Logic Function with transistors

Thread Starter

AD633

Joined Jun 22, 2013
96
Hi,

Consider the circuit of Figure 3 where the transistors operate in a regime switching (cut / saturation).

Complete the table to the side and say what logical function is implemented by the circuit. Justify the values​​, presenting the state of conduction of each transistor.


When we have V1 = V2 = +5 V and +5 V.

\(T1:
VCE1=Vcc-Rc1*Ic
VCE1=10-(5*10^3)*Ic \)

I don't know Ic and i can not determine Ib because i also dont know the value of the base resistance Rb

\(5=(Rb??)*Ib+Vbe \)

Is there any other way for determing Ic?

Thanks
 

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Jony130

Joined Feb 17, 2009
5,487
But why you need Ib and Ic current? You have been told that BJT work only in two regions of operation. The cut-off region and saturation region. So if for example V1 = +5V, T1 is in saturation region Vce1 = 0V And T3 will be OFF (cut-off) regardless T2 state.
 

Thread Starter

AD633

Joined Jun 22, 2013
96
But why you need Ib and Ic current? You have been told that BJT work only in two regions of operation. The cut-off region and saturation region. So if for example V1 = +5V, T1 is in saturation region Vce1 = 0V And T3 will be OFF (cut-off) regardless T2 state.
So its fairly simple to determine what is the logic function implemented by the circuit:

V1 = V2 = +5 V and +5 V

VCE1=0 V and VCE2=0 V

V1 =+5 V and V2= 0 V


VCE1=0V and VCE2=+VCC

V1 =0 V and V2=+5 V


VCE1=+VCC and VCE2=0 V

V1 =0 V and V2=0 V


VCE1=+5V and VCE2=+5 V

So i think that the logic function implemented by the circuit is a NOT.
 

Brownout

Joined Jan 10, 2012
2,390
So its fairly simple to determine what is the logic function implemented by the circuit:

V1 = V2 = +5 V and +5 V

VCE1=0 V and VCE2=0 V
Good so far.

V1 =+5 V and V2= 0 V


VCE1=0V and VCE2=+VCC
That would be impossible

V1 =0 V and V2=+5 V


VCE1=+VCC and VCE2=0 V
Not quite.

V1 =0 V and V2=0 V
VCE1=+5V and VCE2=+5 V

So i think that the logic function implemented by the circuit is a NOT.
1) Why would a NOT function have two inputs?
2) Why are you ignoring t3?
 

Thread Starter

AD633

Joined Jun 22, 2013
96
VCE1=0V and VCE2=+VCC

That would be impossible

Yes it should be, VCE1=0V and VCE2= 0V

V1 =0 V and V2=+5 V


VCE1=+VCC and VCE2=0 V

Not quite.
I did not understand this.Are you saying that it should be

VCE1=+VCC-(Rc1)*(Ic1sat) and VCE2=0 V

1) Why would a NOT function have two inputs?
Yes it would not make much sense :)

2) Why are you ignoring t3?
I forgot about the NMOS.In a NMOS transistor we need to have a positive Vgs potential so that we get a high sinal in the output right?

The logic function is a NAND?
 

Jony130

Joined Feb 17, 2009
5,487
V1 =0 V and V2=+5 V


VCE1=+VCC and VCE2=0 V
I did not understand this.Are you saying that it should be

VCE1=+VCC-(Rc1)*(Ic1sat) and VCE2=0 V
Look at the circuit diagram and notice that if T1 is in cut-off And T2 is in saturation, we have two Rc resistors connected in series.
Vc2 = 0V and Vce1 = ??


I forgot about the NMOS.In a NMOS transistor we need to have a positive Vgs potential so that we get a high sinal in the output right?
No, the NMOS work just like a inverter. Vgs = 5V---> Vds --->0V
 
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