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some questions....

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bogdanfirst

New Member
if i think of a bipolar trnasistor, it has 3 areas.p,n,p or n,p,n that means that they are like 2 diodes connected with the anodes or cathodes together....
but in real life if i connect 2 diodes like this, do they work as a transistor?
and another thing about tolerances....
lets say.....if i got a resistor with a 2% tolerance and one with 5%, 2K and 3K. i connect them in series. i get a resistor 5K, with what tolerance?
2000*0.98=1960; 2000*1.02=2040
3000*0.95=2850; 3000*1.05=3150
so i will have a resistor with the value of 4810-5190, wich means 5K+190R/-190R so i get 5K+/-3.8%
am i right?
and if i have them in paralel......a littel bit more complicated, but do i have it right since now? if yes, then ill continue with the paralel.
 

kinjalgp

Active Member
Two diodes can't make a transistor because of several reasons which are,
1) Doping levels of both diodes are almost same whereas a real transistor has heavily doped emitter, mediumly doped collector and lightly doped base.
2) Physical area of all three semiconductor layers in two diode configuration is same whereas in real transistor collector has the largest area and then comes emitter and base in descending order.
 

Gene

New Member
I don't see anything wrong with your calculations. However, if tollerences are critical to your circuit, recommend you actually measure the resistor circuit you build with a good ohmmeter.
 

bogdanfirst

New Member
yes, thats right, but i was just doing these calcs for curiosity.
i am not sure but if i use the, in paralel, i wouldnt have a tolerance like x ohm +/- y %. i think it will be like x ohm, +y%/-z% i mean +something and - another thing.
am i understood?
 

Gene

New Member
If it couriosity, why not do what you did in the first example? Figure the parallel circuit (1200.048). Then, apply the tollerance to the two resistors and re-calculate the circuit (1161.440). The tollerance is 3.32% (wish I had a good calculator here)!!!
 
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