Re: Simplest current regulator

On Jan 30, 3:25 am, Fred Bloggs <nos...@xxxxxxxxxx> wrote:
bill.slo...@xxxxxxxx wrote:
On Jan 29, 5:07 am, whit3rd <whit...@xxxxxxxxx> wrote:

On Jan 28, 4:03 pm, Phil Endecott <spam_from_usenet_0...@xxxxxxxxxxxx>
wrote:

Of course I can use a zenner with a potential divider in an op-amp
circuit, but I need help to use it in something simpler.

Well, I'd think an LM10 and pass transistor is fairly
simple (half a square inch of circuit board), but there's
another option that hasn't been discussed yet: a current
mirror.

Just run a resistor from V+ to base of two transistors,
both emitters to GND, and connect collector 1 to base,
and LED from V+ to collector 2.

The trick is, the transistors have to be on a common heat sink
and the first has to have much smaller emitter area than the
second (the collector current is scaled by the emitter area,
all else being equal).    LED current is N*((V+) - Vbe))/R
and the 'N' represents the area ratio.

ICs use this kind of trick all the time, but buying transistors in
onesies it's hard to know what N is going to be.

You can buy dual transistors - Farnell list a bunch, of which the
PBSS4240Y looks good - and mimic the effect of emitter area by
arranging the Vbe for one transistor to be higher than that of the
other. At room temperature 86mV of extra bias gives you a 30:1 current
ratio.

You'd use a 300R resistor to draw some 12mA from the battery; 10mA of
that would go into the collector of the Vbe-setting transistor, 1mA
would go through an 82R resistor to the base of the Vbe-setting
transistor, and from there most of that 1mA would go on to the other
terminal of the battery through a 680R resistor across the Vbe of the
Vbe-setting transistor, while the remaining 1mA would provide the base
current for the other half of the dual transistor, which would be
sinking roughly 300mA from the battery via the white LED.

Is this what you're talking about, a Vbe multiplier with current source
setting???:

Exactly. 82R/680R should get you in the ball-park. You'd need to trim
a 200R pot in series with 560R to get exactly 300mA through the LED at
any specific battery voltage.

   View in a fixed-width font such as
                Courier.

.
.
.
.       .------+--------------------.
.       |      |                    |
.       |      |                   ---
.       |    [300]                 \ /
.       |      |                   ---
.       |      |                    |
.       |      |                    |
.       |      |                    |
.       |      |                  |/
.       |      +-------+----------|
.       |      |       |          |>
.      ---     |      [82]          |
.       -       \|     |            |
.       |        |-----+            |
.       |       <|     |            |
.       |      |     [680]          |
.       |      |       |            |
.        ------+-------+------------'
.
.                   ZXTDAM832

                     2X NPN

You will end up with a current hogging problem when the high current
transistor approaches saturation. This will have a double whammy effect
of 1) lowering the Vbe of the Vbe-setting transistor, and 2) increased
base drop due to base resistance of the high current transistor. My
initial estimate will be a LED current roll off to 100mA or less. Then
the current setting has a substantial dependence on Vbatt, maybe 15-20%,
like all other reference-less schemes. It's a good idea but will not
handle low battery.
The LED is almost certainly one of these:http://www.lumileds.com/pdfs/DS25..pdf
They have very substantial dynamic resistance in the 1-2.5R range,
making them easy to drive with a voltage source.

Any circuit is going to stop working when the battery voltage gets too
low; as the right-hand transistor goes into saturation not only will
its Vbe rise but also its current gain will fall.

The purpose-built LED drivers that use a switched inductor to control
the LED current should give a longer battery life than anything we can
get out of a simple bipolar-based circuit.

I wouldn't like to drive even these LEDs from a voltage source - the
resistance sn't all that high, and the forward voltage is temperature
dependent and falls with increasing temperature, which - since the
junctions run fairly hot - offers the chance of developing a nett
negative resistance over periods longer than the thermal time-constant
of the package.

--
Bill Sloman, Nijmegen

--
Bill Sloman, Nijmegen
.

Relevant Pages

  • Re: Current flows from collector to base of PNP transistor
    ... I was surprised, however, to find that when no power source is connected the batteries seem to discharge into the circuit at about 20 mA. ... There is some sort of "parasitic" effect in the transistor that I need to consider. ... So I will waste current that could be going into the battery, and I potentially have more current flowing into the LM334 than it is designed to cope with. ... That shows that the base-emitter junction is a diode, with the current flow in the direction of the arrow. ...
    (sci.electronics.design)
  • Re: 555 - Adjustable duty cycle circuit questions
    ... I have a circuit consuming 3.8mA comtrolling another circuit consuming ... The transistor controls a LM317 regulator set to ... The motor has the required ... but not well from a 9V battery. ...
    (sci.electronics.basics)
  • Re: Help Getting Started - Simple DC Circuit
    ... >> (I apologize for posting this basic circuit question. ... >> the switches will be closed and all 30 LEDs will be glowing. ... >> unplug the battery, as the machine is ok once all the switches are closed. ...
    (sci.electronics.basics)
  • Re: Astable Multivibrator help!
    ... In order to guarantee oscillation, you need to add a diode logic OR function which provides base drive only if one OR the other of the transistors is OFF, like shown below, a lock-up cannot occur because there will be insufficient base drive. ... Oddly, with the two diodes, it takes the circuit about 20 seconds to start up. ... the circuit will fire the LEDs up immediately and start the alternation in one or two seconds. ... Even at 9V, which I assume means an alkaline battery that will spend most of its time in the 7.2-8.4V range, you have the caps charging to ...
    (sci.electronics.design)
  • Re: Help! Capacitive coupled transistor setup
    ... It is the reference circuit from Kodak for their CCD imaging sensors. ... with the emitter of each transistor and its corresponding ... resistor to the base of one of the transistors. ... end of the resistor to V high and observe the LEDS, ...
    (sci.electronics.design)