next time) and your description, it sounds like you had the LED connected from the LS05 output to ground. That's wrong. Connect the LED and resistor in series between
+5V and the LS05 output.
The relay should be connected between +5V and the LS05 output; you don't need a series resistor for a 5V relay. What you do need, however, is a back emf diode connected backwards across the relay coil
Here are the datasheet ratings. If a specific part has a saturation voltage of 200mV at 16mA, it can probably sink say 40mA successfully (0.5V, 20mW), but you'd be breaking rules and using it beyond spec.
Hello everyone, I apologise in advance for my lack of knowledge. I am completely self-taught in electronics so might be missing a basic concept here... I have a 74LS05 IC (Hex Open Collector Inverter) that I want to use to drive a motor (via a relay). I tested it first with a LED and it worked fine (after I realised I needed to put a 3K9 (pull-up?) resistor to the +5V line to get it going). Now when I attach the relay in place of the LED, I don't get enough voltage to fire the 5V relay. I will try a diagram for when it worked with the LED...
Output of 74LS05 ????????????????>?????? ?Vx LED ? ? ? > ? < 3K9(R1) ? > GND ? ? ???? +5V
Now, when I try it with the relay in place of the LED, I don't get enough voltage. I tried various values for R1 - the results were: R1 Vx
100R 2.90 V
820R 0.70 V
3K9 0.16 V
22K 0.03 V
Now, I'm thinking if I go any lower than 100R, the voltage at Vx will permanently be enough to switch the relay, so I'm a bit lost - I don't think my theory is correct. In case you need it, the resistance across the relay coil is 120R. If anyone can point me in the right direction, I would very much appreciate it. Even better, if someone can tell me why the LS05 needs this resistor at all, maybe I would understand it a bit better. By the way, I am testing this with nothing attached to the relay - I just wanted to see it switch before I loaded it. Will I need to allow for the load on the relay in the solution to this problem? I'm thinking not because they are seperate circuits anyway, aren't they?
I haven't got any TTL data books any more, so I don't know the figures, but I suspect that a 7405 won't sink enough current to drive a relay. Guessing at about 5mA sink current, if the resistance of your coil is less than (order of) 1kohm, you're sunk.
Replace it with a ULN2003 or similar (different pinout!).
Incorrect. They will reliably drive a 12V 900ohm relay coil. TI data book sez SN74LS05 will sink 8mA max. Admittedly, that does not compute; the relay would seem to require 13mA for full drive. I have not made any measurements of my relay drive voltages; i put a
100 ohm resistor in series with the relay coil, and then a "snubber" diode from the '05 collector to the +12V. The relays i use seem to close in 1mSec with random bounce for another
1mSec; dropout seems to be about 5mSec (if i remember correctly).
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What advantage can you see to adding the resistor?
Best regards, Spehro Pefhany
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The other part of the answer, which is missed in this, is that the 74LS05, cannot sink enough current to drive this relay. The relay expects to be driven of 5v presumably. As such, it requires just over 41mA to drive it (5/120). The LS05, can only sink up to about 20mA (varies with the version). Hence the ciruit above, is using the transistor to sink the current, and the LS05 to drive this. The reason the LED worked, is that most only require a couple of mA, to give a reasonable output....
Connect 3 '05 gates in parallel: all inputs tied together, all outputs tied together.
Connect the relay between 5V and the output of the '05s. Connect a diode across the relay coil: Stripe (cathode) to 5V; the other end to the outputs of the '05.
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tie the OC output to a 10k pullup resistor to 5v VCC. A relay is much slower than the OC line.
connect the output to a general purpose transistor, an NPN should work, tie the collector to the side of the relay that would be grounded, the other side of the relay to the +12v
the emitter to ground, the base to your output of the '05
I would bet money the 0 to 4 volt swing of the '05 output will drive the transistor, which will drive the relay.
You may need a diode to drain the relay when the coil wants to dump its energy when its off, so put a damper diode across the NPN collector/emitter, the same way a horizontal output transistor is arranged.
You may also need series resistors to limit the current, no need for the relay coil to smoke, so +12v into a series resistor, into the relay + side, then connect the relay - side to the collector, then the emitter to ground
When the relay is energized/released there will be voltage spikes
"Spehro Pefhany" schreef in bericht news: snipped-for-privacy@4ax.com...
Shorter turn-off times of the relay. Cheaper than a zener. Just pick a R with the same DC resistance as the coil. If you don't mind the double current, remove the diode.
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Err... Unless it is an appropriately rated zener diode, the diode belongs antiparallel to the relay coil, not antiparallel to the transistor. If you place the diode antiparallel to the transistor a large voltage transient will still be produced and the transistor or "damper diode" you placed antiparallel to it (whichever has the lowest breakdown voltage) will still be vulnerable to avalanche breakdown and possible destruction. Don't be fooled by the numerous hobbiest schematics that can be found littering the internet that do this. They are doing it incorrectly and consequently their products' reliability suffers.
In a TV the damper diode is needed to provide a conduction path for the yoke coil current which is kind of in parallel with the horizontal output transistor. In your standard transistor-drives-relay-coil circuit, the relay coil (inductive device) is in series with the transistor. This is an important topological difference.
side,
If the appropriate diode is placed antiparallel to the relay coil no voltage spikes during turn on or turn off will be produced by the relay coil. Depending upon the nature of the load and voltages on the relay contacts, it is possible that that part of the circuit could cause voltage transients and electrical noise both at turn on and off (IE: inductive load, when contacts close the contacts bounce and turn on-off several times causing inductive load to produce high voltage and electrical arcs across the relay contacts).
I'm not an expert on TVs, but I'll refer to the schematics of figure one and two of this document:
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They show the yoke coil being in series with a capacitor which together is in parallel with the horizontal output transistor (HOT).
For the relay coil, no. Inductors obey E=L*dI/dt
Where E is the voltage generated (or applied to the inductor), L is the inductance in Henries, and dI/dt is the rate of change of current in the inductor measured in amps per second. Current never changes instantaneously in an inductor. This means current doesn't reverse direction instantaneously either. It must first slowly ramp down to zero, and then change direction slowly ramping back up to a higher value for AC applications. The voltage measured accross the inductor can instantly change, and does instantly change at the moment of transistor turn off, but the current keeps flowing in the same direction.
With the diode antiparallel to the relay coil, and the transistor turns off (pretty much instantaneously), then a new current path is needed. The diode provides this current path. The relay coil thus produces an external voltage of only one diode drop or about 0.7V. Hardly enough transient to even call it a spike at all.
If the relay contacts were fully open, then yeah it would likely take incredibly high voltage. If the relay contacts have just closed, but "bounce" open again slightly, then the air gap will be exceedingly small. It doesn't take much voltage to breakdown this tiny air gap. If you closely study a little two AA cell flashlight (say using 500mA or more lamp) switch in action, you should be able to observe tiny electric arcs (assuming you do this at night, turn off all the lights and cover the lightbulb with something very dark) as the contacts open and close. The voltage is very low at 3V or less, but the gap is small. The effect is much more pronounced and much more observable if you use higher voltages and higher currents (say
12V at 5A). The load need not be inductive.
If the relay contacts are fully closed, and an inductive load is being powered, then at the moment of contact opening an arc will likely form (though this assumes the inductive load doesn't have an antiparallel diode across it for DC applications). Again the voltage produced when the arc is first initiated is very small since the gap is very small. As the relay contacts get farther and farther from each other a higher and higher voltage is needed to sustain the arc. Eventually the inductive energy will all be dissipated in the arc and parasitic coil/wiring resistance and the arc will extinguish.
Right, isnt the diagram showing the flyback in series, with the yoke in parallel? whatever, thats more of a complex inductive application than a relay driver
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So at 3v with even a 1 mil bounce, or whatever bounce where you could see a spark, it would be far more than 3v across the gap thats sparking, wouldnt it?
Right, more complex. The important part though is that current must flow both down through the transistor and at other times up through it. Unfortunately BJTs don't conduct in reverse by themselves (well, the base emitter junction will avalanche at a fairly low voltage, but that degrades the specifications of the transistor) so you have to add the external diode to conduct the reverse flowing current. In the simple relay driver circuit current only ever wants to flow in one direction, down through the transistor.
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Not necessarily. Unless the load is inductive and unclamped it may just be
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