555CN timer - relay or FET on output?

Discussion in 'The Projects Forum' started by mezlo, Jul 16, 2008.

  1. mezlo

    Thread Starter Member

    Jul 16, 2008
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    I'm a novice and it's taken me about a week to get to the point I'm at so hopefully I'm providing enough information.

    I'm setting up a 555CN astable circuit using the schematic shown here. R1 is 270k, R2 150k, and C1 is 10uF. I'm using a 12Vdc source (23A alkaline battery) so I'm measuring around 8Vdc on pin 3. The device I'm powering needs 9-12Vdc and pulls around 82mA @ 12V.

    Since the 8Vdc on pin 3 of the 555CN isn't enough to power my device, I need to add either a relay or a FET. I've done some Googling and am more confused now than when I started. I'm interested in whichever will be the most efficient since both the timer circuit and the end device will run off the same battery.

    In this scenario should I use a relay (reed or electromechanical) or a FET? Also, what ratings should I look for on the relay or FET? I've also seen a few circuits that have diodes as well as a relay in them. Are they needed and if so, what specs should I look for? I buy components at Radio Shack so if you can provide the part number for a suitable relay/FET I'd appreciate it as well (not required, just would be handy).

    Mez

    PS - Can I use a CMOS 555 in this scenario? As I mentioned earlier the less power and more efficient the circuit is the better.
     
  2. Audioguru

    New Member

    Dec 20, 2007
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    A Cmos 555 has a very low output current.

    The typical output voltage of a 555 with a 12V supply and a 82mA load is 10.4V. the guaranteed minimum voltage is 9.75V. Maybe your 12v supply voltage drops when it is loaded.

    The 555 can drive a 2N4401 transistor with a 1k series base resistor and the collector will drop to about 0.1V above 0V when the transistor turns on so the load will get 11.9V.

    If your device is inductive then it will produce a voltage spike when it turns off. A reverse connected diode across it will eliminate the damaging voltage spike.
     
  3. mezlo

    Thread Starter Member

    Jul 16, 2008
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    Thanks for the reply. I'll double check the battery and make sure it's putting out 12v then check the 555 output again. I had read the output was supposed to be Vcc-1.7v so I was a little surprised by the 8V reading myself even though I tested it several times with just my multimeter connected.

    I don't really understand your 2nd and 3rd comments (I'm a novice as I said). Could you either explain how I would hook those into the circuit or provide a schematic showing the transistor, 1k ohm, and diode?

    Thanks,
    Mez
     
    Last edited: Jul 16, 2008
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Radio Shack sells 2N4401 transistors for $0.79 each, but frequently there are several contained in their NPN Transistors (15-Pack) assortment:
    http://www.radioshack.com/product/i...oduct+Type/Transistor&fbc=1&parentPage=family
    That gets your cost per transistor down around $0.18/ea.

    You'll also likely find some 2N3904 and PN2222 transistors in that assortment, and possibly others. The 3904 really "runs out of steam" at about 100mA. A 2222 would work, but the 4401 has higher gain.

    Just to clarify; "connected in reverse" means to connect the cathode of the diode on the more positive side of the coil. The cathode is generally marked with a band towards one end of the body of the diode. You could use a 1N4148, 1N914 or 1N400x (x=1 to 7) diode.

    See the attached schematic. For the 555, only the connection to the output is shown.
     
    Last edited: Jul 16, 2008
  5. mezlo

    Thread Starter Member

    Jul 16, 2008
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    Thank you very much for the connection schematic. I wasn't clear on whether a relay was still involved or not.

    Mez
     
    Last edited: Jul 17, 2008
  6. mezlo

    Thread Starter Member

    Jul 16, 2008
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    OK, I've reassembled the circuit on the breadboard and I'm now getting 10.4V from pin 3 (no clue why I got 8V last night). Since that is sufficient for my device and it only pulls 82mA, can I just connect the device and the reverse diode directly to pin 3 and ground, thereby not using the transistor, relay, and 1Kohm resistor?

    If I do need the transistor and relay, then I have another issue. I'm losing voltage every cycle at the transistor (2N4401) collector. When I put my voltmeter between the collector and Vcc with the diode (1N4148) and relay (12VDC 1A SPDT) disconnected I lose around 20mV each cycle. If I connect the diode and relay and measure the voltage between Vcc and the relay I lose around 50mV each cycle. At this rate I'm below 9V within 10-15 cycles.

    Any ideas what is causing the drain?

    Mez
     
    Last edited: Jul 17, 2008
  7. mezlo

    Thread Starter Member

    Jul 16, 2008
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    Any ideas on what's causing the voltage drain?
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    Your batteries are becoming discharged. As they are used up, their internal resistance increases; thus more and more voltage is dropped across the internal resistance of the batteries themselves.

    Consider using a "wall-wart" type power supply, or better yet, build a bench power supply out of a salvaged ATX form factor computer supply.
    Google "ATX bench supply" for lots of info.
     
  9. mezlo

    Thread Starter Member

    Jul 16, 2008
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    If the issue is the battery then why is only the voltage past the transistor dropping? The voltage at pin3 on the 555 is solid at 10.4V and does not change. The voltage at the collector pin of the transistor starts at ~10.1V steadily drops so that after a minute or so it is below 9V. If I turn the circuit off for a few seconds and turn it back on, the voltage remains low but If I turn it off for a few minutes, the voltage goes back to ~10.1V and starts dropping off.

    This is not a battery issue but rather something with the 2N4401 transistor. I should have tried another one out of the pack but I didn't.

    Mez
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    Hmm. Could be a heating issue. Does the transistor feel warm?

    When the 555 output pin 3 is near Vcc, the collector of the 2N4401 should measure less than a volt.

    When the 555 output pin 3 is near 0v, the collector of the 2N4401 should measure the same as Vcc, as it will be turned off.

    If you have the diode connected backwards, the 2N4401 will rapidly overheat as it will be shorting out your batteries, and your relay will not function properly. The side of the diode that is the cathode has a ring on the body of the diode, and that side should be connected to the Vcc side of the relay's coil.

    You may have the transistor installed incorrectly. The standard configuration for a 2N4401 NPN transistor in a TO92 case is E B C ; that's with the flat side of the case facing you, leads pointing down.
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    I think you have the collector and emitter od the transistor connected backwards. Then the transistor is a 7V zener diode when it is supposed to be turned off.

    With the transistor connected correctly, when the output of the 555 is low at 0.01V then the transistor is turned off and the relay coil pulls its collector voltage up to 12.0V.
    When the outpput of the 555 is high then the transistor is turned on and its collector voltage is about 0.1V.
     
  12. mezlo

    Thread Starter Member

    Jul 16, 2008
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    In the schematic you posted, should the collector or the emitter of the transistor be connected to ground? My guess is the emitter?

    Mez

    PS - What about my first question of whether or not I need the transistor at all since my device will work with 10.4V and pulls 82mA which is within the 555 chip's output limit?
     
    Last edited: Jul 17, 2008
  13. SgtWookie

    Expert

    Jul 17, 2007
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    The emitter should be connected to ground. The emitter in a "traditional" BJT (bipolar junction transistor) is always shown with an arrow in a schematic. The arrow points towards the N junction. Since the arrow is pointing away from the base in this case, it is an NPN transistor. If it were pointing towards the base, it would be a PNP transistor.

    You could run it without the transistor, if the 555 is indeed a bipolar version. If it is a CMOS version, it won't have enough power for the relay's coil. In any event, it's better to blow a $0.18 transistor than a $2.00 timer.

    If you remove the transistor, the ON cycle of your relay will be reversed; if you currently have a long ON time and a short OFF time, removing the transistor will result in a long OFF time and a short OFF time.

    This is because the way the transistor was being used is that of a current amplifier and signal inverter; when there is no current on the base (output of the 555 is zero), the collector is turned off, so the relay is off.

    However, if you connect the output of the 555 directly to the relay's ground side of the coil, when the output pin of the timer is low, the relay coil will energize.

    Whether or not you remove the transistor, you should keep the diode in the circuit. Otherwise, you risk destroying your 555 timer.
     
  14. mezlo

    Thread Starter Member

    Jul 16, 2008
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    When I reversed the transistor, I would get 10.4V at the collector when the 555 was high and 10.7V when the 555 was low. Obviously this wasn't correct either so I ripped out everything past pin3 and started over. After that things seemed to work better.

    That being said, if I don't need the transistor and relay I'd just as soon not use them to reduce the power used by the circuit and therefore extend the battery life.

    I plan to hook the device directly to pin3 with the reverse diode in place.

    Thanks for all the help.
    Mez

    PS - What's the purpose of the 1Kohm resistor? I measure virtually the same voltage on both sides of it.
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    Wait - what is the device?

    The 1k Ohm resistor was to limit the current to the base of the 2N4401 transistor to around 9mA. The 2N4401 has pretty high gain. At a collector current of 100mA, your collector-emitter voltage should have been well under 0.2v, which means you should've seen battery voltage minus the collector-emitter voltage across your relay's coil.

    If there were no base resistor, the base to emitter junction would look like a forward-biased diode in the circuit; basically, it would've shorted the output of your 555 timer.
     
  16. mezlo

    Thread Starter Member

    Jul 16, 2008
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    The device is this Ethernet Link Tester. When connected directly to the battery as shown in the project, the maximum current I've measured with or without a network cable connected is 82mA (using the 10A lead on my multimeter the display read 0.082). Since the 555 outputs up to 100mA, this should be acceptable, right?

    Mez

    PS - If you'll recall, when I first assembled the 555 circuit I measured 8V at pin 3 because I had something wired wrong. If I had wired it correctly and measured 10.4V from the beginning, I would've never asked the original question in the first place. :)
     
    Last edited: Jul 18, 2008
  17. SgtWookie

    Expert

    Jul 17, 2007
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    Ahh, ok. Well, it looks like those devices have a boost converter in them, which is likely an inductive load. Keeping the diode in the circuit is therefore a good idea.
     
  18. mezlo

    Thread Starter Member

    Jul 16, 2008
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    OK, thanks for all of your help with this project.

    Mez
     
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