Parallel zeners?

[MrAureliusR]

The way zeners function still throws my brain for a loop sometimes. Over the Xmas break, I visited my dad who had a coffee maker on the blink. After some tracing, I found that two 12V zeners in series had both failed short (!). After replacing these zeners (actually, replacing each with 2 6.2V zeners in series) it works fine. But during the reverse engineering process, I noticed that these zeners were essentially in parallel with a 5.1V zener. I was really confused, because I figured the 5.1V zener would just clamp the voltage and the (essentially 24V) zener would pass essentially no current.
 
Here's the schematic. The header on the right is coming from the microcontroller board. The 5.1V zener is just creating a supply for the micro.



All I can think of is that the 12V zeners are there to clamp the voltage in case the 5.1V zener ever failed open? The other thing that confused me was the way the motor and heaters were connected across the relay (and I triple checked to see if it was right). It's returning the current through the same resistor that the 5.1V zener uses? It seems strange, though I'm guessing there's a good reason for it. (EDIT: forgot to mention that I'm pretty sure the heating elements were either in parallel with the motor or connected separately, so ignore R4 and R5)

Any and all commentary is appreciated!

[bktemp]

The two stage zener diode circuit is fairly common:
The capacitive dropper C2 acts as a current source. The 24V zener diodes limits the voltage after the rectifier D1/D2 to 24V generating 24V for driving the relay.
The 5.1V zener diodes generates +5V for powering a microcontroller or some other control logic. It has the series resistor R3, therefore it doesn't affect the 24V rail.
The motor return is probably wrong, because it can't work that way.
R6 could be used for sensing if the motor is turned on or maybe for sensing zero crossings.

Typically there is a series resistor for the capacitive dropper circuit, because if you turn the device on when the mains voltage is at peak of the sinewave, a very high inrush current flows for a short time. In your case C3 has probably a high ESR, therefore it couldn't handle the high peak current. So the zener diodes had to conduct the current and failed. I would check C3 if its ESR is <1ohms or just replace it.

[MrAureliusR]

The zener diode circuit is fairly common, but in this case it is a bad design:
The capacitive dropper C2 acts as a current source. The 24V zener diodes limits the voltage to 24V generating 24V for driving the relay.

Right. Because the micro is only seeing the voltage difference directly across the zener, it only sees 5V. I still don't see how it doesn't clamp the main rail down as well. I thought to have a zener work properly it needs to pass a decent amount of current?

The 5.1V zener diodes generates +5V for powering a microcontroller or some other control logic. It has the series resistor R3.
The motor return is probably wrong, because it can't work the way you have drawn.
R6 could be used for sensing if the motor is turned on or maybe for sensing zero crossings.
Typically there is a series resistor for the capacitive dropper circuit, because if you turn the device on when the mains voltage is at peak of the sinewave, a very high inrush current flows for a short time. In your case C3 has probably a high ESR, therefore it couldn't handle the high peak current. So the zener diodes had to conduct the current and failed. I would check C3 if its ESR is <1ohms or just replace it.

I can't work on the coffee maker anymore. And I understand capactive dropper circuits, I just hadn't seen zeners used like this before.

I'm telling you -- the circuit is exactly as shown. I even have a picture of the backside of the PCB with the components annotated if you'd like to see it.
Here it is:


The motor connections are the very bottom left large pad, and the very large bottom right pad. You can see the only path for the current to flow is back through that same resistor (it's not going through the tiny wires over to the microcontroller board.)
Just want to say, this isn't my first reverse engineering rodeo, so to speak.

[MrAureliusR]

Now that I'm looking at it again, it's possible that I missed another wire going from the motor directly back to the AC connections? That's the only thing I can think of.

[sleemanj]

Right. Because the micro is only seeing the voltage difference directly across the zener, it only sees 5V. I still don't see how it doesn't clamp the main rail down as well. I thought to have a zener work properly it needs to pass a decent amount of current?

Ohm's law tells us that if current is passed through a resistor, it drops a voltage, or conversely if voltage is across a resistor it passes current.

The 5v1 zener is not across the 24v clamped rails,  the combination of a resistor and the 5v1 zener is. 

Start with 24v, subtract the 5v1 and the rest is by necessity across that resistor.

[bktemp]

The zener diode circuit is fairly common, but in this case it is a bad design:
The capacitive dropper C2 acts as a current source. The 24V zener diodes limits the voltage to 24V generating 24V for driving the relay.

Right. Because the micro is only seeing the voltage difference directly across the zener, it only sees 5V. I still don't see how it doesn't clamp the main rail down as well. I thought to have a zener work properly it needs to pass a decent amount of current?

R3 has 1.5k. (24V-5.1V)/1.5k=12.6mA. Therefore C2 has to supply more current to power both the relay and the 5V supply.
15nF for C2 seems to be wrong, because this gives an output current of less than 1mA. Also based on the size of the capacitor it is probably more like 470nF or 1uF or so. 1uF gives about 18mA (assuming 110V @ 60Hz input voltage).

[MrAureliusR]

The zener diode circuit is fairly common, but in this case it is a bad design:
The capacitive dropper C2 acts as a current source. The 24V zener diodes limits the voltage to 24V generating 24V for driving the relay.

Right. Because the micro is only seeing the voltage difference directly across the zener, it only sees 5V. I still don't see how it doesn't clamp the main rail down as well. I thought to have a zener work properly it needs to pass a decent amount of current?

R3 has 1.5k. (24V-5.1V)/1.5k=12.6mA. Therefore C2 has to supply more current to power both the relay and the 5V supply.
15nF for C2 seems to be wrong, because this gives an output current of less than 1mA. Also based on the size of the capacitor it is probably more like 470nF or 1.5uF or so. 1.5uF gives about 25mA (assuming 110V @ 60Hz input voltage). That looks more plausible.

It's quite possible I read that wrong. I always find the convention of the 3-digit numbers for capacitors confusing. Resistors? No problem. But I always forget if you start at picofarads or nanofarads. For example, is 471 470pF or 470nF?

The cap probably said 153, and I started from pF.

[MrAureliusR]

The zener diode circuit is fairly common, but in this case it is a bad design:
The capacitive dropper C2 acts as a current source. The 24V zener diodes limits the voltage to 24V generating 24V for driving the relay.

Right. Because the micro is only seeing the voltage difference directly across the zener, it only sees 5V. I still don't see how it doesn't clamp the main rail down as well. I thought to have a zener work properly it needs to pass a decent amount of current?

R3 has 1.5k. (24V-5.1V)/1.5k=12.6mA. Therefore C2 has to supply more current to power both the relay and the 5V supply.
15nF for C2 seems to be wrong, because this gives an output current of less than 1mA. Also based on the size of the capacitor it is probably more like 470nF or 1uF or so. 1uF gives about 18mA (assuming 110V @ 60Hz input voltage).

Bang on. And to think I only did Thevenin and Norton last semester 

[T3sl4co1l]

BTW, the circuit most likely failed from a surge condition.  You shouldn't leave this crappy product plugged in 24/7 if you want long lifetime; alternately, add a resistor in series with C2, maybe 100 ohms 1W.

Tim