Supply voltage is 5V - 9V.
Above is the schematic of my frens project.
Description : It will start GREEN, YELLOW then RED. When RED is ON, Buzzer is ON. The LED will repeat again, GREEN, YELLOW, RED........
After drawing the PCB, the problem my fren got was the buzzer is always ON whereas the LED is not even ON once. Is the any problem with the schematic??
I thought for an NPN to be ON, the Vbe should be > 0.7V rite?
In my case, R5 = 1k, R6 = 10k. When the output is 5V, voltage drop accross R5 = 0.454V, whereas R6 = 4.545V. So Vbe = 4.545V - 0.7V will be in ON state rite??
Of course, 10kΩ-1kΩ combination at 5V supply will not turn NPN on ..
Also, you have 1N4148 diodes, so the control voltage is ≈ 4V ..
Maybe 3kΩ-1kΩ voltage divider is better option ..
But, if everything works well you shouldn't need a voltage divider .. a 10kΩ resistor between diodes and the NPN's base should be enough ..
Better check where is the voltage comming from ..
Hi,
Remove R4, connect pin 11 to pin 15 of 4017 and short JP1.
I agree to swap the 10K and 1K resistor. Of course check if the clock o/p of the 555 (pin3) is there and of suitable rate (not too fast).
You need only ~0.7V to turn on the transistor at ~1mA base current. Also the transistor must always in OFF position, so 1K at the base should pull the base low enough to turn it off.
When op fron the counter is high, enough voltage drop across the 1K resistor to turn on the transistor. It's a voltage divider stuff you hv to experiment the values yourself.
i know that it need ~0.7V to turn on the transistor, but if i put 10K at the bottom and 1K on top, when the output is 5V, voltage drop accross 1K = 0.454V, whereas 10K = 4.545V. So Vbe = 4.545V - 0.7V will be in ON.
But if i put 1K at the bottom and 10K on top, will it actually ON???
really confuse now......