On the LM317, the main current flows from IN to OUT, not out of the ADJ pin. the ADJ current is < 100uA and is negligible. The 1.0 /1W resistor on the output pin is what sets the current limiting. It does this by maintaining a 1.25V drop across that resistor. Since the ADJ pin is sensing the voltage drop, it will regulate to whatever voltage is needed at the output to keep the 1.25 volt drop across the 1.0 ohm resistor. 1.25V across 1 ohm is 1.25A.
R8 is not in parallel with R7. Far from it.
When the CC pin is pulled to ground, then Q2 provides a low impedance path to ground, and this pulls the ADJ pin to ground. The output of the LM317T would go to 1.25V, and this essentially shuts off the charging, because the diode D3 would be reversed biased then (assuming it shutdown the charging because it reached the charge goal) The 240 ohm resistor is correct. You don't want much higher than this, or the voltage drop will be too large for sensing the output voltage, and the current limiting will be inaccurate. To visualize this, just calculate the voltage drop of 100uA across 240 ohms vs 100uA across 10k ohms, and you will see. This voltage drop across the 240 ohm resistor will add to the error in regulation, so you don't want it to be too large.
Also, if pin CC is tied to ground, and Vcc = 5V, then the current through R6 and into the base of Q2 is approximately 1.4 mA, so depending on the voltage across the collector and emitter of Q2, a collector-emitter current of approximately 140 mA (assuming a typical gain of 100 for the 2N3904) could flow into pin CC, but where would this come from if the max current the ADJ pin can source is 100 uA, does this mean current could flow out of Vout, through R7 and R8 and then down through Q2? and if so, the current could be much larger than the 10 mA sinking current capability of the bq2002 indicated from its spec sheet. I think I’m going very wrong somewhere.
When CC is grounded, the transistor Q2 will saturate, and the ADJ pin will be at 0 volts (not exactly, but close). The current flowing out of the ADJ pin is internally limited to no more than 100uA. No more will flow from ADJ. That diagram also has a 10k R15 from the Vin (max 25V) to Vout. So if Vout is 1.25V and Vin is max 25V, then the current through R15 is (25-1.25)/10k = 2.4mA. the R8+R7 are in series to ground through Q2. So 1.25/(R8+R7) = 1.25/241 = 5.19mA . Add this to the 2.4mA from R15 and the total going into the collector of Q2 is 5.19mA + 2.4mA + 100uA = approx 7.6mA. If the hfe of Q2 is 100, then the base of Q2 is only needing 7.6mA/100 = 76 uA. This is all flowing into the CC pin of the control chip, and it does not exceed the 10mA sink capability.
Finally, the data sheet says that Vcc is regulated to 5V +- 5% from an external DC input, from JP4 on the top left of schematic which is recommended to be between 10V and 25V. But from the schematic, I can’t understand how the regulated 5V is obtained? I can’t see how it’s obtained from the LM317T and since the DC input can vary, is another voltage reg IC needed for this?
Vcc comes from the input, regulated by D1, which is a 1N751. That's a 5.1V zener diode. Q1 is there to connect the DC input to the zener only when a battery is attached, so it doesn't consume power when there is no battery. With a battery attached, a small current flows from the DC input jack, into the emitter of Q1, out the base, into the battery positive terminal, out the battery negative terminal and to ground and back to the DC ground, turning on Q1. This supplies 5.1V Vcc to the circuit. If you unplug the battery, Vcc also turns off.
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