Roff
Well-Known Member
I started by somewhat arbitrarily deciding to use 8 TIP127's because it keeps the current below 2A (1.875A) which is well below the 3A spec for minimum beta=1000. In fact, the beta typically peaks at Ic≈2A (see Fig. 1 in the Fairchild datasheet).Thanks for the responses guys! The schematic in Roff's post is exactly what I was envisioning, but was second-guessing myself. Let me see if I have this straight (I reposted the schematic for easy reference).
**broken link removed**
Q2-Q9 each need ~1.4v B-E bias, which is provided when ~14mA flows through R1.
R2-R9 provide parallel transistor compensation. Assuming each transistor is equally sharing the 15A load, that means each transistor is passing 1.875A.
R2-R9, being 0.22Ω will have 0.4125v dropped across each @ 1.875A.
So really, each transistor needs 1.8125v bias (B-E of 1.4v and 0.4125v on R2-R9), which means 18mA will have to flow through R1 to bias each of those transistors (1.8125v/100Ω).
Since the hFe of each PNP is ~1000, that 18mA should be good for 18A total on all transistors.
Do you think I should maybe drop that 100Ω resistor to something like 47Ω to drive the bases a little harder? I just don't want to assume or count on each PNP having 1000 hFe (I know the datasheets says 1000 min gain, but I like to over-engineer).
Also, I see that R2-R9 are 0.22Ω. Can I use 0.1Ω so there is a little less v drop on them? Is there a rule of thumb to determine the "proper" size resistor in this application?
With 15A total collector current, the maximum total base current will be 15mA (min beta=1000). With Vbe typically 1.7V@1.9A (Fig. 2), and another 0.4V across the emitter ballast resistor (0.22Ω), the drop across R1 will be ≈2.1V (simulation says 1.9V). Thus, the current through R1 will be 21mA. Adding this to the maximum base current of 15mA, the Lm317 max current is 36mA. In the sim, the actual total base current is about 4mA, and the LM317 current is about 23mA, due to beta being higher than 1000 and Vbe being less than 1.7V.
You don't "drive" the bases in this circuit. They draw what they need, due to the feedback.
If you make the emitter ballast resistors 0.1Ω, variations between transistor Vbe's will result in more unequal distribution of currents from on transistor to another.
I just looked at the graph of the pulsed current and eyeballed the duty cycle as being about 40%. I just now checked it on the simulator, and it measures 6.1A. 0.707 has nothing to do with the duty cycle here. It is mostly determined by the dropout voltage of the circuit. In fact, when your batteries are fully charged (3.6V per cell), the duty cycle (and average current) drops by a significant amount.Why only 6A? I would think it would be closer to 10A (15A * 0.707), or am I way off here?