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LED, need smaller power consumption.

Hi All,

Circuit:

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Result(LED current, change V2 from 0V to 2.4V) :

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I wish Q62 consume lower current, but it will effect LED current.

I hope LED current can larger than 10mA ( due to small beta of Q64 when variation ) I wish consume less than 30uA in Q62, does it possible?

It seems that I can't use FET here, because the Vto almost

3.0V........... it is hard to drive... in my application, I have only 2.4V and 2.8V supply, am I right?

Best regards, Boki.

Reply to
Boki
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Yes.

No. The SI2312DS gives good Vgs (Vgs = 1.8 V at Id = 4.0 A).

Vdd +--------+----+--------------------+ | | | | .-. | .-. | 1Meg| | V | | V | | - | |1Meg - '-' | '-' | | | | | | |----+--------+ | | | | | | ||-+ | ||-+ | ||

Reply to
John - KD5YI

Mac =E5=AF=AB=E9=81=93=EF=BC=9A

No... customer ask 10mA .... I want to limit Q62's current but keep LED current is 10mA...

room temp...?, it seems that I limit Q62's current will effect LED current.

1=2E I have to use lower Vth (ex: 0.7V) MOS ( because my low power supply ), it seems a little expensive than 2N7000. 2=2E is there any single inverter chip ?

Thank you very much! Best regards, Boki.

Reply to
Boki

John - KD5YI =E5=AF=AB=E9=81=93=EF=BC=9A

This one looks really good.... do you know the price ?

Reply to
Boki

Change your newsreader to display text in a fixed font such as courier.

I forgot the LED series resistors as Mac pointed out.

If leakage is a problem in the first FET, change the drain resistor to 100k.

The SI2312DS is $.57 for one, $.35 ea for 3000 from Mouser. However, see James Arther's (dagmargoodboat) reply to your post regarding subject

2N7000's Vgs. He pointed out the BSS138 at $.05 or so. And, you don't need the high current capability of the SI2312DS.

John

Reply to
John - KD5YI

0k.

I got it! Thank you very much for your information!

Best regards, Boki.

Reply to
Boki

I'm sorry Boki, I didn't mean to confuse you. I was commenting on John - KD5YI's circuit, not yours.

The FET will start to leak a little from drain to source if it gets hot. This can be significant when you have a 1 M resistor in series. But now that I look at the circuit, I see that the LED will keep the drop across the resistor within acceptable bounds, regardless of leakage. Also, once you put in a current limiting resistor for the LED, the 1 Meg pullup on the gate of the second transistor is unnecessary, I think.

Yes. Check various major semiconductor manufacturers websites. TI and National come to mind. You can also try to search at digikey. You can also put a couple of inverters in parallel to get more current handling.

--Mac

Reply to
Mac

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