I am confusing myself on calculating values for making a Darlington pair from two transistors.
Having trouble calculating the base resistance.
I want to connect the second transistor to a transformer,
If my load resistance is 3.2 ohm (transformer) and my supply voltage is 9 Volts, then my load current is 2.81 amps.
I have attached a picture in a word doc. format. It is a bit crude but i hope it will make it clear. The 555 timer output Feeds into NPN transistor Q1 and another transistor Q2. Q2 feeds the transformer T1.
So Q2 has to have an Ic greater than 2.81. maybe 4A transistor
Question
1) Does T1 need to be a high power NPN. If it is connected on the same wire as Q2, then it will also be receiving 2.81 A.
However, i have read that T1 in a Darlington is a low power NPN. therefore, can someone correct my understanding please.
2) How do i calculate the base resistor. R1. Do i use R1=0.2*load resistance*Hfe . Formula from (http://www.kpsec.freeuk.com/trancirc.htm)
The Hfe would be: Hfe of Q1 * Hfe of Q2.
Therefore, if Hfe of Q1 is 40, Hfe Q2 is 40 for example. Hfe = 1600.
R1 would therefore be 0.2*3.2*1600=1K resistor.
3) If there is a better method or I’m doing something wrong please advise.
thanks
Having trouble calculating the base resistance.
I want to connect the second transistor to a transformer,
If my load resistance is 3.2 ohm (transformer) and my supply voltage is 9 Volts, then my load current is 2.81 amps.
I have attached a picture in a word doc. format. It is a bit crude but i hope it will make it clear. The 555 timer output Feeds into NPN transistor Q1 and another transistor Q2. Q2 feeds the transformer T1.
So Q2 has to have an Ic greater than 2.81. maybe 4A transistor
Question
1) Does T1 need to be a high power NPN. If it is connected on the same wire as Q2, then it will also be receiving 2.81 A.
However, i have read that T1 in a Darlington is a low power NPN. therefore, can someone correct my understanding please.
2) How do i calculate the base resistor. R1. Do i use R1=0.2*load resistance*Hfe . Formula from (http://www.kpsec.freeuk.com/trancirc.htm)
The Hfe would be: Hfe of Q1 * Hfe of Q2.
Therefore, if Hfe of Q1 is 40, Hfe Q2 is 40 for example. Hfe = 1600.
R1 would therefore be 0.2*3.2*1600=1K resistor.
3) If there is a better method or I’m doing something wrong please advise.
thanks
