# confused with a npn transistor collector current

#### terrakota

Joined Feb 8, 2005
67
Hi, i'm studing the basics of transistors but i'm a little confused about the collector current.

what i understand is that the collector current is controlled by the gain ratio of the base current, ok if i have a base current of 1 ma with a transistor of 100 dcbeta
i will have a 100 ma collector current, so if i increase/decrease the second loop voltage the current will remain the same? if i insert a resistor with diferent values in this loop the collector current will change?

how i know the values that i need for the resistor out from the collector current? how i determine the secont voltage source?

thanks for your help

#### Brandon

Joined Dec 14, 2004
306
Originally posted by terrakota@Nov 25 2005, 12:38 PM
Hi, i'm studing the basics of transistors but i'm a little confused about the collector current.

what i understand is that the collector current is controlled by the gain ratio of the base current, ok if i have a base current of 1 ma with a transistor of 100 dcbeta
i will have a 100 ma collector current, so if i increase/decrease the second loop voltage the current will remain the same? if i insert a resistor with diferent values in this loop the collector current will change?

how i know the values that i need for the resistor out from the collector current? how i determine the secont voltage source?

thanks for your help
[post=11935]Quoted post[/post]​
(Assuming all NPN)

loop voltage? You have Vb, Vc, Ve, Ib, Ic, Ie.

With xtrs, there is no simple method. You need to solve it for the situation. Start with the constant voltages. Grounds, Sources, and voltage drops.

You know that Vb-Vc=0.7 with the Xtr on so you assume a 0.7 drop there. If you know Ve (i.e. its one of the constants) then you know Vb now. If Vb is a constant voltage, assume the xtr is on, and set Ve=Vb-0.7. If Vc is the constant, then assume the xtr is in saturation then Ve=Vc-0.2.

If there is a resistor between a node on the xtr (b,c,e) and a constant voltage, the voltage at that node now is variable and will flutuate between the closest source and the appropriate voltage drop depending on the leg of the xtr.

Example.

V+=10v, V-=-5v, Vc connected to Rc connected to V+. Vb connected to Rb connected to ground. Ve connected DIRECTLY top V-.

So, since Ve connector directly to V-, Ve is a constant. It is -5v. Done.
Assumiung that the xtr is on, then Vb-0.7= Ve, therfore Vb=-4.3v Since Rb is connected to ground and Vb is -4.3v, then Ib=4.3/Rb.

Ic = Ib* β

A. (V+)-Vc/Rc=Ic

Now, this is where an issue could pop up. In this relationship, there is a chance that Ic is so large that you get an inconsistant value for Vc. An inconsistant value for Vc in this case would be anything less than -4.1 volts. Why -4.1? You assume that the xtr is in saturation, which means fully on. When fully on, Vc=Ve+0.2 so Vc would need to be -4.1 volts, but you ONLY do this, when you solve A and find that the Vc value u get is illogical. If you need to assume saturation, then you KNOW that Vc is SUPPOSED to be, you know what V+ is, now you need to RE-SOLVE for Ic using the saturation voltages. The reason is that it just can not push out the current. Anything extra that flows into the base that can not be utilized is just shunted to the emitter.

When Ic ≠ Ib* β its a good sign that the xtr is saturated.

If xtr goes into saturation, then Vc-Ve=0.2.

NOw, if you want to solve for resistors, you just do the same thing, but you solve for the resistors instead of voltages. You will need to figure out what voltages you WANT before hand and then Vb, Vc, Ve will be set and you can solve for Rc, Rb, Re.