233 has a 5 % tolerance to start with, and the datasheet gives a zener impedance of 7 Ω maximum. That means that a 12 mA difference in zener current will give a 84 mV difference, that's another 1.4 %.

What you should do depends on how you are driving the NTC thermistor.

If you are sensing temperature using a voltage divider formed by an NTC thermistor and a fixed resistor, then it will definitely work better to use a voltage divider for your switching threshold also. Both dividers should be connected to the same voltage. You can consider using capacitors in parallel with the resistors to improve the bandwidth of the voltage dividers if the voltage has a lot of ripple.

But if you are sensing temperature using an NTC thermistor driven by a constant current source, then you would be better off using a stable reference voltage for your switching threshold.

A hybrid option would be to create a low noise voltage rail (could be 3.3V from an LDO) to drive your NTC thermistor divider and your comparator divider. If the main voltage rail is very noisy, or if you need the utmost in precision this may provide more reliable operation.