I have an embedded computer. It had digital I/O. Per the manufacturer, "When the DIO pins are configured as outputs, they can source 4 mA or sink 8 mA and have logic swings between 3.3V and ground."
OK, I prefer to work as source. So... How do I get 4 ma @ 3.3V to control a 5V relay that needs 100 ma to operate?
ISTR something about transistors, but it's been a few years since my electronics classes....
Right now I need to be able to power LEDs, but eventually I need to siwtch 120 VAC and 24-48 VDC at a few amps load, and I thought I could use relays for that. But this has thrown a wrench into the works....
"Captain Dondo" schreef in bericht news: snipped-for-privacy@corp.supernews.com...
Yan,
A transistor solution shown below. The NPN are general purpose transistors like BC547. Don't forget the diode otherwise your transistor will be blown. An 1N4148 or similar will do. A much simpler solution is using a FET. A BSS295 or similar will do.
That's more like what I imagined for the OP, since that turns on the relay when the I/O is sourcing. However, assuming that the TIP120 is a Darlington, this also tosses away some extra of the very few volts available at +5V for the relay.
I agree. For a really low Vcesat at pretty high currents, the TIP42A is decent. But in this case at 100mA, I think the common 2n3904 won't be too bad.
What bugs me and what drove me to offer two BJTs as a solution is that the OP probably has no real idea what the output voltage is at when sourcing 4mA. The OP sees the specification, but I wouldn't be surprised to see a drop of almost a volt at that current (200 ohms in the FET, or even more, isn't unusual.) So I'd generally rather add another BJT just to keep it as "no problem, at all" and just not worry about the question. Plus, it does put that nasty relay inductor just a little bit further away from the I/O. :) Well, that's how I was thinking, anyway.
"Jonathan Kirwan" schreef in bericht news: snipped-for-privacy@4ax.com...
You're right. The transistor solution shown above inverts the output signal which might not be wanted. The FET solution does not invert. As you have 4mA source available, you may use a single transistor as shown below. You - or the OP - only need to keep an eye on the type of transistor choosen. Its current amplification should be high enough to drive the transistor fully into saturation at the 100mA collectorcurrent required. So not every general pupose transistor will do. In both transistor solutions the transistor that drives the relay should have not too high a VCEsat, so to keep enough voltage for the relay.
"Jonathan Kirwan" schreef in bericht news: snipped-for-privacy@4ax.com...
May work but my main objection is the 2N3904 is used as a lineair amplifier. The base current though the 2N3906 may not be high enough to drive the transistor into saturation. Used in another - simpler - circuit, the 2N3904 may do it all on its own. (See my other posting.) I'd remove R4 as it has no use and lower R3 to let's say 1k so to use the available control current. Then I'd remove (short) R2 but place a similar or even somewhat lower resistor between the base of Q2 and the R1/Q1 node to limit the base current for Q2.
Okay. I think that's a valid criticism to consider. I happen to like keeping Q1 out of saturation, myself. But let's look at the details and see where that goes.
The arrangement of R3 and R4 yields about 90% of the 3.3V (current required is minimal and won't pull down the I/O pin) or say 3V at the base of Q1. This means that about .6V or .65V less than this will appear at the emitter of Q1. That alone suggests about (3 - .65)/1k or 2.35mA collector current for Q1. With the collector voltage no more than the harder-driven Q2 Vbe below 5V, or let's say about 4.15V, this means about 0.85V/10k or 85uA through R1. This leaves most of the 2.35mA or let's say at least 2.2mA for base drive on Q2.
With 100mA drive for the coil, even if 200mA for a short time, the
2.2mA will probably be okay as base drive for it. It's a relay though with inductance, so it probably will be a ramp up on the current until the DC ohms takes over.
It seems okay.
I think I agree. I just don't like pulling the mA out of the I/O, I guess.
I added it for the specific scenario that the circuit might be tested without an I/O actually connected. I just wanted to make sure there was a little bit of pull-down to keep Q1 off in such a case.
Well, not unless you also kill R2.
Ah, there you killed it. But then you are sucking that current out of the I/O. I suppose that's okay. I just prefer overkill. For a one-off (which is all I ever try and make), anyway.
A problem you created by deleting R2, which I had in the circuit in the first place and yielded a nice, poor-man's controlled current sink for the base drive.
I got to thinking some more about this. Why are we restricting Vcc to 5 volts? It may be an artifical restriction. Later on he will have 120VAC and 28-48 DC that he wants to switch with a relay. Nothing says he can't use those (or any other) sources to power the relay. He can use a single bjt/fet/darlington as a driver stage. If there's no common ground, or if the 5V is all that's available, he could drive an opto with the driver stage. Opens up the choice of relays, too.
NPN transistor (eg BC337) emitter to ground colector to relay coil other end of relay coil to +5V supply, diode (eg 1N4001) with the cathode (banded end) to +5V and the other end to the collector, base of the transistor goes to a 1K resistor and the other end of the resistor goes to the chip's output pin.
for switching the 120AC it may be easier to use a triac and a MOC3010 opto-isolator.
for the DC 48V is kind of high for a commonly available relay to switch (most 240VAC relays are only good to 30VDC) a power MOSFET may be better suited.
Perhaps - but connecting an opto directly? What protects his embedded computer? Use a driver! The 4K7 (or whatever) base resistor protects the computer from shorted LED or bjt, and keeps the current minimal during normal operation.
How many outputs would you have total? You can use a darlington ULN2804? has 8 of these in a single chip which will drive loads of 500 ma per output, with freewheeling diodes available internally, and very easy to interface to. You can drive the LED's off this, or the relays, or both.
I use these all the time for TTL to Relay Drivers. My control board Outputs are rated at 6ma.
Richard
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