In my cirucit, you will see I have a voltage divider on the base of each transistor to turn it off.
Without the voltage divider, the transistor will not turn off.
Is that enough proof?
Why don't you actually build the circuit and prove it will work, before putting it on the web?
In my cirucit, you will see I have a voltage divider on the base of each transistor to turn it off.
Without the voltage divider, the transistor will not turn off.
Is that enough proof?
Why don't you actually build the circuit and prove it will work, before putting it on the web?
Nope! It's not more proof than mine, which you don't accept.
I normally simulate circuits or parts thereof using Proteus ISIS before designing the circuit board. The actually built circuits either proved the positive simulation results or the flaws (as already shown during simulation).
On the other hand it is a matter of politeness to prove a statement, and not force others to prove your statement was wrong.
Your statement clearly said: "That circuit won't work."
So it's your turn to prove your statement!
My proof is as follows:
counter output H, timer output H - UB=0.872017V,
counter output H, timer output L - UB=0.372356V, which turns the transistor off.
Refer to the voltage probes at the base resistor shown in the attached simulation screenshots.
The circuit with the CD4017 doesn't flash the LED four times because its outputs Q0 to Q3 light the LED continuously then it turns off for two clock cycles before it is reset.
The diode connected to the base of the transistor won't turn it off but it might cause the LED to dim slightly. A Schottky diode might work if it is not too cold.
A trained monkey would use a Schottky diode. But, then again, he might say the voltage drops of the diode and the low from the 555 necessitates a voltage divider on the base of the buffer transistor. Trained monkeys are very smart, you know.
In the meantime I have put it on my website, under: Spot The Mistake, to help my readers design circuits that WORK.
A trained monkey would use a Schottky diode. But, then again, he might say the voltage drops of the diode and the low from the 555 necessitates a voltage divider on the base of the buffer transistor. Trained monkeys are very smart, you know.
In the meantime I have put it on my website, under: Spot The Mistake, to help my readers design circuits that WORK.
The circuit with the CD4017 doesn't flash the LED four times because its outputs Q0 to Q3 light the LED continuously then it turns off for two clock cycles before it is reset.
exactly that with the difference that my circuit uses two LEDs to flash alternating.
It's no problem to add more LEDs, but the design must be changed for higher current transistor drivers for the LEDs.
A 9V Alkali battery won't do the job for long. In the video 12 LEDs per channel are used. Blue LEDs normally have a forward voltage of 3.5V, which means 6 parallel strings of two series LEDs must be used. The current flow will be ~120mA at full power, depleting the battery quickly.
I can't build the three flasher circuit at this time since I don't have a 4017 on hand. Local shops do not provide CMOS ICs and I must travel to Bangkok to get some.
exactly that with the difference that my circuit uses two LEDs to flash alternating.
It's no problem to add more LEDs, but the design must be changed for higher current transistor drivers for the LEDs.
A 9V Alkali battery won't do the job for long. In the video 12 LEDs per channel are used. Blue LEDs normally have a forward voltage of 3.5V, which means 6 parallel strings of two series LEDs must be used. The current flow will be ~120mA at full power, depleting the battery quickly.
I can't build the three flasher circuit at this time since I don't have a 4017 on hand. Local shops do not provide CMOS ICs and I must travel to Bangkok to get some.