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Voltage Signal Circuit Zener

Discussion in 'General Electronics Discussion' started by rc_rs_ss, Aug 17, 2011.

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  1. rc_rs_ss

    rc_rs_ss

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    Aug 17, 2011
    Hey Everyone- I have been researching on ways to send out an electric signal to trip a relay to shut off my battery charger. This is the circuit I found online::

    Battery+(Resistor)(Zener Diode)(Relay Input)(Battery-)

    I need this circuit to send a 8.2 or 10Volt 15mA signal to my Relay when the batteries reach 235 volts.
    Following are the SPECS for the Input signal on my Relay:::

    Control Voltage Range 3-32 Vdc 90-280 Vrms (60Hz) 18-36 Vrms/Vdc
    Max. Reverse Voltage -32 Vdc - - - - - -
    Max. Turn-On Voltage 3.0 Vdc 90 Vrms 18 Vrms/Vdc
    Min. Turn-Off Voltage 1.0 Vdc 10 Vrms 4.0 Vrms/Vdc
    Nominal Input Impedance 1500 Ohms 60K Ohms 9.0K Ohms
    Typical Input Current 3.4mA @ 5 Vdc, 20mA @ 28Vdc 2mA @ 120 Vrms, 4mA @ 240 Vrms 3mA @ 24 V

    Is there any way I can make this circuit work and with which diode??
    Any help on this would be great Thanks!
     
  2. rc_rs_ss

    rc_rs_ss

    26
    0
    Aug 17, 2011
    I read something about putting a voltage divider in the circuit. Would that still be accurate enough if I dropped the voltage down to 10? any suggestions?
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,473
    2,819
    Jan 21, 2010
    What I would do is use a comparator to compare a reference voltage to a resistively divided fraction of your 235V battery voltage. The output of the comparator could then be used to drive a transistor which could switch your relay.

    Naturally, I would be pretty careful with the 235VDC (which could give one a deadly shock).

    I would also consider a small amount of hysteresis because the relay is likely to chatter as the voltage on the battery hovers around the switching point.

    The problem with using just a resistive divider like you suggest is that it either won't give you the current you require and/or it will drastically change voltage as you load it. With zener diodes, you have to deal with the issue of them having somewhat variable and "soft" knee voltages.
     
  4. rc_rs_ss

    rc_rs_ss

    26
    0
    Aug 17, 2011
    Thank you, this helps a lot! I thought the Zener diode wouldnt let any current through to the Relay until the voltage at the diode reached the 10 volts for the 10 volt rated diode. Does this not solve the chattering issue from the relay or am I confused about this concept? Ideally I want the batteries to be anywhere from 230 to 240. Does give it the extra leaway? I am going to look more into the comparator and transistor method tomorrow. Thanks again!
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,473
    2,819
    Jan 21, 2010
    In theory, yes. However in practice they will start to conduct at a lower voltage and their conductivity will increase rapidly as the voltage does.

    Depending on the actual current required to pull in the relay, it might close at the zener voltage plus 6, or 6.5, or 8, or 9.

    In addition to that, it will pull in quite slowly. If you are switching a load that draws significant current you may well destroy the contacts.

    Also the full current for the relay needs to flow throught the zeners. So if the relay requires 50mA, the Zeners will need to dissipate over 10W of power -- and that's a lot (especially considering it's coming out of your battery)

    Like almost anything, there are literally dozens of ways to solve this problem. A comparator is simply the first one I came up with. It is flexible, but also not trivial in operation. It also requires a power supply
     
  6. rc_rs_ss

    rc_rs_ss

    26
    0
    Aug 17, 2011
    I have been researching the comparator idea and am trying to create a parts list so I can get this rolling. The input control specs on my solid state relay has a 3.4mA @ 5 Vdc option and a 20 amp 28 volt option. I believe this means it can be anything between those marks correct? It also has an input impedance of 1500 ohms(not sure what it means).
    I am going to use the Lm339 comparator and a transistor to drive my relay. To make my charger stop charging at 240 volts I need to drop this voltage down to be compatible with the comparator. can anyone help me with any of this? I know this probable seems simple but I am having a bit of trouble with it.
    Thank you again-
     
  7. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    Look at the LM3914. This is a LED bar or dot driver, it contains a voltage reference and ten comparators. An indication of battery voltage could be given and, when the last LED is lit, a small amount of current could be taken and amplified to switch your relay.
     
  8. rc_rs_ss

    rc_rs_ss

    26
    0
    Aug 17, 2011
    I am reading datasheets that say the current drive is programmable. Does that mean it is programmable to be used with 240 volts or do I still need the voltage divider?
    Do i need a computer program and arduino or something for this? Or is it programmed a different way??
     
    Last edited: Sep 25, 2011
  9. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    You will need a resistor and zener diode (12V) to provide a power supply for the chip and a voltage divider to provide a sample of the measured output voltage.

    When they say that the LED current is programmable, they mean it can be set. This is done with a single resistor. If you go for the dot mode, the power supply needs would be quite low.

    You will need a transistor at the LED output and a high voltage MOSFET to drive a high voltage relay. I have a IRFR320 (400V, 1.7A) taken out of a dead compact fluorescent lamp, doubtless they can be bought. Another source may be computer switch mode power supplies.

    Relays with high voltage DC coils seem to be rare but you could use a 230V AC relay with a series resistor.

    Duke
     
    Last edited: Sep 25, 2011
  10. rc_rs_ss

    rc_rs_ss

    26
    0
    Aug 17, 2011
    So should I use 12 volts for the reference voltage also? The relay coil control can trip with 3 to 32 volts by the way. Thank you so much for the help so far.
     
  11. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    The weather is nice here at the moment so I should be out in the garden.
    The LM3914 should be supplied with about 12V with a resistor and zener.
    The internal reference voltage is low but they provide a means of raising it with an internal op-amp and resistors. I would set it for something less than 12V, say 8V.
    You then need a voltage divider to give say 7.5V at your required switching level.

    I did not read your post properly and did not realise that you had a solid state relay. It is likely that this could be driven directly by the 3918 in place of a led or in series with a led.

    If you ae serious, I could refresh my memory on the 3819 and do a few resistor calculations but I must go and mow the lawns now. Last time before winter?

    Duke
     
  12. rc_rs_ss

    rc_rs_ss

    26
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    Aug 17, 2011
    When the grass stops growing you can substitute the lawn mowing with snow shoveling! And Windshield Scraping!
    If you want to do some calculations that would be Great! If its not to much trouble that is. I do have a separate 12 volt power supply that powers my cooling fan, unless they need a common ground. Thanks so much for the help, I have been trying to figure this out for awhile.
     
  13. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    I attach a diagram of what I thought could work. The voltage is divided by 50 to make 250V=5V at the chip. Making the reference 5V supplying the resistor chain and placing a 40k resistor to ground, gives a range of 4V to 5V which would indicate from 200V to 250V. Since there are 10 LEDs these would change every 5V. Select which LEDs you need to drive your relay.

    A power supply is needed and I chose 12V. The feed resistor will need to be about 10W, quite a large power loss.

    You could try the simple resistor- zener- relay at little cost but the zeners I have seen have a tolerance of about 5% so some try and mod would be necessary.

    Duke
     

    Attached Files:

  14. rc_rs_ss

    rc_rs_ss

    26
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    Aug 17, 2011
    Thank you! I am going to order the parts this weekend. Ill let you know how it works.
     
  15. rc_rs_ss

    rc_rs_ss

    26
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    Aug 17, 2011
    Should I get all half watt resistors besides the 10 Watt feed resistor? And then what watt for the 12 volt zener?
     
  16. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    The current drawn is
    10 mA LED (perhaps change scaling resistors)
    10 mA relay
    10 or 20mA zener depending on whether relay is energised
    Total 30mA

    6.8k at 223V = 33mA, 7.3W
    zener 20mA at 12V = 240mW, use 0.5W
    The 1M sense resistor dissipates 50mW only but I would go for 0.5W resistor to get better voltage capability.
    You may want to adjust the resistors around the 3914 to get standard values.

    The 7.3W loss in the feed resistor seems a lot. If you try the simple zener relay solution the loss will only be there when the voltage is high and power is available.

    You may be able to reduce the power consumption by using a 12V switch mode power supply. Some modern wall warts are switch mode and presumably rectify the input AC to get DC to switch. The wart should then be able to run on DC. A norrnal transformer wall wart would go pop on DC.
     
  17. rc_rs_ss

    rc_rs_ss

    26
    0
    Aug 17, 2011
    Ok, I am going to try the zener diode and divider trick first. If I get a 12 volt zener diode I will need 12 volts going to the zener when the supply voltage is 235. Correct? This means I need one 18.58k resistor and 1k resistor. Maybe adjustable resistors to fine tune it??
    Does this sound like it will work?
     
  18. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    This is unlikely to work since you are dividing down voltage difference with the voltage.
    You could try seven 33V zeners (BZX79 0.5W) in series to drop most of the voltage without reducing the voltage difference greatly. The details I have seen is that they have a tolerance of 5% so it may be necessary to consider what the effect of this will be.
    I would test the zeners and relay with an adjustable power supply.

    I am not too clear on the spec of the relay or whether the voltage will be protected from going above 235V which could cause excess current. The current drain will be very low on low voltage and rise rapidly as voltage rises above the zener voltage.
     
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