I feel really stupid for asking this question because I'm sure it's ridiculously simple, but......
I have an IC that has an output that is active low with a 10mA current limit. How do I drive an LED with this? I assume I need a small MOSFET, but I'm not sure which (NPN or PNP?), or how to wire it.
Thanks in advance, and sorry for the ignorance. 😱
I have an IC that has an output that is active low with a 10mA current limit. How do I drive an LED with this? I assume I need a small MOSFET, but I'm not sure which (NPN or PNP?), or how to wire it.
Thanks in advance, and sorry for the ignorance. 😱
Actually active low simply means that when you send a true state to it, it goes low.
So, if it is a reset out line and you issue a reset command it goes low.
The way that you're asking the question suggests what is called an open collector, or open drain output. This means that the output is not driven in both directions only to ground and this is why these are active low. The designer supplies an external "pull up" resistor, or simply connects the load from the pos supply through a current limiting resistor. OC outputs were often used in bussed configurations where they provide a wired OR logic connection. But they also have many other uses.
Some OC outputs are rated for 30V or more and allow 5V logic to interface to higher voltage external circuitry such as 24V relays.
So the question is, does this output have a source current, you've stated the sink current as 10 mA. Really, it doesn't matter, you can wire it as described above to take advantage of the higher sink current that most outputs provide. If it does source current it will just drive it to the off state, and it doesn't matter.
Let's say you have a +5 supply, and the LED on voltage is 2V, then there will be 3V on the current limiting resistor, ignoring the sat voltage of the output device. Select R to provide whatever operating current you want for the LED as long as it is less than 10mA: might want to limit that to 8 for some margin.
Pete B.
So, if it is a reset out line and you issue a reset command it goes low.
The way that you're asking the question suggests what is called an open collector, or open drain output. This means that the output is not driven in both directions only to ground and this is why these are active low. The designer supplies an external "pull up" resistor, or simply connects the load from the pos supply through a current limiting resistor. OC outputs were often used in bussed configurations where they provide a wired OR logic connection. But they also have many other uses.
Some OC outputs are rated for 30V or more and allow 5V logic to interface to higher voltage external circuitry such as 24V relays.
So the question is, does this output have a source current, you've stated the sink current as 10 mA. Really, it doesn't matter, you can wire it as described above to take advantage of the higher sink current that most outputs provide. If it does source current it will just drive it to the off state, and it doesn't matter.
Let's say you have a +5 supply, and the LED on voltage is 2V, then there will be 3V on the current limiting resistor, ignoring the sat voltage of the output device. Select R to provide whatever operating current you want for the LED as long as it is less than 10mA: might want to limit that to 8 for some margin.
Pete B.
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Hi Pete,
Sorry I didn't elaborate enough, it is in fact an open-drain charge status output for a battery charging IC.
If I wanted to drive a LED to full brightness with say 30mA, I would then need to use a MOSFET correct?
Thanks
Sorry I didn't elaborate enough, it is in fact an open-drain charge status output for a battery charging IC.
If I wanted to drive a LED to full brightness with say 30mA, I would then need to use a MOSFET correct?
Thanks
I really suggest keeping it simple, there are LEDs that are happy with 2mA. If you must have 30 then use a 10K pullup on that output and just drive a plain old PNP transistor as an emitter follower with the series R-LED in the emitter circuit to +5V. 2N2907 or PN2907 should be fine or anything with a 500 mA rating.
Note my edits for stronger drive.
Pete B.
Note my edits for stronger drive.
Pete B.
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I will keep it simple then, which will allow me to keep the PCB at a micro 5.6mm x 7.6mm.
Thanks again for the help.
Thanks again for the help.
It doesn't; probably because it's such a simple thing that anyone even capable of using the IC should already understand. 😱
It's the MAX1555.
It's the MAX1555.
I have used that chip before. Just connect a LED in series with a current limiting resistor of about 470R-1k from the input to the status output pin. (If you need to use both power inputs, add a pair of 1N4148 diodes.)
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