2017-08-19 Emitter Base Breakdown Can Damage Transistor 

From YT reply to Neilgn  2017-08-20

Take any NPN transistor and measure the gain, and write it down.  Then put a 1k resistor in series with the transistor’s emitter and connect the emitter to positive and base to negative of a power supply.  Adjust the voltage until 2 or 3 volts are across the 1k resistor.  The voltage across the E-B junction should be more than 5 volts, maybe 7 or 8 V.  Now remove the transistor and measure the gain again.  When I did this years ago, the gain went from over 200 to less than 150.  I don’t know what causes this, but I’ve never seen any transistor that doesn’t do it.  When I did this, I cut off the leads of the transistor afterwards, because I was concerned about the damage to the transistor. 

The JT is only about 50 to 60 percent efficient, so if you have just a single current limiting resistor, it and the LED could be more efficient than the JT.


2 Responses

  1. Stephen A says:

    I’m getting higher efficiencies at 6 volts in (4xAA) into 5s2p white LED’s (about 15 volts out). Ever tried that? Where the LED string comes off in between L1 and Q1, I use a schottky, and then a capacitor, usually a ceramic disk 10n at least.

    • admin says:

      The efficiency of a Joule Thief is its biggest disadvantage. The circuit has to switch hundreds of milliamps at very low voltages that most transistors can’t handle. The demands on the transistor are extreme.

      It will be more efficient if you reconnect the LEDs in parallel and use AA cells for 4.5V supply, with current limiting resistors. I calculated and measured the led brightnesses with a light meter.

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