I know that 12v goes in and 8.4v comes out, but I'm not sure what the roles of the elements in the middle and what process go through. Please explain this process easily. Please.!!!
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I know that 12v goes in and 8.4v comes out, but I'm not sure what the roles of the elements in the middle and what process go through. Please explain this process easily. Please.!!!
The LM317 is a regulator IC that produces a constant voltage. The output voltage is determined by R3(470),R4,R2.
Since R4 is a variable resistor, you can turn this and adjust the output so that 8.4V comes out.
Lithium rechargeable batteries are 4.2V ±0.05V per cell, so the voltage must be set correctly.
Q1, R1 are circuits that determine the maximum charging current. TR turns on when Vbe (base emitter voltage) becomes 0.65V. That is, voltage is applied across R1 by the current flowing toward -. V = IR is a voltage. Currently, R = 1 ohm, so V = I. When this voltage is over 0.65V, Q1 turns on, and then the current decreases as the output voltage of LM317 goes down. Simply put, controlling the current is not controlling the real current, but lowering the voltage to lower the current.
On the left, Q2, R5, and DS1 are indicators that indicate that they are charging. The role of R6 and D1 is important.
TR said that it turns ON when Vbe exceeds 0.65V. And the forward threshold voltage of D1 is also 0.65V.
Therefore, the base voltage of Q2 becomes D1 voltage (0.65V) + R1x I, so even if there is almost no charging current, Q2 continues to be displayed as charging. So, I put R6 to lower the voltage at both ends of D1.
In this way, the voltage across D1 becomes a little lower, and when a lot of current flows, the charging display is displayed, and when the charging current decreases to a certain extent, the LED turns off.
In other words, the lower the R6 resistance value, the dull the charge display sensitivity, and the higher the R6 resistance shade, the more sensitive the sensitivity becomes.