Need help with using transistors as a switch with 555 timer

Status
Not open for further replies.
Why only use the current from just the third transistor? Why are the others not included in that calculation?
 

Why only use the current from just the third transistor? Why are the others not included in that calculation?
The 6V zener diode supplies the Cmos 555 that draws a VERY low current and the third transistor. The other two transistors are not powered from the 6V zener diode, instead they are powered from 12V.
 

So I use one power supply that powers both the chip and lights. The power going into the chip is limited by a resistor leading into the zener diode.

- - - Updated - - -


I had planned on using 10 sets of 5 LED's (50 total) in series. Will the chip you mentioned handle the output to the transistors since I have drastically increased the current needed? Are the calculations you made for the resistor into the transistor the same for the other three into the other two transistors?
 

So I use one power supply that powers both the chip and lights. The power going into the chip is limited by a resistor leading into the zener diode.
No.
The chip does not eat all the power that is available. It uses a very small current. The resistor feeding the zener diode provides about 5mA to a 5mA zener diode and a few mA to the third transistor.

The transistors do not care about LEDs, instead they care about current. The datasheet for almost EVERY transistor shows that it saturates well (turns on) when its base current is 1/10th the collector current.

Each series string of LEDs uses 20mA and since there will be 10 strings then the collector current in the output transistor is 200mA.
Then the base current of the output transistor is 20mA which is also the collector current of the middle transistor.
Then the first transistor (previously called the third transistor) will have a collector current of 2mA and have a base current from the Cmos 555 of only 0.2mA which it can easily provide.
It is easy to calculate the resistor values.

Did you properly calculate the power supply voltage?
If an LED has a typical forward voltage of 3.2V then its maximum might be 3.6V and all your LEDs might be 3.6V. Then 5 LEDs in series need 3.6V x 5= 18V plus about 4V for their current-limiting resistor so the total supply must be 18V + 4V= 22V.
The forward voltage of the LEDs might be less than their typical voltage and might be 2.8V so calculate the value of the current-limiting resistor with a 22V supply, five 2.8V LEDs and about 25mA.
 

I never knew you had to account for voltage for the resistors leading into the LED's. Why is that? The LED I want to use has a typical forward voltage of 3.4 volts @ 20ma.
 

You cannot buy a "typical" LED. You get whatever they have. It might be 3.0V or it might be 3.8V or what is the voltage range listed on the datasheet.
Maybe a huge manufacturer already bought all the typical ones.
The current-limiting resistor must have some voltage across it so that the circuit will work with any forward voltage in the range listed.

Look at the voltage-current graph on the datasheet for your LED. If its current is 20mA when its voltage is 3.4V then the current is probably 10mA when its voltage is 3.3V and it changes when the temperature changes. Its current might be 40mA when its voltage is 3.5V and as it is heated by the high current then the current keeps increasing (thermal runaway). The series current-limiting resistor with some voltage across it helps to prevent that.

Or you can connect some LEDs and test them so that you can select a suitable current-limiting resistor for each string.

Cheap Chinese flashlights use the internal resistance of a poor quality battery to limit the current and they use kids to test and group all their LEDs so that in one flashlight all the LEDs have the same forward voltage.
 

Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…