Hi,
I have a circuit that runs off a 5VDC regulator. I need to drive 2 relays, and the ones I have are 6Vdc, but ofcause they switch fine at lower voltage.
The problem is, to switch the relay I have a transistor, and then a diode to prevent back current, after the transistor and diode I am left with 3.8V which is very close to the limit of what the relay will accept, so its not that reliable.
Is there a way to limit that loss, or do I have to run at a higher voltage to start with ?
/Jan
Don't put the diode in series - put it in parallel with the relay coil, in reverse polarity, of course.
Good Luck! Rich
Use an additional 8V reg just for the relay.
Also do what Rich said, the transistor VCEsat should be around 0.2 - 0.4V so you should have way more than 3.8V left for the relay.
If you can use an adjustable regulator, you can sense the 5 volts at the relay, thus avoiding the voltage drops.
Chuck
----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
As has been mentioned, the diode should not be in series with the coil but reverse parallel.
To keep the transistor voltage loss to a minimum, drive the base with plenty of current to as to saturate the transistor ( say use a ratio of Ic/Ib of 10:1 ) and select a transistor with a low saturation voltage Vce(sat).
What's the coil resistance btw ?
Graham
Thanks all, I connected the diode across the relay coil legs and it works.
Sounds right with the 0.2-0.4 loss from the transistor, plus the 0.7V from the diode = 3.8.
I cant run the circuit at much higher voltage, since theres a mcu accepting at most 5.5V.
But it works with the parallel diode, thanks.
/Jan
What voltage is available up stream of the regulator? Using that voltage (with an additional series resistor, if needed, to drop the extra voltage) will keep the regulator output cleaner and give you a certain activation.
Eeyore skrev:
100ohm between the coil legs acording to my DMM. The base of the transistor is being driven from a mcu, so it can get atmost 20mA./Jan
John Popelish skrev:
Theres 9-14V or so before the regulator, from a 12V 20Ah battery drawing
7.5A at peak, my circuit is only using about 60mA at peak, and
you have the diode in the wrong place, (should be parallel with the relay in the non-conducting direction)
if you can use the power from before the regulator to run the relay that would help.
Bye. Jasen
The suggestion seems to be that if you run the relay off a separate supply path, you reduce the risk of switching transients on the mcu rail - a dropper resistor from the unreg supply would be fine and if you included a small electrolytic this would charge to the full supply when the relay is off giving a brief burst of higher voltage making the relay snap to more positively, you could also reduce the overall relay current to the minimum safe holding value without compromising pull-in. With the voltage margins you stated, my suggestion of a separate 8V regulator would require a LDO type.
It would also eliminate about a half watt of heat into the 5 volt regulator (move it to the series resistor), possibly eliminating the need to heat sink it.
Now that's useful.
Such measures can help reduce the build-up of hot spots.
Graham
Excellent advice.
Yes, it can cause latch up.
Graham
Config your circuit using a common emitter for the transistor. the collector goes to one side of the coil. the other side of the coil to your 5 VDC. the base with a resistor in series goes to your switching signal.
Place a protection diode across the relay coil. , the cathode (line side) will go to the 5VDC+ of the coil, the anode to the collector side of the coil.. Coils will release high voltage energy if allowed to discharge at a rapid rate. Coils discharge in reverse polarity and thus the diode makes for a component to do the job.
Also, it's good practice to put a bypass diode from the input to the output of the regulator. many regulators do not like voltages much higher on the output side than on the input side. if you place the diode with it's cathode to the input and anode on the output. current will only flow when the output is higher than the input and thus force caps on the output to discharge. etc..
-- "I\'m never wrong, once i thought i was, but was mistaken" Real Programmers Do things like this. http://webpages.charter.net/jamie_5
John Popelish skrev:
My regulator dont need heatsink, its a small plastic version, rated at max 100mA, the relayas draws 50mA each, but wont be on at same time, and the rest draws a few mA, so it will do :)
I have the bulky 7805 laying around too, that needs to be bolted to the copper.
/Jan
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.