# Can someone check if i did this Transistor problem right?

Discussion in 'Homework Help' started by dacrazyazn, Apr 14, 2009.

1. ### dacrazyazn Thread Starter Member

Mar 30, 2009
22
0
Alright, the problem is...

the parameters of the transistor are β=50 and Vγ=0.7v. Find ic and Vce for Vs=0.8V and Vs=2V

Using what i know for transistor in an active mode,
ic=β*ib
Vbe=Vγ

so for Vs=.8V
i did --> ic=β*ib=50*(.8-.7)/(10*10^3)=.5mA
for Vce = Vcc-ic*Rc=15-(5*10^4mA)(10*10^3)=10V

I did the same for Vs=2V and i got 6.5mA for the ic and no Vce because it is saturated.

Please let me know if i did this right. I was thinking that the 20kΩ resistor that is parallel with the Vcc would effect something. Thanks!

File size:
134 KB
Views:
36
2. ### PRS Well-Known Member

Aug 24, 2008
989
36
Ah, but in the second case how do you know the transistor is saturated? You didn't show that and your teacher surely wants you to show that. It's easier to think about this circuit if you redraw it so that it is a simple inverter with Rc=10k and RL=20k. Vcc is then 15 volts. Vo=Vcc*RL/(RL+Rc) and this is readily seen by voltage division.

Remember that Vce(sat)=.2 volts. (you didn't use this standard parameter for BJTs) You should calculate Ic(sat) as (Vcc-Vce(sat))/Rc or, (15-.2)/10000=1.48mA

It should strike you from this that there is simply no more current available from the source than 1.48mA. This is the meaning of saturation. Since 6.8mA exceeds this amount the device is indeed in saturation. In the first case it was not in saturation and the output was, indeed 10 volts.

3. ### dacrazyazn Thread Starter Member

Mar 30, 2009
22
0
How would you redraw that circuit as you said it? because this is my first hw on transistor, im still confused on what the purpose of transistor is? real life applications..

But as for the saturation, i did have the work on my paper and it turned out that ic is greater than ics. and to compute ics i did Vss/Rc. therefore showing saturation.

As for the problem overall, did i do it right then? and so that 20Kohm resistor does not effect this circuit right? Thanks!

4. ### PRS Well-Known Member

Aug 24, 2008
989
36
The transistor is a switch that is ten volts or zero volts (on/off). It is used in so many applications I don't know where to begin since you can use it to turn just about any maching on and off electronically.

You cannot ignore the 20k resistor. It combines with the 10k Rc to form a voltage divider. Without the 20k resistor, the voltage drop across Rc while the xistor was off would be 15 volts. On and off in that case would be 0 and 15 volts.

Sorry, but I have no way to draw, except Paint, and I'm just lousy at it. Perhaps someone else will redraw the circuit for you in its standard form.

5. ### dacrazyazn Thread Starter Member

Mar 30, 2009
22
0
Can you find a pic or describe what the inverter is like? i tried to google for inverter and it just show me power supply inverters...

6. ### Engr1001 New Member

Apr 15, 2009
3
0
This calculation seems incorrect
for Vce = Vcc-ic*Rc=15-(5*10^4mA)(10*10^3)=10V I think it ignores the 20k to ground.

The theven equivalent for the 15v and 10K / 20K resistors is 10v in series with 10K //20k or 6.7K. Vce is 10v- 6.7Kx0.5ma = 6.7v

The transistor is an inverter because a positive voltage at the input causes the collector voltage to decrease (because the transistor is turned on). If the voltage at the input approaches zero then thr collector voltage increases. Therefore the Vce acts inversly to Vbe.

7. ### PRS Well-Known Member

Aug 24, 2008
989
36
I hope it shows up. This is the first time I have tried to put a picture up in this forum.

File size:
18.7 KB
Views:
26
8. ### PRS Well-Known Member

Aug 24, 2008
989
36
Egnr1001, look at the way I came up with 10 volts. What do you think?