Sunday, July 18, 2010

NOT gate using a transistor

A simple circuit of NOT gate which is realized using bipolar transistor is given below.



BC549 is a low frequency, low power general purpose npn transistor designed for switching applications. In the above circuit, transistor is in common emitter configuration. Main property of this configuration is that it provides 180 deg phase shift. So if the input is high the output will be low. Biasing is done in such a way that the operating point is near the origin of the transfer characteristics. Thus when a high voltage is given to the base, the transistor enters into the active mode and quickly get into saturation. Thus a current will flow through R2 and R3. Due to this the collector voltage goes low as result of the drop across the R2. Now if a low voltage is given to the base, transistor is cut off and no current flows through R1and R2. Thus the collector voltage is almost Vcc. This is how it works as a not gate.

Input and the output of the circuit given above are TTL compatible.

Equivalent circuits are shown below



7 comments:

Anonymous said...

Sir,
I have a doubt. Why there is a decrease in voltage drop when current flows through R2 and R3? I have read current is proportional to Voltage drop, according to ohm's law
Logesh

Anil C S said...

Ya,its true that, as more current flows through through a resistor the voltage drop across it increases. Now consider the circuit given above. '

For our discussion assume that the transistor given above act as an ideal switch. Now when the input is low, transistor is off and no current flows through the resistors R2 and R3. The equivalent circuit is added in this post.

Then the output voltage is given by V0 = Vcc-IxR2. Since the current I is zero V0= Vcc=5v.

Now when the input is high the transistor is on. Thus the current I flows through R2 and R3. Refer the equivalent circuit given above for this case. Current I=Vcc/(R1+R2) (Note: This is valid only when the transistor is ideal).

Then the output voltage

V0=Vcc-IxR2 = 5-(Ix10k)= 5-( (5/(10k+1k))x(10k) )

Thus V0= 0.4545V.

You can also find V0 from the equivalent circuit shown above as follows

V0= VccxR3 / (R2+R3).

In the above analysis the ohms law is valid. Hope your doubt is clear. If you are still having doubt, you can contact me through technoburst@gmail.com

Anonymous said...

sir plz tell what happens in pn junction of transister in not gate

Anonymous said...

why npn is used?? why not pnp??

naran pindoria said...

just working as a switch.

Sarah Martin said...

Great Post!

Unknown said...

The only problem - Voutput is 2.28V when Vin=5V. Should remove R3 or Increase R1..

Post a Comment

Twitter Delicious Facebook Digg Stumbleupon Favorites More

 
Design by Free WordPress Themes | Bloggerized by Lasantha - Premium Blogger Themes | coupon codes