Hi,

Im having some trouble understanding some transistor operations. Currently, im trying to master the BJT NPN transistor. Therefore, i just would like to know some stuff 8) .

If a transistor is operating in the saturation region, and a load is placed at the emitter, the voltage drop across the load is according to 'Vbb - Vbe' or 'Vcc-0.2V?
Currenly im quite confuse between the emitter follower and the transistor acting as a switch.

Thx for any explanation.

 

Hi Fabbie,
In order for an NPN transistor to conduct between its collector and emitter, its base voltage must be higher than its emitter voltage by a diode voltage drop which is Vbe. A transistor will saturate with a very low voltage between its collector and emitter depending on the amount of collector current and base current. The emitter follower 's emitter load resistor will have a voltage across it of the base voltage minus Vbe, even when the base voltage equals the positive supply voltage (it usually doesn't get that high because the base current through its base drive resistor will have a voltage drop). With the base voltage equal to the positive supply voltage, the emitter is at Vcc minus Vbe and therefore the transistor isn't saturated and can be very fast. In order for an NPN emitter follower to saturate its base voltage must be higher than the positive supply voltage. Some audio amps use a "bootstrap" capacitor from its output to the center-point of its base drive resistor for the upper NPN emitter follower output transistor, to increase the base voltage for positive voltage swings so that it saturates, providing max output voltage (and therefore max power).

 

I dont get one thing though, if we are dealing with digital signals to the transistor base, is it possible for the base voltage to be same as the Vcc?
An example is the PIC, the I/O pin output voltage during HIGH is approximate to the Vdd, which is around 5V. Does this mean that the Vcc should be less than the base voltage of 5V to fully saturate?

All the while through looking at the Q-point analysis, i thought that as long as we have a high base current, we would able to saturate the transistor

 

basically to drive a transistor into saturation you have to forward bias both the base-emitter and the collector-emitter junctions.

its not the base current that controls the behaviour of the transistor. its the base-emitter voltage Vbe that determines the behaviour of the transistor. but ofcourse there is a relation between the base current and the collector current because both of them are proportional to each other.

 

so just to confirm, if i have a Vcc value of 12V and a base voltage of 5V, does this mean the transistor is not saturated?
this means that the emitter voltage equals to 5V-0.7V?
Furthermore, the purpose of Vcc is only to control the collector current, am i right?

 

Hi Fabbie,
That's right, if the transistor has a Vcc of 12V but a base voltage of 5V, the transistor is not saturated and its emitter voltage is approx. 5V minus 0.7V. It's an emitter follower, the emitter voltage "follows" whatever the base voltage does.

Vcc does not control the collector current. Since the base voltage is controlling the emitter voltage, the emitter current is also controlled by the base voltage. In a high gain transistor, the collector current is practically the same as the emitter current, so any change in Vcc has no effect on collector current. The emitter follower is a fair example of a "variable current source" at its collector. If the base voltage is fixed, it becomes a "constant current source". The convention is that a PNP emitter follower is a current source at its collector and a NPN one is a current sink.

 

nice explanation audioguru. i would recommend fabbie to read the following article

http://amasci.com/amateur/transis.html

 

That should read "basically to drive a transistor into saturation you have to forward bias both the base-emitter and the base-collector junctions. Collector-emitter is not a junction.

 

Are you thinking that the PIC is TTL, or are you talking about driving a bipolar transistor with the output of a PIC? The PIC is, of course, made in CMOS technology.

 

sorry, that was a typo!!!!

 

What i mean was using the PIC output to saturate the transistor.

Moreover, mind looking at this circuit

How isit the Ib=1.936mA? From using the Ib=Vbb/Rb, shouldnt it be 160ma?

Im just testing the transistor saturation in this case. Furthermore, is the voltage at the Ve correct? If saturation do occur, shouldnt it be approximately the same as the Vcc?

Attached Images

untitled_111.jpg (19.6 KB, 557 views)

 

actually Ib = ( Vbb - Vbe ) / Rb in base-biased amplifiers where Vbe is 0.7 for silicon transistors. but this is the case when the emitter is grounded which is not the case here.

i think you should tell us what you are trying to do. if u just want to turn on or turn off a trasistor from a PIC output then see the page http://www.winpicprog.co.uk/pic_tutorial_extras.htm

 

Im basically trying to make a H-bridge driver. The emitter resistor u see there is actually acting as my load(in other words motor). But to make this driver i need a thorough understanding on the transistor.

In any case, using a resistor to act as my motor is that correct or wrong?

 

okay now its easy to see what you are doing. and you are right in that you need to have a gut level understanding of BJTs to efficiently design an H-Bridge driver. you can find alot of info on the internet for H-bridge drivers. following are some articles i found by searching on google that would be helpful in developing the theory for you.

http://www.bobblick.com/techref/proj...e/hbridge.html
http://www.robotroom.com/HBridge.html
http://www.dprg.org/tutorials/1998-04a/
http://www.4qdtec.com/h.html

i hope that helps

 

Thx u guys for all the reference given!

I just have one question, if i intend to OFF a PNP transistor, is it alright to leave the base-emitter junction floating?