# Homework Help: Math Biasing

1. Mar 16, 2014

### Duave

Can someone please tell me if I have solved the problem? Is the math correct? Were the assumptions correct?

1. The problem statement, all variables and given/known data

Show that:

I(B) = 44 uA
I(C) = 4.4 mA
V(CE) = 1.2 V

https://scontent-a.xx.fbcdn.net/hphotos-prn1/t1.0-9/1011057_10151937081760919_165466716_n.jpg

2. Relevant equations

V(B) - V(BE) = V(E)

I(B) = V(B)/R(B)

I(B) = [Beta][I(B)]

3. The attempt at a solution

The dotted lines were placed for clarity.

Part A

V(B) = 5V
..................
V(BE) = 0.6V
...................
V(B) - V(BE) = 5V - 0.6V
..................................
V(B) - V(BE) = 4.4V
..................................
R(B) = 100 x 10^3(ohms)
..................................
I(B) = [4.4V/100 x 10^3]
..................................
I(B) = 44 uA
..........................
I(C) = [Beta][I(B)]
.........................
I(C) = (100)(44 x10^-6)
.................................
I(C) = (4.4 mA)
.....................................
V(C) - [I(C)][R(C)] = V(CE)
.....................................
10V - (4.4 x 10^3(A))(2000(ohms)) = V(CE)
...........................................................
10V - (8.8(V)) = V(CE)
...............................
1.2V = V(CE)
...........................

Part B

Find Beta

(Beta)[I(B)] = I(C)
.........................
I(S){e^(V(BE)/V(T)] -1} = I(C)
..........................................
I(S){e^(V(BE)/V(T)] -1} = (Beta)[I(B)]
.....................................................
Beta = [I(S)/I(B)]{e^(V(BE)/V(T)] -1}
....................................................
Beta = [{[V(CC) - V(CE)]/[R(C)]*[e^{V(BE)/VT]}/I(B)]{e^(V(BE)/V(T)] -1}
.......................................................................
Beta = [{[V(CC) - V(CE)]/[I(B)][R(C)]
.....................................
Beta = [{[10V - 1.2V]/[44 x 10^-6(A)][2000]
...........................................
Beta = 100

2. Mar 16, 2014

### Staff: Mentor

Looks good. But why do you calculate beta (with something calculated in (A)), if you can use it in (A)?

3. Mar 16, 2014

### rude man

Where do you assume beta = 100? It can't be calculated except by assuming the answers a priori.
.

4. Mar 16, 2014

### Duave

@mfb

I made a mistake. I know that beta is unitless.

My question to you is, if I erase the (A), would the statement then be COMPLETELY correct?

Thanks again

5. Mar 16, 2014

### Duave

@ Rude Man

I understand what you mean. I was to answer the question below:

Suppose that the circuit shown in this thread goes into saturation. What is the MINIMUM value of β which would cause saturation?

Thanks again

6. Mar 16, 2014

### rude man

In that case beta needs to be a bit more than 100.

7. Mar 16, 2014

### rude man

.

No. A beta of 100 will not saturate the transistor.

No. See above comment.
It takes more than 4.4 mA to saturate the transistor.

I'll leave it here for the time being ...

8. Mar 16, 2014

### Staff: Mentor

@Duave: All the confusion here comes from the missing problem statement.
Please post the full, exact problem statement here, otherwise we keep on guessing what you are supposed to do.

9. Mar 16, 2014

### Duave

I have to solve for I(B), I(C), and V(CE)

The whole problem statement for Part (a) is Literally:

"
Show that:

I(B) = 44 uA
I(C) = 4.4 mA
V(CE) = 1.2 V
"

I have to show calculations that prove that I(B) = 44 uA, that I(C) = 4.4 mA, and that V(CE) = 1.2 V.

Seriously, that's all that I was given.

The whole problem statement for Part (b) is Literally:

"Suppose that the circuit shown in this thread goes into saturation. What is the MINIMUM value of β which would cause saturation?"

Thanks again

10. Mar 17, 2014

### rude man

OK, we caw summarize this as follows:
part (a) beta = 100 and your computatiuons are correct.
part (b) did you ever find the minimum beta?

11. Mar 17, 2014

### Duave

@rude man

Thank you so much for your response:

Right now β = 100

iC/iB = β
..............................................
iC = 4.4 mA
..............................................
iB = 44 uA
..............................................
4.4 mA/44 uA = β
..............................................
100 = β
..............................................

if VC drops to VCE then,
..................................................................
VC = 1.2V
..............................................
IC(1.2V) = VC/RC
........................................................................
IC(1.2V) = 1.2V/2000(ohms)
........................................................................
IC(1.2V) = 0.6mA
........................................................................
if IB is fixed, then
........................................
IC/IB = β
.............................................
0.6 mA/44 uA = β
........................
13.64 = β
........................

So, the minimum β = 13.64

Is this correct or am I still missing soemthing? Thanks again.

12. Mar 17, 2014

### Jony130

For this circuit Ib = (Vin - Vbe)/Rb = (5V - 0.6V)/100k = 44μA
The maximum collector current is equal to
Ic_max = (Vcc - Vcs(sat))/Rc ≈ Vcc/Rc ≈ 10V/2K ≈ 5mA

So if BJT current gain Hfe (β) will be larger than 5mA/44μA = 114 the BJT will enter saturation region.

13. Mar 17, 2014

### Duave

Thanks Jony130

I can understand the max value you that you calculated, but does my minimum look okay or do I need to fix that?

14. Mar 17, 2014

### Jony130

But what this minimum Hfe value will represent? Because if you want Ic = 0.6mA for fixed base current (44μA), then you need to use a BJT with a Hfe equal to 13.64.

15. Mar 17, 2014

### Duave

Jony130,

So does the collector branch always determine the minimum Hfe (β) that will cause saturation? That's right isn't? In this circuit, The collector determined the minimum saturation value.

16. Mar 17, 2014

### Jony130

17. Mar 17, 2014

### rude man

Part b assumes eveything in the circuit stays the same except beta can rise.

So to compute the min. needed for saturation, use the same base current, multiply by beta, and let that product = the collector current needed to reduce the collector voltage to zero. Then solve for beta.