Electronic – Transistor not supply enough current

ledtransistorszener

I am trying to run a 100W 32V Led with a DC-DC boost converter from a 6 volt battery, and I want to make it so it will shut off when the battery voltage gets low. So I am using a TIP122 transistor with the collector connected to the ground input to the Boost converter, The emitter connected to the battery ground, and the base connected to a 5.6V zener diode. The LED takes 3.3A, but even with a base current of 60mA, the transistor is only supplying 600mA.

Am I doing something wrong?

schematic

simulate this circuit – Schematic created using CircuitLab

Best Answer

The things you are overlooking and how, in a nicely ordered list are as follows:

  1. The LED takes 3.3A. 100W at 6V is 16.7A. That's assuming 100% efficiency, which is much more likely to be 80%, which would make 20.8A at 6VDC input. But even in "perfect" conditions you are trying to "melt" your transistor that has a maximum current capability of 5A.
  2. Transistors do not only have base current and amplification, but also base voltage and saturation voltage. The latter two for a darlington like the TIP122 are incredibly high. Just look at the datasheet.
  3. You are, I assume calculating the base current, because the 5.6V zener in the 1N47-- series (which is NOT the 1N4736A, that's the 6.8V one) has it's approximate voltage at 45mA, but with a reasonable margin.

To fix the situation, on 6VDC you are going to not want to use a Darlington, but another scheme, such as just a power transistor driven by another PNP transistor, so they can both saturate and actually get the saturation voltage of 0.5V or below, in stead of 2V (leaving ONLY 4V for your unit, i.e. 100W at 80% conversion over 4V gives 31.25A !!!).

Or by using a MOSFET setup.

And then make sure you properly calculate all the set-points for the transistors and zener diodes using their datasheets.

You might also want to think about some hysteresis. If you turn off the battery drain at 5.5V, for example, then the battery will lose it's drain and jump up in voltage. This can cause:

  1. Oscillations

or more likely:

  1. A transistor or MOSFET that starts hanging between on and off dissipating a BUCKET of energy to regulate the current slowly down to keep the battery voltage at the set-point.

To blatantly advertise myself, a lot of steps about transistors can be found here:

https://electronics.stackexchange.com/a/174703/53769

(they do ignore saturation voltages, because that makes it more difficult, but for a Darlington it's important: They will just always waste 1.5V to 3V depending on the type, no matter what you do).

And more self advertising (since it might help) an answer about adding switching hysteresis to MOSFET switches, so they turn on at one voltage and off at a voltage a little below that, to prevent stuttering:

https://electronics.stackexchange.com/a/132634/53769

(You can leave out the diode, that one came from the OP, but was never actually explained)

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