To drive a LED, the current should less than 25uA at turn off state. ( The supply voltage now is about 2.8 ~ 3.3V )
At turn on state, the driving current should be larger than 7mA to avoid low beta of transistor, in this situation, the supply voltage is about 2.4~2.6V.
The Vf of LED is 1.9 ( somebody's measurment result, no other informtion ..)
--------------------------------------- Some of my other questions:
Hi, Boki. You seem to be impervious to advice here.
Yesterday, Mr. Fields offered you a perfectly good one-chip solution for your problem. Looking at the post through Google Groups, the link was cut, but it's available if you click Show Options and click "Show Original". You can then cut&paste.
Look at the LTC1440, as suggested by Mr. Fields. With one IC and two
1% resistors, your problem is solved. Total supply current is less than 4uA over temp, plus whatever you need for the resistive voltage divider (should be less than 2uA). The device is optimized to source current, but will sink 8mA with an output voltage of less than 300mV.
If you want to solve your problem, try Mr. Fields' solution. Go to the Linear website and look at the data sheet. Your circuit is the Typical Application on the bottom of page 1. Choose other 1% resistors to get the trigger voltage you need.
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Considering your skill level, you're not getting under 15uA off-stste with discrete transistors. Linear has already (re)invented the wheel. And since you've been casting around here, nobody's offered to do your job for you.
If there's real money involved, and the extra dollar is that important, there are many good EEs who will be happy to consult with you about your problem. Subcontract one.
It sounds like you're a Pioneer/Ranger campaign contributor who's lucked into a really good Homeland Security contract. I wish you well.
When Vcontrol is low, the FET turns off, and no current flows. When Vcontrol is high (around 2.5 V), the FET has a very low resistance (be sure to pick a FET which has a low Vth), and can be viewed as a short circuit. So now the voltage at the BJT's emitter will be one diode drop below the voltage at its base. So, R3 will have an essentially constant current through it, which means the LED will have an essentially constant current. Note that the BJT is NOT saturated in this application, so you can count on fairly high Beta.
If you choose R1, R2, and R3 optimally, you can probably get a reasonably good circuit. Maybe 25 Ohms for R3, then pick R1 and R2 so that the base is at around 1 Volt. So, R2 = 1k, and R1 = ~1.5 k might work.
I would try to keep Vcontrol/(R1+R2) > 10 * 7 mA / Beta_min if you know what I mean.
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