[moved] Why Zener getting heat at no load in Capacitor power supply arrangement..

Hi,
With this PS, trying to operate 2 relays(5v, spdt, rated coil current 80mA for each) with pic mcu.
It is noticed that at no load(when the relays not being operated) the zener(5.1v, 1.3watt) is geeing heat.
And if the relays are kept turned on the zener is ok.
Can you suggest what should i do.
The ckt of PS is attced here..
tnx

Re: Why Zener getting heat at no load in Capacitor power supply arrangement..

Hi,

Because the zener is in shunt configuration.
The less current the load draws the higher the current through the zener.

Klaus

after putting series Resistsnce (Rs) in series with zener and o/p is taken out across zener, the same thing is happening (ie, at no load the zener as well as Rs GETTING HEATED.
AS PER MY CALCULATION,
1) Max current through zener= bener wattage/zener volt=1.3/5.1 = 0.255amp,
2)Min Value of Series resistor, Rs= (12-5.1)/0.255 = 27ohm, and wattage = 27x0.255x0.255 watt= 1.75watt
so i have used to check, Rs with 18, 47 ohm of 2watt and 50 of 5watt. but no improvement found to minimise the heat... both zener and Rs GETTING HEATED..
tnx

Your calculation and concept of power are wrong.

The Zener is in shunt configuration - it 'soaks up' the excess current that isn't flowing into the load and in doing so it dissipates heat. The less load you place on it, the more it has to dissipate and the hotter it gets.

1) Max current through zener= bener wattage/zener volt=1.3/5.1 = 0.255amp,

Click to expand...

Gives you the amount of current that would make the diode dissipate 1.3W, that isn't what you want. The intention is only to ensure the total load current maintains 5.1V after dropping through the input capacitors.

Calculate it like this:
1. Find your total maximum load current, that is the LED, relays and anything else added together and activated.
2. Ignoring the voltage drop in the bridge rectifier, work out the RESISTANCE you would need to drop the incoming AC to 5.1V at the current from step 1. (230 - 5.1)/I
3. Work out the capacitor value with equivalent reactance, this will be your input capacitance. Xc = 1/(2 * pi* f * C)

If you remove the Zener altogether that would then give you 5.1V at the output under full load.
The problem is the voltage will rise if the load current drops, that is where the Zener comes into play.

4. Find the total minimum load current, that is with the circuit working but in a state where it's current is as low as it can be. Relays not energized etc.
5. Subtract result 4 from 1 to get the difference between maximum and minimum.
6. Choose a 5.1V Zener rated to carry the current from step 5.

For example, if your circuit draws 100mA maximum and 40mA minimum, the Zener would need to pass 60mA which means a rating of at least (5.1 * 0.06) = 0.3W

As your components are getting too hot, it suggests the input capacitors are too high in value and letting more current than necessary flow through them. Note that this kind of simple circuit has NO safety isolation and must, including whatever it supplies, be completely insulated. Do not connect the output to any other equipment. The input capacitors should be rated for AC use and the 1M resistor across them should be rated for high voltage as well.

Where is Rs in your schematic? If it is the resistor in series with the LED, the value should be (5.1V - Vf)/Iled. For example for a standard red LED with Vf = 1.6V and I=10mA it would be (5.1 - 1.6)/0.01 = 350 Ohms.


Brian.

STILL_Zener getting heat at no load in Capacitor power supply arrangement..

Tnx for reply..
I have calculated as u suggested #4. And calculatiob sheet and schematics are also here.
But still ZENER(5.1v/1.3 w) getting hotter with calculated Film cap(~2.5ufd) is used.
If 2.5ufd IS REPLACED WITH 1.0 ufd notice that zener heat is less than 2.5ufd used.


tnx

Would u Pl suggest what i am missing here.
Actually I want to run with this supply 12f675/ 2 nos relay(80ma coil current for each)/1 led(10ma)
....

You have calculated about 1 W zener diode power dissipation with 2.5 uF capacitor. The zener can be expected to get quite hot, what are you wondering or complaining about?

If 2.5ufd IS REPLACED WITH 1.0 ufd notice that zener heat is less than 2.5ufd used.

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What else?

There should be a minimal series resistor (50 to 100 ohm) for the capacitor + rectifier to limit surge currents when connecting the circuit to the mains.

I agree, the calculations look correct and it works as expected. You will find that if the LED and relays are operated, the Zener goes cold, that is how a shunt regulator is supposed to work.
Note that shunt regulators are not energy efficient, by design they always consume the maximum power your load needs, even if it isn't needing the power at the time.

Brian.

If you have read the zener diode's datasheet carefully, which I suspect you have not, a zener's maximum power rating can only be achieved with a particular thermal configuration.

For instance; the Micro Commercial Components 1N4733 indicates that the power dissipation can only be met with a thermal resistance from the junction all the way to ambient of less than 100K/W.

#7,
Yes, zener gets cold if total load is operated.

I am getting confused, let me clear first___________________
As I know here Zener is used to maintain constant voltage 5.1v - at NO LOAD and LOAD condition.
In my case,
At LOAD condition, maintained 5.1 and NOT GET HEATED----------OK.
But at NO LOAD condition,
(1) if zener NOT used with the ckt ---voltage gets rise-------- OK.
(2) WHILE Zener used with ckt --- voltage kept at 5.1, but available current(allowed to pass thru film CAP while calculated), in my case 200ma will flow thru zener.
And this 200ma will bear with ZENER without getting HOTTER & that is why 1-watt zener has choosen to operate the condition.
As (1watt/5.1)=196ma=~200ma.

Is it not correct concept??
If My question was, how to operate those load CURRENT for 2-relay ( say 200ma total) with min current 10ma and max 200ma without ZENER HOT

So what would be the best way to operate such CURRENT with film cap. and how much max current can be handle with film cap arrangement..
thx both of u..

Did you understand my previous message? You require proper thermal management of the diode to properly dissipate the heat.

For thru hole diodes, this means keeping the leads to the board as short as possible and have a pair of copper planes on the board (one on each lead) which will act as a heatsink.

If My question was, how to operate those load CURRENT for 2-relay ( say 200ma total) with min current 10ma and max 200ma without ZENER HOT

Click to expand...

Short answer, you can't. The Zener will always dissipate heat, the best you can do is as Schmitt Trigger states, its to remove the heat as efficiently as possible.

The whole idea of that kind of power supply is the voltage dropping is done in the reactance of the input capacitors where in theory, no heat is dissipated. Essentially, the capacitors are the top half of a potential divider and the load is the bottom half. It is expected therefore that if the load current varies, the output voltage will vary too, in fact with no load it would rise to around 320V! To maintain a constant voltage, the current through the capacitors has to be constant and the Zener is there to ensure that. Whatever current DOESN'T flow into the load is passed by the Zener instead, as one increases, the other decreases and vice versa.

If you really want low heating you have to look at different kinds of power supply, in particular SMPS but the complexity increases quite a lot. An SMPS also offers you an isolated output which is much safer than a transformerless design.

Brian.

Hi,

In my eyes the "low heat" solution is to use a series voltage regulator instead if the shunt (zener) solution.

Maybe the most popular: 7805 type voltage regulators.

Klaus

because a cap dropper psu circuit is a current source - if you remove the load ALL the available current goes thru the zener - heating it - you have to design for this - unless the load will ALWAYS be there ...

Vmains / Xc = current available, + 20% to allow for a safety margin, this current x 5v1 = watts at no load, so 100mA gives 510mW - use a 2 watt zener with big pads for head dissipation ...

#13,

so 100mA gives 510mW - use a 2 watt zener with big pads for head dissipation ...

Click to expand...

This is too much margin, Why this 4 times more to choose zener, is there specific reason behind that??

https://www.edaboard.com/showthread.php?386831-Supply-5V-5mA-from-the-240VAC-mains-without-SMPS
the above is also about cap droppers...there is a simulation there that you can adopt for your use. Its in the free ltspice.

With a cap dropper you just have to decrease your series capacitor until you can just about supply the max load required...then your dissipation will be minimised in the no load condition.

So get it on the simulator and iteratively reduce the capacitor till you can no longer supply the load at max load and minimum vin.
Also, as said b4, remember to add a resistor for inrush limiting