# Zener Diode Parallel With Capacitor Problem

#### decep370

Joined Nov 22, 2008
3
Good day to all,

I have a DC starter having 300 VDC as input supply and a 1.1 Kohm resistor connected in series to it. After the 1.1 Kohm resistor are 04 x time relays connected in parallel with a 24 VDC Zener diode to regulate the voltage. There is also a 470 uF 40V Capacitor connected in parallel with the Zener diode. When the whole circuit is power up, the time relays are getting 24 VDC and then raising to 64 VDC instead of a constant 24 VDC.

Is the 40V capacitor suppose to charge up the time relays at the same time when the zener diode is regulating the voltage at 24 VDC?
Can the capacitor be taken out so that the voltage to the time relays be maintain at 24 VDC?

Attached is the circuit diagram and very much thanks for the help

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#### AnalogKid

Joined Aug 1, 2013
9,639
What is the time relay part number and manufacturer?
What is the current for each relay?
How loing does the voltage sit at 24 V before rising to 64 V?
What is the part number for the zener diode?
What is the power rating for the resistor?

ak

#### WBahn

Joined Mar 31, 2012
26,398
What is the 1.1 kohm resistor rated at? You are asking it to handle about 70 W. If it can't do that, then it is probably heating and becoming something other than a 1.1 kohm resistor.

#### decep370

Joined Nov 22, 2008
3
Thanks for the replies, the time relays are ABB CT ERS.21.
When the DC starter is switched on, I measured the voltage across the 1.1 Kohm resistor is about 275 VDC. This means that the time relays should only have 24 VDC. However, the voltage across the relays shoot up from 24 VDC to 64 VDC in about 2 secs. Once the system is cut off, the voltage across the time relays takes about 5 secs to go down from 64 VDC to 0 VDC. This is not suppose to be happening as it will cause the time relays to keep activating although power supply is cut off. Thinking of removing the capacitor from the circuit but no understand why the voltage can go to 64 VDC when there is a 24 VDC zener diode in parallel.
Thanks.

#### WBahn

Joined Mar 31, 2012
26,398
Again, what is the power rating of that 1.1 kΩ series resistor?

What is the part number for the Zener? If you are really getting 64 V across it, then it would be dissipating some serious power.

If it is taking 5 s for the cap to bleed down, then the time constant is probably about 1 s or so. That would correspond to an initial discharge current of CV/T = 30 mA, which is a fraction of what your relays should be pulling (though it is well within an order of magnitude). How fast do you need the relays to shut off?

#### decep370

Joined Nov 22, 2008
3
Sooty for the late reply as I need to go back office to get the info.
The Zener diode is 1N2986B and the resistor is able to handle 120W. The time relays have to cut off by one sec or the alarm relay will activate the alarm. Thanks

#### WBahn

Joined Mar 31, 2012
26,398
If you don't need the capacitor, then ditch it. Otherwise either reduce it by quite a bit (say to 100 uF) or put a suitable bleed resistor in parallel with it.

How good is the heat sinking on the diode? It is spec'ed to have 12 °C/W from case to junction, so with 6 W of power being dissipated you are looking at 72 °C of rise from whatever your case temperature is being held to. That means that the case temperature can't exceed about 100 °C, which means your heat sink from ambient to the case is going to need to be less than about 10 °C/W to give you any kind of margin. Ideally you will only be dissipating about half that power in the Zener, but it's a good idea to allow for the case if all of the current end up going through the Zener.

But that wouldn't explain the kind of voltage you are seeing. It might explain 26 V give or take, but not 64 V.

Your relays nominally pull about 24 mA. So try replacing them with a resistor (or resistors) that will pull about that much current (x4) and see what your Zener voltage does.

Since the Zener has a positive tempco, you could put multiple ones in parallel and see what effect that has (but I wouldn't hold my breath that it will solve your problem).