Help me in calculating base resistance for transistor

[rahulb]

Hi,

I am making a circuit for a led chaser. the leds should glow one by one.

I am testing the circuit with one led for each transistor later I will add array of leds to each transistor.

I want the base resistor value to make the transistor turn on. Leds will have resistors to limit current to them.

Problem is : first led is glowing continuously, second one is not glowing and third one is glowing dim but continuously. they should glow and turn off one by one.

Voltage used is : 12v DC


datasheet link : https://bit.ly/32yOfpg

thanks

 

[73's de Edd]

Sir rahulb . . . . .

He say . . .
I have not looked up the logic of the 4017 to see if it connected correctly.
If IC2 is not powered, it may be using power from the 555.
I say . . . .
That connection is being circuit confirmed at the 2 o'clock position from the IC2 on the schema . . . if it was physically done ? .
I second the motion on the adequate HF bypassing of power busses . . . close in to their pins!


Methinks that wherein your problem doth seem to lie, is that direct input of the pin #3 output of the 555 is . . . . TOTALLY . . . .1,271and 99/10,000ths % . . . swamping the pin #14 clock input circuitry within the 4017.

Try changing your circuitry to the BOX add ons.

That additionally gives you a visual indicator of your 555 logic output states and then creates an attenuation of the clock input signal to the 4017.

NEW SCHEMA ILLUSTRATION . . . . .


73's de Edd . . . . .


I was just outsides and saw something fly --------> by me, that had SIX legs . . . . . I counted 'em 1-2- 3- 4- 5- 6 . . . . . it was GREEN . . . . . and was repeatedly jumping higher than my head . . .
it was a big 'ole grasshopper with the hiccups.


.

.

 

[duke37]

I have not looked up the logic of the 4017 to see if it connected correctly.
If IC2 is not powered, it may be using power from the 555.
Try a capacitor across IC2 supply.

 

[Ylli]

Edd's modifications are good ones. I'd just like to add my suggestion that you add 0.1 uF bypass capacitors on the power pins to ground at each IC - shortest possible leads.

 

[Audioguru]

Of course the +12V should have a supply bypass capacitor to GND with short leads.
With the supply bypass capacitor added then the 4017 will work perfectly when the pin 3 output of the 555 directly clocks it without any added resistors.

 

[rahulb]

the supply bypass capacitor on pin 3 of timer 555?

 

[Harald Kapp]

the supply bypass capacitor on pin 3 of timer 555?

By all means: no. Pin 3 is the output. Placing a capaictor there will degrade the slope of the clock signal.
Power supply bypass capacitors are always placed between the power supply connection of an IC (GND <> V+ GND <>Vcc etc.). Use leads as short as possible.

 

[Harald Kapp]

and then creates an attenuation of the clock input signal to the 4017.

The attenuation is not required as both 5555 and 4917 operate from the same 12 V supply.
5 % attenuation (220 k vs. 10 k) will not have a menaingful influence on the clock signal.

Visualizingthe clock signal by the LED is a good debug option.

What kind of 4017N chip do you use?
The older CD4017 series could work from a 12 V power supply. But I cannot find a CD4017 with Suffix N.
The 4017s I find with suffix N are from the 74HC series and these are specified for a max. operating voltage of 6 V only. Using one og these chips on 12 V will have destroyed the chip.
If that is the case, operate the circuit from a 5 V supply. You can use a 7805 type voltage regulator to create 5 V from the 12 V. Use 12 V only for the LEDs.

 

[rahulb]

The attenuation is not required as both 5555 and 4917 operate from the saem 12 V supply.
5 % attenuation (220 k vs. 10 k) will not have a menaingful influence on the clock signal.

Visualizingthe clock signal by the LED is a good debug option.

What kind of 4017N chip do you use?
The older CD4017 series could work from a 12 V power supply. But I cannot find a CD4017 with Suffix N.
The 4017s I find with suffix N are from the 74HC series and these are specified for a max. operating voltage of 6 V only. Using one og these chips on 12 V will have destroyed the chip.
If that is the case, operate the circuit from a 5 V supply. You can use a 7805 type voltage regulator to create 5 V from the 12 V. Use 12 V only for the LEDs.

no, actually the 4017 chip no. is HCF4017, I could found it in eagle so added that one.

 

[rahulb]

Of course the +12V should have a supply bypass capacitor to GND with short leads.
With the supply bypass capacitor added then the 4017 will work perfectly when the pin 3 output of the 555 directly clocks it without any added resistors.

hi,

what should be the value for bypass capacitor?

 

[Tha fios agaibh]

As Wlli said, .01uf between pins 8 and16 of ic2 and another bypass cap on pins 1 and 8 of ic1

 

[Harald Kapp]

actually the 4017 chip no. is HCF4017

O.K., that should be good for 12 V.

 

[WHONOES]

As I have pointed out before, digital device output levels are only guaranteed to be 0.8V for a logic "0" which is easily enough to turn on a BJT. In this sort of circuit you should always add a pull down resistor to the base emitter junction to ensure that it is below the value at which it could switch on which may be as low as 0.5V. Not to do so is poor design. In the instance of the presented circuit, I would suggest a 1K resistor.

 

[Audioguru]

As I have pointed out before, digital device output levels are only guaranteed to be 0.8V for a logic "0" which is easily enough to turn on a BJT. In this sort of circuit you should always add a pull down resistor to the base emitter junction to ensure that it is below the value at which it could switch on which may be as low as 0.5V. Not to do so is poor design. In the instance of the presented circuit, I would suggest a 1K resistor.

The 4017 is ordinary Cmos that has "rail-to-rail" output voltages when not loaded. The datasheet of a CD4017 and most other ordinary Cmos logic Ics lists a maximum output logic low voltage of 0.05V when it is not sinking any current. The base of an NPN transistor as a load does not source current when its voltage is only 0.05V. Then a base resistor to ground is not needed.

Your "low of 0.8V" is when a logic output is driving an old LS-TTL input emitter that pulls up with 0.4mA.

 

[rahulb]

please tell me if my resistor value of 2.2 k to the base is fine or I have to change it

 

[Audioguru]

The datasheet for most little and medium-power transistors shows that for it to saturate well then its base current should be 1/10th the collector current. Then when the supply voltage is 12V the base resistor of 2.2k has 11.3V across it which calculates to be a 5.1mA base current for a 51mA collector current. actually, the CD4017 has a voltage loss of about 1V when its output high current is 5mA so the base resistor will have 9V across it which calculates to have a current of 9V/2.2k= 40.9mA.You do not know how much collector current you need so when you find out how much it is then you can simply calculate the base resistor value for the transistor to saturate well.

You do not say the color and voltage of your LEDs. If it is a 2V red LED then the 180 ohm current-limiting resistor has 10V across it and calculates to have a current of 55.6mA so the 2.2k base resistor is a little high for a single red LED. Use 1.8k instead. But most ordinary LEDs burn out with a current as high as 55.6mA.

 

[rahulb]

The datasheet for most little and medium-power transistors shows that for it to saturate well then its base current should be 1/10th the collector current. Then when the supply voltage is 12V the base resistor of 2.2k has 11.3V across it which calculates to be a 5.1mA base current for a 51mA collector current. actually, the CD4017 has a voltage loss of about 1V when its output high current is 5mA so the base resistor will have 9V across it which calculates to have a current of 9V/2.2k= 40.9mA.You do not know how much collector current you need so when you find out how much it is then you can simply calculate the base resistor value for the transistor to saturate well.

You do not say the color and voltage of your LEDs. If it is a 2V red LED then the 180 ohm current-limiting resistor has 10V across it and calculates to have a current of 55.6mA so the 2.2k base resistor is a little high for a single red LED. Use 1.8k instead. But most ordinary LEDs burn out with a current as high as 55.6mA.

actually I need to connect a lot of leds. thats why I am using transistor.

4 leds in series and 34 such series in parallel.

So, if one series consumes 10 mA then 34 series will have almost 340 mA.

The resistor 180 ohm I chose for 4 led in series.

So, you are telling that base current should be 1/10 , then will I need 34 mA on base?

thanks

 

[Harald Kapp]

you are telling that base current should be 1/10 , then will I need 34 mA on base?

Exactly.
A 4017 wil have a hard time churning out 34 mA. I suggest you use darlington transistor which have a much higher gain so a base current of 1/100 is sufficient. OR use MOSFETS which do not require a base current at all.

 

[Audioguru]

Again you do not say the color or voltage of the LED. If it is a white, blue or modern green LED then its voltage could be 3.5V and four in series need 14V plus at least 2V for the current-limiting resistor. They will not light up with a supply of only 12V.

With a 12V supply the Texas Instruments datasheet for a CD4017 shows a minimum output current of only 7mA when the output is struggling to reach 8V. But the typical output current is 14mA. With the voltage loss of 4V and a current of 14mA then the heating is 4V x 14mA= 56mW. The maximum allowed heating for one output is 100mW so 56mW will make it pretty hot.

Don't use a transistor or a Darlington. Use a little Mosfet instead.

 

[rahulb]

Again you do not say the color or voltage of the LED. If it is a white, blue or modern green LED then its voltage could be 3.5V and four in series need 14V plus at least 2V for the current-limiting resistor. They will not light up with a supply of only 12V.

With a 12V supply the Texas Instruments datasheet for a CD4017 shows a minimum output current of only 7mA when the output is struggling to reach 8V. But the typical output current is 14mA. With the voltage loss of 4V and a current of 14mA then the heating is 4V x 14mA= 56mW. The maximum allowed heating for one output is 100mW so 56mW will make it pretty hot.

Don't use a transistor or a Darlington. Use a little Mosfet instead.

sorry for the mistake. I am using red, green and yellow 5mm leds. one color for each of the three outputs.

I checked the forward voltage of green because I have the confusion about it. It comes around 2.1 to 2.2 volts.

thanks