2013-03-09 Light Deactivated Joule Thief

I built this up – tack soldered it together – before I drew the schematic.  My intention was to make a circuit that was very low on power when the light deactivated it, and to use parts that the neophyte experimenter would already have.  I wasn’t concerned about complexity or cost, but the three extra transistors, resistors and LED are relatively inexpensive.  As I said, the experimenter may already have them so it wouldn’t be any added expense.

Some experimenters connect a CdS photocell across the base to emitter of the Joule Thief transistor.  This shunts all of the 0.9 milliamp base current from the 1k to negative when light hits it.  Thus during the daylight hours, the circuit is still drawing nearly a milliamp, which is much less than the 50 to 80 mA during normal operation, but it is still substantial and would discharge the battery over a few months.

My choice was to not use a CdS photocell because they are much more expensive and difficult to obtain than a LED,  resistors and transistors.  Because the LED puts out so little current, it must have more than one transistor to multiply the current up to enough to turn the other transistors on or off.  Also, the sensor LED’s current output must be inverted so that it shuts off, rather than turns on.

Another characteristic of a CdS photocell is that the resistance is somewhat linear with light hitting it.  It doesn’t have a “knee” like a diode has.  As the light hitting a LED gets brighter, the output voltage finally gets to the point where it will reach 0.6V and turn on the transistor.  This helps make the deactivation more decisive.  With the CdS photocell, the resistance could decrease in a bright ambient light, but the circuit could continue to run, and all the current through the photocell is just wasted.  There is a way to change this, it’s called a Schmitt Trigger.  But this just adds complexity to the circuit with no significant benefits.

Circuit function

Normally the feedback winding would be connected to the positive, here it is connected to the collector of a PNP transistor.  During normal operation with no light on the sensor, the sensor LED and Q1 are open, and the 470k turns on Q2 and its collector current turns Q3 fully on, so it is as if the feedback winding is connected to positive.  The Joule Thief operates normally.

When the ambient light hits the sensor LED, the small current generated turns on Q1, which shunts all of the current from the 470k to negative.  Q2 turns off, and Q3 also turns off, and no longer lets current enter the feedback winding.  The Joule Thief cannot operate without the current to the feedback winding, and it turns off and the LED goes dark.  The amount of current drawn while the circuit is disabled is only microamps – just the current through the 470k.

How well does it work?

The circuit works just fine.  When there is no ambient light, the voltage drop across the Q3 is only .036 volts.  That’s low enough to make it look like the feedback winding is connected directly to the positive.  There is no discernible difference in the LED brightness when I short across the Q3 – if you short Q3 E to C, it’s just a regular Joule thief.  The light sensor is not very sensitive and has to be held about a foot away from my white LED task light to turn it off.  But this is very dependent on the type of LED used and the gain of the transistors, especially Q1.  In my case I used an amber LED that did not put out much light, and I think that it would be more sensitive if the LED was a superbright red LED.  The circuit will usually turn off when it’s outside during daylight, due to the light being much brighter than the indoor lights.  There’s a lot of room for experimentation with different transistors and colors of the LED to get various light sensitivities.  The JT is just a run-of-the-mill everyday conventional JT, so the LED’s light output is dependent on what and how it’s made.


3 Responses

  1. Paul says:

    In defence of the simple LDR.
    My experiments with LDRs suggest that they work well as a switching base to emitter shunt. With a JT that draws ~50 mA input current when on, light shuts it down to ~0.4 mA when off. I have a 3.3 K base resistor installed. That means a 100:1 ratio of off to on current which seemed good enough to be practical. Surprisingly perhaps, the threshold is quite crisp. With the larger surface area ones, I get good sensitivity and that makes up for a bit higher drain when off, I think. As an added bonus, I find that the light level that extinguishes the JT is lower as the battery drains lower, so it becomes more economical with its light output as the battery dies.
    I noticed at dusk with a saggy battery (i.e. voltage varying with current), it started to flash at about 1 Hz. That could be useful or irritating. Paul

    • admin says:

      I forgot to mention that when the base bias resistor is increased, the idle or standby current also goes down. 1.5V across 3.3k is only 0.45 mA, which would take several months to drain the AA cell. The high gain transistors such as the BC337-40 make this possible.

    • admin says:

      One more thing… I blogged the problem with the high gain BC337-40 in my “.. not a Joule Thief” blog under the Update at the bottom.
      http://rustybolt.info/wordpress/?p=4479

Leave a Reply

Your email address will not be published. Required fields are marked *

© RustyBolt.Info/wordpress
CyberChimps