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Given the circuit below, I want to discuss some issues concerning the way the circuit works. Here is the circuit: enter image description here

Let me report what is wirten in the handout:

The circuit furnishes an initial voltage of 2.5 voltes per cell at 25C° to rapidly charge the battery. The charging current decreases as the battery charges. and when the current drops to 180mA, the charging circuit reduces the voltage to 2.35v, leaving the battery in a fully charged state. The LM301 compares the voltage across R1 to 18mv set by R2. The comparators output controls the voltage regulator, forcing it to produce the lower float voltage when the battery charging current drops below 180mA. The 150mV difference between charging and float voltages is set by the ratio of R3 and R4. The LED shows the state of the circuit.

Let's take the case of a 6V battery. To charge that battery, usullay it is suggested to charge it with a Voltage around 7.5V (2.5v per cell as it is mentionned). LM350 or LM317 have a 1.25 voltage reference. So how do we acheive a starting charging voltage of 7.5V from this circuit (When I did my calculation I did not end up with 7.5v but more with Vref=1.25v and Iadj=50uA).

The handout says that "LM301 Compares the voltage across R1 and R2", so the OPAMP here plays a role of a comparator (The output is either High or Low), am I right? In case I am right, how can this OPAMP be Low (sink current to allow the upper LED to be ON ) since its output (PIN 6) is taken form an emitter of a npn transistor (From the datasheet of the LM301).

How does the PNP transistor contribute to the selection of charging voltage (2.5V per cell of 2.35v per cell) ?

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2 Answers 2

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the LM301 datasheet shows and Sziklai pair Q12,Q11 pulling down on the output via R10 and R11, it is these components that pass the bulk of the LED current when the output is low. enter image description here

the PNP transistor is controlled via the pair of 4.7K resistors by the op-amp that is cating at a comparitor, the trasistor is either off or fully conducting

the 1.5K will pass about 10mA when 12V is selected or 5mA when 6V is selected , that will add about 1V or 0.5V to the voltage on the 100 ohm resistor, which itself lifts the whole LM315 regulator. (these figures are all estimations they are inaccurate but should be in the right ballpark.)

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  • \$\begingroup\$ Thanks Jasen, but if you look closer to the circuit of the OP AMP LM301, its output (PIN6) to R10 or Q16, it is directely connected to the emitter of Q15 and R11, if it is connected between R10 and Q16, it would be easier for me to understand it. \$\endgroup\$
    – learn design
    Dec 24, 2019 at 6:12
  • \$\begingroup\$ yeah, it's a long-ish path through those two resistors to Q11, but they are low resistance. \$\endgroup\$
    – Jasen
    Dec 24, 2019 at 6:15
  • \$\begingroup\$ Why do you say that 1.5K will pass about 10mA? if the transistor works as a switch, the maximum current that will pass trough 1.5K is (Vout-VCE)/1.6k, that will be less than 10mA in case of Vout=12V \$\endgroup\$
    – learn design
    Dec 25, 2019 at 21:12
  • \$\begingroup\$ Vout is 15V ("2.5 per cell"), but i said that because it was close I didn't want to do any hard arithmetic. \$\endgroup\$
    – Jasen
    Dec 25, 2019 at 21:37
  • \$\begingroup\$ Ok Jasen thank you very much, just one last thing, what would Vbat be in case of cell discharged, i.e when the battery is concidered discharged ?? \$\endgroup\$
    – learn design
    Dec 25, 2019 at 21:46
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The schematic shows 18mV across R2 which has a current sink from the LM334. The PNP can pull up the Vadj to turn off the 3 terminal regulator.

This voltage ladder across the 3.3 Ohm resistor must match the voltage drop across the current shunt 0.1 Ohms

Simulation seems to agree with description

I had the pot originally tuned to 760 Ohms below but later dropped it to 500 Ohms in link in comments for better performance at low battery.

enter image description here

This is way easier than computing it.

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  • \$\begingroup\$ Could you please illustrate the voltage ladder and how doest it set the Vout by some calculations, because I did not come out with something convenient, as I said in the question, I get Vout higher than the specified one!!!! So could you clarifier with some calculations ?? \$\endgroup\$
    – learn design
    Dec 24, 2019 at 6:25
  • \$\begingroup\$ sure I can do that in Falstad \$\endgroup\$
    – Tony Stewart EE75
    Dec 24, 2019 at 6:57
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    \$\begingroup\$ It charges more than 180mA at low voltage after I reduced the pot then when the voltage rises to 7.1 set by a voltage source here, the threshold matches ~ 180mA tinyurl.com/uyk4vp7 \$\endgroup\$
    – Tony Stewart EE75
    Dec 24, 2019 at 7:09
  • \$\begingroup\$ please let's focus on the first sentence of your answer "The PNP transistor can pull up the Vadj to turn off the 3 regulator". I think the PNP transistor is there just to add a certain voltage when it is fully ON (When the transistor is ON Vout=2.5v per cell, when it is OFF Vout=2.35v per cell) but I don't think that the transistor will turn OFF the regulator. \$\endgroup\$
    – learn design
    Dec 27, 2019 at 7:46
  • \$\begingroup\$ When Vout-Vadj drops below 1.25 the 3 terminal regulator is current starved emitter follower internally thus cuts output when PNP shunts that. \$\endgroup\$
    – Tony Stewart EE75
    Dec 27, 2019 at 7:59

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