I'm making an LED flasher as an indicator, for use on a 12 to 15v supply. Directly driving an LED using a TS555CN timer IC. around one second off and a few milli-second on, just long enough to spot, so as to minimise current consumption. LED will be between +ve and output, so sinking current, but I'm struggling to track down data on how many milliamps this IC can safely sink on the output pin. Anyone know?
If he LM555CN is the same, then according to High Level Output Current: 50 mA Low Level Output Current: - 200 mA Pd - Power Dissipation: 600 mW
A search using the part number and the keyword 'datasheet' reveals many hits. A random one reveals an absolute maximum I-out of +/- 100mA but recommended operating conditions of output sink current 10mA and output source current of 50mA. The significance of the recommended operating conditions is that it may well prove difficult to produce a reliable practical circuit using much higher currents, though you are allowed to sink more if you can get everything else right, including dissipation and circuit function. So I recommend not going much above the recommended 10mA.
If the LED is a modern one, rather than a 1970s one from the junk box, 10mA continuous is likely to be blindingly bright. So I wouldn't worry about needing the maximum current if it's just an indicator. You could also probably pulse the LED harder for a shorter length of time if needs be too.
Theo
They're not the same. The TS555CN is a CMOS low current device. See Mouser's entry for the right part or my post above.
I thought you could get leds that did this already? Brian
If it were really important to have a higher peak current for a very short time than the device could sink then you could use a capacitor which is then discharged more slowly by the active device. But of course the more average power used the less the OP's design criterion of low power is met.
If you want to sinc more, surely 1 transistor should be able to suffice. Brian
There are but they tend to wink rather than a brief pulse flash that the OP wants.
There is (was?) a dedicated LED flasher chip that used a charge pump and capacitor to very effciently produce a bright pulsed LED. LM3909?
Note that's a "bright pulsed" mid 90's LED. A modern LED will be extremely bright...
Use a bipolar transistor as a buffer ?
NPN
PNP
Circuit diagrams of this quality, redo your Hfe and/or bias calcs. Today, there's hardly a situation where I'd use a 2N3055, but at one time, I used to drool at the chance to own one. The Hfe is kinda low on those. Like driving a nail with a dull hammer.
Also, consider whether the result "inverts" or not, and adjust your 555 design accordingly.
Another way to do this, is with a CD4538, with the two units tied tip to tail. CD4000 series are current limited in behavior, and sink or source
10mA on a good day. Definitely needs a bipolar buffer. These are flexible enough, you can make any waveform you want, with free Q and Qbar outputs for a free inversion.(This is a PDF file, add a file extension.)
Page 6 here, gives an idea how to emulate a 555 using the two monostables in a 4538. You have more options with stuff like this.
Fairchild Semiconductor Application Note 138 May 1975 "Using the CMOS Dual Monostable Multivibrator"
Yes, not guaranteed to start or some such. Look at the logic table carefully, when selecting an input method, to make sure you've covered that off.
I'd probably just grab a 2N2222 or similar from the bin, for a start at a buffer.
And with the efficiency of modern LEDs, you might be surprised how far 10mA goes. With a lens, such a LED could be annoying.
Paul
I have some "high efficiency" (ultra bright) LEDs running at just under
100uA and they are bright enough to be seen on an overcast day from 5+ metres away. These 5mm LEDS are specified at around 15000mcd brightness at 20mA but with much reduced brightness at my operating 100uA. I'm running two LEDs continuously from 4 off AAA alkaline batteries which last 3 to 4 months before require changing.
If the OP has 15V to work with, the series resistor wastes most of the energy.
You could put five 2.5V white LEDs in series, for 12.5V drop. Leaving only 2.5V for the dropping resistor and the Vcesat. And then get five times the light. Like, point the LEDs in different directions so you can see the indicator from more orientations.
An opportunity-rich project, suitable for breadboard first.
My first 555 project, was adding two yellow LEDs to a frisbee, so you could throw the frisbee at night.
Paul
I see that is the correct device. That states the 50/-10mA output as you noted. Any idea why it should say the Power Dissipation is 1250mW? That figure doesn't appear in the data sheet pdf link on the webpage.
anyone who puts 1,25W through a DIL 8 pin package is in for some magic smoke. 125mW is more likely
Supply current max 250uA times VCC=5V gives 1250 uW (page 9).
Maybe the units got screwed up.
1.25mW is pretty good.Paul
No idea. Although you can derive a figure from the thermal characteristics, maximum operating temperature and an arbitrary assumption about case temperature.
Maximum dissipation is typically a marketing figure, based on totally unrealistic heat sink arrangements. With power devices they tend also to give you a realistic figure based on credible heatsinks. But with a CMOS device generally you are using it to save power. (Except for high frequency use where you would still not be able to approach the headline maximum dissipation but it might get quite hot.)
Actually you cant.
I tried and at best I came up with 1W on a very good thermally conductive PCB, for 25C ambient.
And as experienced designer in these matters no kit that didn't expect to get to around 50C inside a cease ever survived.
So you have at best 100C to play with, for a 150C junction temperature and the spec says IIRC 125C per watt - so around 800mW at the most. But its a pretty useless spec anyway since its unlikely that a chip that can only source 50mA at 5V - 250mW - into an external load, is going to be able to dissipate 750mW unless its output is completely shorted.
Actually they normally give two totally hopeless figures - maximum dissipation on an 'infinite heatsink' and maximum dissipation in free air, both at 25C.
To work out actual permissible dissipation you need to know the thermal resistance of the heatsink - or PCB if that is what its mounted on - and the likely ambient temperature.
During my time designing power amps in Africa, a black heatsink in the sun could reach 80C before even being switched on...and ambient temps in discothèques were often up in the sweaty 40s.
In the end we simply built them and raised the thermal cutout switch point until they didnt pack up in normal use or die from heat either.
I didn't say it was a useful figure - like most marketing its only purpose is to sound good. I just said you could derive it! How about
40degC/W junction to case thermal resistance, case temperature 70degC, max junction temp 150degC? That gives you 2000mW. Nonsense, I agree.HomeOwnersHub 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.