Values needed for Emitter Follower Voltage Regulator

[joemmonster]

Hi, im a newbie here and i am currently making a circuit for our school fair project. the only problem is i want to decrease the voltage of my battery from 9 volts to 5 volts without changing the current... then i searched through the internet and found out that the simplest way and efficient is to use an emitter follower voltage reducer/regulator like this:

http://hades.mech.northwestern.edu/...ransistors#Emitter_Follower_Voltage_Regulator

but i dont know what are the values needed for the resistor and zener diode, and type of transistor should i use.
if ever, i want the circuit to produce about 350 mA current, 5 volts...

thanks..

 

[chuckey]

Your transistor has 9V in and 5 V out at .35 of a amp. so power in it is 4 X .35 = 1.4 Watts. So select a 2 watt transistor and put it on a heat sink if required. The next thing to do is to work out the voltage required on the base, for a single transistor = 5.8 V for a darlington transistor = 6.6 V, this is the zener diode voltage. The base current = 350 mA/gain of transistor, 100 for single, 3000 for darlington. So the current flowing down the resistor = this figure + 1mA extra for the zener diode to remain working., So the resistor will drop 9 - (the zener voltage) at a current (see above), so divide one by the other ( R = V/I) to get R. Or else buy a 5V stabilizer IC, LM317/7805 ? or similar.
Frank

 

[joemmonster]

thanks, they said that a 5v regulator will do better.. and now, i decided to go with it..

 

[jeffrey samuel]

Is not darlington pairs stability lesser than that of LM7805 I am curious that is it

 

[Tahmid]

Is not darlington pairs stability lesser than that of LM7805 I am curious that is it

Yes. Voltage drop in transistor(s) will change with change in current, and thus power dissipation. So, a dedicated regulator chip such as LM7805 is better.