Help with TI bq2002 IC battery charger circuit - Page 1
Forum section: electronics / beginners

author: jmk62
author_rank: Newbie
timestamp: May 11, 2013, 02:38:05 pm
content: Hi guysI'm new to this site, and new to electronics in general. I'm trying to teach myself by building some simple hobby projects. The first is a 2 cell NiMH battery charger and I'm planning to use the TI bq2002 IC. There was a previous post in this site regarding this application and there was some helpful information posted. Seehttps://www.eevblog.com/forum/beginners/battery-charger/. I followed a link posted in the discussion to a spec sheet for a battery charging system developed by TI and I want to adapt it to my application. The link to the spec sheet ishttp://www.ti.com/lit/ug/sluu007b/sluu007b.pdf. If anyone has some spare time and the knowledge to help me out, that would be fantastic.I’ll be referring to the circuit on page 3 of the link to the spec sheet posted above where the LM317T is used as a current regulator. I can’t seem to understand how the LM317T is configured to achieve this. A quick search for the specs of the bq2002 IC show that pin CC is an open drain output, which when switched to a high impedance (I assume floating) then it should enable charging current to flow, and when connected internally to ground inhibits charging current. If pin CC is floating, then current from the ADJ pin can’t pass through the high impedance path along Q2, so will pass through R8 and hence charge the batteries. I’m guessing R8 doesn’t affect the supply current from the LM317T as when its connected in parallel with R7 across Vout and ADJ, the equivalent resistance is approximately 1 ohm and therefore the output current is still approximately 1.25 A flowing through R7. But how is current charging of the batteries inhibited when pin CC on the bq2002 IC is tied to ground? I guess any current that flows from ADJ (which the LM317 spec sheet indicates is between 50 and 100 uA) will flow through Q2 and not R8 if the path through Q2 has a significantly lower impedance path, so does this mean there is effectively an open circuit between Vout and ADJ pins on the LM317 and hence no current flows out preventing charge flowing to the batteries? If so, is the 280 ohm resistor large enough to achieve this, or is my thinking way off the mark?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.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?

author: codeboy2k
author_rank: Super Contributor
timestamp: May 11, 2013, 05:10:00 pm
content: 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.QuoteWhen 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.QuoteVcc 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.

author: jmk62
author_rank: Newbie
timestamp: May 11, 2013, 06:14:24 pm
content: Thank you so much codeboy2k. Perfect explanation. Took me a few read through's and some quick maths but I'm much more confident now on my understanding of the circuit. I really was way off before your reply so many thanks for your time and clear explanation. I've never seen a diode being used as a voltage regulator so that idea didn't even cross my mind but just had a quick read through some articles on it and the idea seems pretty neat. Right, now to order some parts. Many thanks again. I can now enjoy my weekend.

author: codeboy2k
author_rank: Super Contributor
timestamp: May 12, 2013, 06:17:53 am
content: You're welcome. Keep at it, learn by doing, and it's great that your trying to understand it better by asking questions.Zener diodes are useful and cheap regulators and voltage references, and are even used inside the LM317T as a reference.They can be used as a reference voltage to make a better regulator than would exist if used alone.They can be used as a voltage controlled switch.They can be used for overvoltage protection.They can be used for input clamping (a form of overvoltage protection)They can be used in place of a resistor, to drop a voltage from an input terminal, when you have a wide input current range or an unknown input current. A resistor would drop different voltages based on the current across it, and a zener will always drop the zener voltage across it, to some maximum power handling capability (Watts = Volts Dropped x Current) of the zener diode.Zeners are fun. Coming up with new ways to use them is fun

author: mansi1234
author_rank: Newbie
timestamp: November 29, 2015, 10:09:27 am
content: Can anyone tell what is the use of j4 in the link