Need help in understanding the working of battery level indicator

Question

I have got hold of a battery level indicator on net. I works but I am not able to understand how it is functioning. I am new to electronics and I just dont want to build a circuit from a diagram. I want to understand the working clearly. I have tried a lot but cant understand how to start.

The circuit works. Not very well but it works. When the battery voltage = 12V all LED light up When Batt vol = 10 v only 4 light up with 8V three light up ...and so on

Although the range is quite large but still it works.

Now I am not sure how this is working. I am including a image of the circuit diagram

From where should I start analyzing this. I hope I should start from the right most transistor. But not sure. If someone can help me understand this, it would be very helpful for me.

The circuit can be simulated in https://www.falstad.com/circuit/ by importing the below text

$ 1 0.000005 1.3241202019156522 45 5 50
t 352 320 400 320 0 1 -4.33543346557741 0.6206926084618494 100
t 544 320 592 320 0 1 0.6196395528457068 0.6622667997764897 100
t 736 320 784 320 0 1 0.6567047725079855 0.6835911623784181 100
t 896 320 944 320 0 1 0.6720640836582178 0.6956916987466071 100
r 336 144 336 240 0 10000
r 704 144 704 256 0 10000
r 192 64 192 208 0 10000
r 192 256 192 384 0 10000
r 416 192 416 128 0 1000
r 608 224 608 144 0 1000
r 800 224 800 160 0 1000
r 960 240 960 160 0 1000
162 416 80 416 128 2 default-led 1 0 0 0.01
162 608 80 608 144 2 default-led 1 0 0 0.01
162 800 80 800 160 2 default-led 1 0 0 0.01
162 960 80 960 160 2 default-led 1 0 0 0.01
w 192 64 416 64 0
w 416 64 416 80 0
w 416 64 608 64 0
w 608 64 608 80 0
w 608 64 704 64 0
w 704 64 800 64 0
w 800 64 800 80 0
w 800 64 960 64 0
w 960 64 960 80 0
w 192 208 192 240 0
w 192 208 304 208 0
w 304 208 304 144 0
w 304 144 320 144 0
w 320 144 336 144 0
w 704 256 704 320 0
w 704 320 736 320 0
w 400 304 400 192 0
w 400 192 416 192 0
w 544 336 544 320 0
w 592 304 608 304 0
w 608 304 608 224 0
w 704 336 704 320 0
w 784 304 800 304 0
w 800 304 800 224 0
w 960 240 960 304 0
w 944 304 960 304 0
w 784 336 896 336 0
w 896 336 896 320 0
t 1152 320 1184 320 0 1 0.6825152937173588 0.7046520376169576 100
r 1184 160 1184 240 0 1000
162 1184 80 1184 160 2 default-led 1 0 0 0.01
w 960 64 1184 64 0
w 1184 64 1184 80 0
w 1072 336 1152 336 0
w 1152 336 1152 320 0
w 1184 240 1184 304 0
w 1184 336 1184 432 0
w 1184 432 976 432 0
w 976 432 976 448 0
w 976 448 944 448 0
w 944 448 704 448 0
w 704 448 192 448 0
w 192 448 192 384 0
w 192 256 192 240 0
w 192 64 64 64 0
w 192 448 48 448 0
w 704 144 1072 144 0
r 1072 176 1072 288 0 10000
w 1072 144 1072 176 0
w 1072 288 1072 336 0
v 80 272 80 160 0 0 40 12 0 0 0.5
w 64 64 64 160 0
w 64 160 80 160 0
w 80 272 80 368 0
w 80 368 48 368 0
w 48 368 48 448 0
w 336 144 704 144 0
w 336 240 336 272 0
w 336 272 336 320 0
w 400 336 544 336 0
w 336 320 352 320 0
w 944 336 1072 336 0
w 592 336 704 336 0

Answer 1

Imagine that all the NPN transistors behave as a controlled switch. The switch contacts are between collector and emitter. When the switch closes, the LED turns on.

That switch is controlled by the Base-Emitter voltage \$V_{BE}\$. \$V_{BE}\$ needs to be more than 0.7 V for the switch to close and the LED to turn on. That \$V_{BE}\$ input of the transistor behaves very similar to a diode (it is a diode!). This means that when \$V_{BE}\$ < 0.7 V there is (almost) no current flowing between Base and Emitter. When more than 0.7 V is applied the Base-Emitter operates in forward mode and it just "eats up" all current that is fed to it. This means the voltage will stay close to 0.7 V.

Note how all the Base-Emitter connections are all in series. If nothing else was connected then to make all 5 \$V_{BE}\$ = 0.7 we would need to apply 5 x 0.7 V = 3.5 V.

But there are some 10 kohm resistors as well. I will call the transistors Q1 to Q5 from left to right. The one on the right will pull up the voltage on the Base of Q5 (on the far right). This means that Q5 will be the first to see 0.7 V across its Base-Emitter when the battery voltage increases.

As the battery voltage further increases the \$V_{BE}\$ of Q5 will stay 0.7 V. In a similar fashion the other 10 kohm resistors will apply voltage to the other Base-Emitters. As the Base of Q5 is 0.7 V, there will need to be 1.4 V at the base of Q4 before it will start to conduct. In the same way Q3 will need 2.1 V at its Base. So from right to left each transistor will need a higher battery voltage to start conducting.