From: wthurt on
I would like to use the LM334 with 9V batteries to produce an
approximate 3mA constant current through a diode. But I am a little
confused by the device schematic. This device has three leads (V+,
ADJ, and V-) --- I have put a 25 ohm resistor between ADJ and V- to
select the current I want. But I'm not sure where to put my diode?
where does the constant current flow?
From: Jasen Betts on
On 2010-05-30, wthurt <wthurt(a)gmail.com> wrote:
> I would like to use the LM334 with 9V batteries to produce an
> approximate 3mA constant current through a diode. But I am a little
> confused by the device schematic. This device has three leads (V+,
> ADJ, and V-) --- I have put a 25 ohm resistor between ADJ and V- to
> select the current I want. But I'm not sure where to put my diode?
> where does the constant current flow?

point (A) or point (B) whichever is most suitable.

|
(A)
|
|V+
-----
|LM |R
|334|----.
| | |
--- 25R
|V- |
+------'
|
(B)
|




--- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: John Fields on
On Sat, 29 May 2010 22:08:10 -0700 (PDT), wthurt <wthurt(a)gmail.com>
wrote:

>I would like to use the LM334 with 9V batteries to produce an
>approximate 3mA constant current through a diode. But I am a little
>confused by the device schematic. This device has three leads (V+,
>ADJ, and V-) --- I have put a 25 ohm resistor between ADJ and V- to
>select the current I want. But I'm not sure where to put my diode?
>where does the constant current flow?

---
Into V+ or out of the junction of Rset and V-, so you can put your
diode between the supply and V+ or between the junction of Rset and V-
and ground.

From: whit3rd on
On May 29, 10:08 pm, wthurt <wth...(a)gmail.com> wrote:
> I would like to use the LM334 with 9V batteries to produce an
> approximate 3mA constant current through a diode.

A resistor and two-transistor current mirror will give you constant
current
in the 0.2 to 5V range of Shottky through LED diodes... use a 2.7k ohm
resistor. Unless you need high accuracy, the IC is just an expensive
bit of overkill.