Reed Relay Help for Dummy

Discussion in 'General Electronics Chat' started by CoachKalk, Sep 27, 2011.

  1. CoachKalk

    Thread Starter Member

    Sep 20, 2011
    139
    2
    UGH! I was obviously wrong when I thought I understood the way a Reed Relay (I guess any relay, in general) works. Or I have the setup horsed up!

    I have been trying to get up to speed on a few different components for projects I am tossing around and I found a project online that, for the most part, was an exact match.

    I will try to somehow get the circuit I tried attached to this thread, but it really seems so simple I must be missing something obvious.

    9V Supply
    100k R
    Photoresistor
    2N4401
    Reed Relay

    The R1 and Photoresistor in series (Voltage Divider Set-up Right?)feeding the base pin of the 4401. When lit, the transistor is "off".
    When dark, the V across base increases and triggers the transistor. I started with a simple LED circuit connected through the collector/emitter to see if I had the transistor pins understood. Worked like a champ. Light source on - LED off. Light source off - LED on. Hurray!

    I thought I could set the same exact LED test circuit up - through the relay. Intead of directly turning on the LED, I thought the transistor would trigger the relay connection - thus completing the LED circuit.

    So, a few questions about this experience.
    1. When using a transistor, is the circuit "dead" whenever the base pin is not receiving the necessary trigger voltage? Meaning, you cannot set up a LED indictor using the Emitter pin to show the condition of the "base"?
    2. I thought it would be easy to find, but no luck. Can someone just confirm the pins of a reed relay. Which pins "close" the other pins?
    3. General question about relays - the circuit that activates the relay is separate that the circuit completed by the relay? The completed circuit may be sig. different (as long as within the relays specs) than the activating circuit?

    I know these are probably very simple concepts, but I want to make sure I get corrected if my non-technical understanding is off base.

    Thanks - help any way you can. I know it is very difficult with out the exact circuit I tried - I will work on getting it added here.

    Steve
     
  2. jwilk13

    Member

    Jun 15, 2011
    228
    12
    Do you have the part number for the relay? That would help with the pinout information. The ones I'm familiar with are either 14-pin DIP packages (with only 8 pins) or a SIP package with 4 pins. Check the datasheet for this part to see if one of those configurations matches what you have. The only other type of "reed relay" I've seen requires an external magnet to close the switch, which I'm guessing you aren't using.

    And yes, schematics are always helpful ;)
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Attaching complete and accurate schematic(s) is key to receiving a prompt and (usually) accurate reply. Well-drawn and documented schematics eliminate virtually all questions that would be necessary if described via text, as even relatively simple circuits require a good deal of typing to describe, and it takes a lot more thought to mentally compose a schematic from the text than seeing a schematic drawing.

    Ahh, is this a plug supply or a PP3 "transistor" battery?
    NC, NO or both? What is the operating voltage and coil resistance or current?

    OK, so you have R1 going to +9V and the photoresistor going to GND.

    This is where the words go wrong.
    Did you mean that you have the LED's cathode connected to the transistor's collector, anode to +9v, and the transistors' emitter to ground? Or was the anode on the emitter, cathode to ground, and collector to +9v?

    Did you have a current limiting resistor on the LED so that you don't burn it up?
    100k is quite a large value for a base resistor when you're trying to drive a relay coil.

    With a 9v supply, and assuming Vbe will be 0.7v, (9v-Vbe)/100k = 8.3/100k = 83uA (microamperes) current, which is normally enough to saturate a transistor that is driving an 830uA load. However, relay coils need much more current than that.
    Transistors are controlled by current; a relatively small current through the base-emitter junction controls a relatively larger current through the collector. When using a transistor as a saturated switch, you use 1/10 the desired collector current through the base.

    If the base current is not sufficient for the desired current, Vce (voltage on the collector referenced by the emitter) will increase (depending on how low the base current is), up to the point where when Ib=0, Vce = the supply voltage.

    Reed relays are available in quite a number of configurations. Do you have a manufacturer and part number, or a photo, or a link to a website that has a photo or specifications? Usually these types of things are documented in a datasheet provided by the manufacturer.

    Usually. However, you can make a relay "latching" by having the relay contacts power the coil once a momentary switch has energized the coil temporarily.
    Sure.
     
  4. colinb

    Active Member

    Jun 15, 2011
    351
    35
    Should he include a flyback protection diode across the reed relay coil?
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    If the coil is being switched by a transistor connected to a CdS cell, the cell will be so slow that a flyback diode won't be needed. It's never a bad idea to use one though.
     
  6. CoachKalk

    Thread Starter Member

    Sep 20, 2011
    139
    2
    I have attached the picture I went from. This is a DIY from the internet I found. I was trying to light a LED just to see if I could get the setup down (not a start/stop watch).

    The Reed Relay is OMR-C-105H. 4 Pins
    No part number was given so maybe that is the problem right out of the gate.

    What you see is what I have/tried - minus the stopwatch. I'm not going to lie, I didn't evaluate the circuit at all (not that I would have been overly successful had I tried at this point it sounds), but I jumped in.

    I thought I could just use the setup and test across the relay to see if it was triggered (before I came up with LED indicator plan).

    Yes - I did use a resistor to protect the LED.

    "Did you mean that you have the LED's cathode connected to the transistor's collector, anode to +9v, and the transistors' emitter to ground?" - YES

    "With a 9v supply, and assuming Vbe will be 0.7v, (9v-Vbe)/100k = 8.3/100k = 83uA (microamperes) current, which is normally enough to saturate a transistor that is driving an 830uA load. However, relay coils need much more current than that.

    Transistors are controlled by current; a relatively small current through the base-emitter junction controls a relatively larger current through the collector. When using a transistor as a saturated switch, you use 1/10 the desired collector current through the base.

    If the base current is not sufficient for the desired current, Vce (voltage on the collector referenced by the emitter) will increase (depending on how low the base current is), up to the point where when Ib=0, Vce = the supply voltage."

    Hmmm ... more homework on transistor basics I think for me!

    [​IMG]
     
  7. colinb

    Active Member

    Jun 15, 2011
    351
    35
    You could use a Darlington transistor (AAC book topic) (WP article) so you could drive the relay coil with more current from your high-impedance control source.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Here is a wiring diagram for your relay:
    http://www.te.com/catalog/common/images/PartImages/1v99015e.jpg
    Basically, the two pins that are furthest apart are the contacts; the other two are the coil.

    Didn't find a datasheet, but here's a catalog page:
    http://www.te.com/catalog/pn/en/2-1440000-2

    The coil is 5v, and its' resistance is 250 Ohms. 5v/250 Ohms = 20mA.

    Usually, you need to get the voltage across the coil to at least 70% of the rating for the relay to engage. However, let's just go with the 5v and 20mA for now.

    You're powering it from 9v; so you will need to limit the current through the relay coil; otherwise you may burn up the coil.

    (9v-5v)/20mA = 4/0.020 = 200 Ohm resistor needed in series with the coil.
    Now for the base current needed; well, 1/10 of 20mA is 2mA, you're powering the circuit with 9v, Vbe will be ~ 0.8v, so (9v-0.8v)/0.002 = 8.2/.002 = 4.1k Ohms; that's how low R1 needs to be. But, that's not going to work very well, as your CdS photoresistor would have to be REALLY well-lit in order to turn it off.

    As colinb mentioned, you can use a Darlington transistor to amplify the current through the junction of R1 and the photoresistor/CdS. Darlingtons generally have a Vce of 0.7v to 1.3v or so when sinking current, depending on how hard they are turned on, and how much current they are sinking.

    Here's an alternative to a Darlington; Q1 amplifies the current available from the junction of the resistor & photoresistor/CdS.
    Q2 amplifies the emitter current from Q1; R2 limits Q1's collector current and therefore Q2's base current. The base current needed for Q2 I've already calculated as 2mA. Since I added another PN junction in there, Vbe will now be ~1.3v instead of 0.7v-0.8v. So, (9v-1.3v)/2mA = 7.7/.002 = 3,850 Ohms. 3.9k is the closest standard value.

    R4 is a "base return" resistor. It ensures that Q2 is turned off unless Q1 is sourcing current.

    R5 is the relay coil current limiting resistor I mentioned before. 1/8 Watt rating or higher would be fine.

    [​IMG]
     
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