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Weird relay problem..


Guest Kasamiko
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Guest Kasamiko

I just finished working with my simple Power-On Delay circuit to protect appliances with compressor in case of sudden power loose and resume..
It is a simple 555 monostable circuit that drive a relay after 5 minutes delay..
Now my problem is that everytime I attached this device to my freezer it fuction as expected but will triple the time to cool my freezer!! ::) ::)
First I suspected that I have a dirty relay contacts and my freezer can't get the voltage it requires..but monitoring the voltage while the freezer motor is running it's all normal.. ???
Did I missed anything?? ??? ??? ???

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Kasamiko,

If you have the freezer motor running normally there should not be a problem with the time to cool your freezer? Are you sure it takes longer time than normal? There must be something else wrong with your freezer. One thing comes to mind: does this thing delay for 5 min every time the thermostat is about to switch on the freezer?

Ante ::)


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Guest Kasamiko

If I'll plug my freezer directly to the AC outlet, I'll got ice in a matter of 2 to 3 hours..and if I'll used this device It can't produce ice the whole day!! ???
One thing I noticed that if I'm not using this device everytime the motors run there is a rumbling noise and a little vibration but if I used it motor runs smooth and no noise..

"does this thing delay for 5 min every time the thermostat is about to switch on the freezer? "
NO, it delays the power, I mean ALL the power going IN the freezer ONLY during power failure to protect the compressor..

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Kasamiko,
1) Maybe something is happening to the "freon" during the delay.
The motor works hard, but normally, without a delay. But it runs smooth and quiet (and doesn't do its job) after the delay.
Perhaps the "freon" developes a gas bubble because it warms-up during the delay, at a location where it should be liquid, so that the compressor doesn't have any liquid to pump.
2) Maybe condensation forms on the cooling coil during the delay, which quickly freezes when it starts running, blocking its air-flow.
3) Try a 10 or 15 minute delay.

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Guest Kasamiko

I simulated the 5 minutes delay by unplugging it for 5 minutes and plugging it again directly to AC mains without problem at all. I was thinking of a corroded relay contact so it can't deliver the normal voltage to the freezer.. I'm planning to use a solid-state relay..We'll see the result.. ;) ;)

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Kasamiko,
Doesn't your voltage monitoring prove that the relay is OK?


First I suspected that I have a dirty relay contacts and my freezer can't get the voltage it requires..but monitoring the voltage while the freezer motor is running it's all normal.

Maybe the mains voltage is low before or after a power failure which causes a "freon" bubble with a delay.
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Kasamiko,

If it works fine when pluged directly to the wall you must have some resistance in the circuit. Have you measured the voltage before and after your relay? Does it maybe have burned contacts or is it in some way not suitable for this job?

Ante ::)

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Guest Kasamiko

Kasamiko,

If it works fine when pluged directly to the wall you must have some resistance in the circuit. Have you measured the voltage before and after your relay? Does it maybe have burned contacts or is it in some way not suitable for this job?

Ante ::)


I've changed the relay 3 times!! ::) ::) 35Amp. power relay..
That makes me say "WEIRD"
Solid-state relay is my last bet.. ;D ;D
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Guest Kasamiko

Kasamiko,
Please post a schematic of your monostable, maybe it is re-triggering (cutting power) when it should be driving the relay continuously.
If you power the coil of the relay from a battery as a test then the relay contacts will be proven.

I will try it..and I'll post the schematic shortly..
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Guest Kasamiko

@MP
Can you recommend a better way to try the relay?? I'd tried driving it directly by the 555 but there are instances that the chip can't trigger it. The relay coil is about 180R.

@Ante
I haven't tried it out..Buty based on the schematic I posted, I'll remove the transistor and short the Collector to Emitter, then I'll use the power from the device itself..

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Info on false triggering:

Copied from

http://home.cogeco.ca/~rpaisley4/LM555.html

"RESET" And "CONTROL" Input Terminal Notes
Most of the circuits at this web site that use the LM555 and LM556 timer chips do not show any connections for the "RESET" and "CONTROL" inputs for these devices. This was done in order to keep the schematics as simple as possible.

When the "RESET" terminal is not going to be used it is normal practice to connect this input to the supply voltage. This is especially true when the CMOS version of these timers is used as the inputs of these devices are very sensitive.

In many cases the "CONTROL" input does not require a bypass capacitor when a well regulated power supply is used. It is good practice however to place a 0.1 microfarad or larger capacitor at this terminal.

The venerable LM555 timer chip and its twin brothers the LM556 have been a cornerstones of model railroad electronics but the sensitivity of the trigger input gives rise to many false triggering problems. The addition of a 470K ohm resistor and a 0.1uF capacitor at the trigger input (Pin 2) will provide a time delay of 1/20th of a second from the time the input goes to zero volts until the trigger threshold is reached (1/3Vcc). This can eliminate false triggering in most cases and if the problem persists the size of the capacitor can be increased.

The following schematic shows two additions to the basic 555 timer circuit. One reduces the trigger sensitivity and the other will double the output pulse duration without increasing the R1 and C1 values.

LM555Helpers.GIF
555 Timer Helpers Schematic
The addition of a capacitor to the trigger will not work for short output pulses as there is also a short delay in the recovery of the trigger terminal voltage.

The second 555 timer helper will extend the timers output duration without having to use large values of R1 and/or C1. By connecting a 1.8K ohm resistor between the supply voltage and pin 5 of the 555 timer chip the output pulse duration will approximately be doubled.

To achieve long output times electrolytic capacitors are used for C1 and the value of R1 may be as high as 1 Meg. However with high resistance values for R1 the leakage current of the timing capacitor (C1) becomes a significant factor in the operation of the timer.

The circuit will run much longer than expected and may never time out if the leakage current is equal to the current through the resistor at some voltage. Tantalum capacitors could be used as they have very low leakage currents but these are expensive and not available in large capacitance values.

The boxed in area of the drawing shows the internal circuit at pin 5 of the timer with the 1.8K resistor added. The voltage at pin 5 will be increased from 0.66Vcc to 0.84Vcc which is equal to the voltage across the capacitor after two time constants. This allows the same output time to be achieved with a smaller resistance or capacitance value thus reducing the error caused by the capacitor leakage current. Conversely, for a given value of R1 and C1 the output time will be doubled. (One time constant is equal to R1 times C1).

The trigger voltage level of the timer will also be increased with the addition of the resistor to pin 5 but this should have no effect for most applications.

This is not an ideal solution to solving long duration timing situations but will be OK for times of less than five minutes.

RESET And CONTROL Input Terminal Notes

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Hotwaterwizard,
That's a great way to modify the 555's internal divider in order to double its time duration to up to 5 minutes.

Kasamiko,
The 555 doesn't do a very good job at directly driving relays. Its voltage ouyput, with a 15V supply and a 100mA load, is guaranteed to be only 12.75V, which is a 2.25V drop. This drop is more at lower supply voltages. The LM555 data sheet is here:
http://www.national.com/ds/LM/LM555.pdf
So it may give only 9.5V into your 12V relay, causing intermittent inactivation.
Therefore your transistor relay driver is needed.
What is the value of the transistor's base resistor?

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Kasamiko,
22K is too high a value for the base resistor of a 9013 transistor that must saturate when driving a 67mA load. The guaranteed current gain of a 9013 is only 64 @ 67mA.
Usha marks a gain-bin code on their 2SC9013 transistors. They even claim that their transistors are used in radios that you can drink, and you won't get sick! (See their funny typo error)
Their 2SC9013 data sheet is here:
http://www.ushasemi.com/pdf/Transistors/2SC9013.pdf
Although they show it saturating very well with Ic/Ib = 10, if your 9013 has a gain of 64 then 1mA should be OK . Use 10K for the base resistor.

Is the supply 12.7V with the relay activated?

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Kasamiko, in answer to your previous question, I find that when using transistors as switches by saturation, it is much better to use them in a sink mode rather than as source mode. Because of this, I use a PNP type of transistor. It has less effect on your supply voltage and all you need is to ground the resistor at the base to activate it. I also like using the transistor above the relay and keeping the coil of the relay grounded. After you have used your circuit for a length of time, it will be normal to have dirt build up on the circuit board. In the method that you have used, you can easily find arcing at the relay contacts after some time because of the voltage always present on the coil. This does not occur when the relay coil is kept at ground potential when it is not in use.

Here is the relay switch that I always use. If you use it, you will need to have your circuit send a low to activate the relay instead of a high.

MP

post-555-1427914162455_thumb.jpg

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MP,
1) Your terms "sink" and "source" are reversed.
Sink is used when a device pulls-down a load to ground.
Source is used when a device pulls-up a load to B+.
2) How can the PNP have less effect on the supply voltage than an NPN, when their current amounts are the same?
3) How can relay contacts arc, because of a low voltage at the coil?
4) Does the measly 12V attract dirt to the relay like the thousands of volts do at my TV's screen?
Doesn't the high mains voltage at the relay contacts attract more dirt than the 12V?
5) Your PNP driver needs an inverter stage, so use an NPN instead.

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audioguru,
1. I have not reversed my terms. I was refering to what he is doing to the base or in other words, what his 555 circuit will be doing to cause the relay action. Not a reference to what the transistor is doing to the relay. This is a proper use of the term. By your own definition, the 555 is sourcing or sinking as I have described. Those of us who use microcontrollers are used to this reference as the microcontroller sinks or sources the base of a transistor, and thus the term sinking and/or sourcing.

2. The device triggering for a PNP depends upon a ground at the base resistor. The present design depends upon the correct voltage and the correct current to be available to the base resistor. It is also dependent upon a clean signal. The present circuit is also more prone to noise problems.

3. When you break the field, you have this problem. Not something that you would expect from theory, but something most hands-on technicians see everyday in the field.

4. If this equipment gets the use in the environment like I think it is going to be used, I do not think that it will have to rely on static build up to get it dirty. I think that it will get it's share of dirt. Why keep a voltage potential on the relay coil just waiting for a short to ground? It has been my experience from field work that any device that sits at a voltage potential waiting for a short to ground will eventually find it. Just give it some time.

5. Why add an inverter? Just make the signal a low pulse. You wouldn't even need the 555. You could use almost any device with an RC time constant control.

MP

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