Need LED driving circuit. ( <15uA and >8mA)

[Boki]

Dear All,

Here is the requirment.

1. To drive a LED, the current should less than 25uA at turn off state.
( The supply voltage now is about 2.8 ~ 3.3V )

2. 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:

1. Why beta will be changed?
2. The Vf of LED is normal ?

Best regards,
Boki.

 

[Chris]

Dear All,

Here is the requirment.

1. To drive a LED, the current should less than 25uA at turn off state.
( The supply voltage now is about 2.8 ~ 3.3V )

2. 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:

1. Why beta will be changed?
2. The Vf of LED is normal ?

Best regards,
Boki.

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.

http://www.linear.com/index.jsp

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.

Git R Done
Chris

 

[Boki]

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.

http://www.linear.com/index.jsp

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.

Git R Done
Chris

Hi Chris,

Chip solution is good, but my project is mass production.

I think I have to consider discrete solution first..., am I right?

Thanks.

Best regards,
Boki.

 

[Chris]

Chip solution is good, but my project is mass production.

I think I have to consider discrete solution first..., am I right?

Thanks.

Best regards,
Boki.

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.

Git R Done
Chris

 

[Mac]

Dear All,

Here is the requirment.

1. To drive a LED, the current should less than 25uA at turn off state.
( The supply voltage now is about 2.8 ~ 3.3V )

2. 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.

What transistor?

The Vf of LED is 1.9 ( somebody's measurment result, no other
informtion ..)

---------------------------------------
Some of my other questions:

1. Why beta will be changed?
2. The Vf of LED is normal ?

Best regards,
Boki.

You can drive the LED with a current source. You can use a FET in series
with the whole mess to keep the off-current basically zero.

Here is a circuit that might work:


VCC
Vcontrol-----+------+ |
| | |
| \ V LED
| / R1 -
| \ |
| / |
| | /
| +------------|
| | \>
| \ |
| / R2 \
| \ / R3
| / \
| | /
| | |
| +--------------+
| __| d
| ||
+-----------------|| N-channel
||__ s
|
|
GND

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.

--Mac

 

[Boki]

Mac 寫�：

What transistor?


You can drive the LED with a current source. You can use a FET in series
with the whole mess to keep the off-current basically zero.

Here is a circuit that might work:


VCC
Vcontrol-----+------+ |
| | |
| \ V LED
| / R1 -
| \ |
| / |
| | /
| +------------|
| | \>
| \ |
| / R2 \
| \ / R3
| / \
| | /
| | |
| +--------------+
| __| d
| ||
+-----------------|| N-channel
||__ s
|
|
GND

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.

--Mac

It is professional and excellent!
Thanks.

Best regards,
Boki.