Remember the V(CE) vs I(C) curve and consider an NPN transistor. Thus forward biasing simply means there is positive voltage on the p-type junction relative to an n-type junction.
Thus, in our case forward biasing the BC junction means having a voltage V(BC) > 0.
Now, when collector voltage V(C) drops below the base voltage V(B) and forward biases the collector‐base junction (when V(BC) > 0):
The strong electric field which opposes the movement of the charged particles into the collector is now weakened.
There is now a numerous charged particles into the collector, hence current increases and the current gain factor, β,
decreases.
This corresponds to a logical more of "0" where the transistor acts as a switch, showing a low resistance path for current conduction.
The things you are overlooking and how, in a nicely ordered list are as follows:
- 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.
- 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.
- 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:
- Oscillations
or more likely:
- 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)
Best Answer
Good.
Then the transistor will drive into saturation.
There are many times when this is useful - most notably in the NPN low-side switch. Here we want the transistor to behave like a switch so we inject a base current high enough to ensure that the transistor saturates so that the collector-emitter voltage, VCE is as low as possible. This eliminates variation in load current due to variations in the transistor gain and also ensures that power dissipation in the transistor is at a minimum.
Figure 1. A typical NPN low-side switch can be driven into saturation by making R1 low enough. Source: LEDnique by the author.
To ensure saturation it is normal to assume a much lower current gain than the hfe paramater might suggest. 10 to 20 is typical. See the linked article for a worked calculation.