# Simple Transistor Circuit

1. May 6, 2010

### I_am_learning

1. The problem statement, all variables and given/known data
I have A transistor Circuit As shown in the figure. I want to find out the current in all branches that is I(b), I(c) and I(e).

2. Relevant equations
As far as I know, the equation I(c) = β * I(b) is only valid when I(b) is less than certain critical value called the saturation current. Above it, I(c) don't increase with I(b) but is constant (as shown in the graph in the figure.) and Is equal to the saturation current given by
I(c) saturation = Vcc / (Rc + Re)

3. The attempt at a solution
I attemted a solution as shown in the figure. I would like to know if it is correct?

If its correct, then One thing is bothering me too much.
If we calculate the voltage across the base and collector terminal, it comes out to be
V(cb) = (Vcc - 1mA * 12K) - (Vcc - 1.4mA * 5K)
= -5V

i.e. Collector base junction is forward biased and 5v is across its terminals. I think for forward bias, the junction acts just like a regular diode, so, a forward voltage of 5V and a reverse current of 1mA is certainly not allowable!! What am I getting wrong?

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2. May 6, 2010

### I_am_learning

Why is it that I have no response?
Am I missing something?
Is my question not in the right place?

Either wrong or right, I should be informed at least something!

3. May 6, 2010

### vk6kro

The base current is a large part of the emitter current, so you need to allow for the extra voltage across the emitter resistor.

I did a simulation of this and got the following values:
Ve = 6.63 V relative to the negative rail.
Vb = 7.43 V
Vc = 6.65 V

Ib = 1.15 mA
Ic = 0.69 mA
Ie = 2.2 mA

Vbe is 0.8 volts and the transistor is very saturated with a Vce of only 0.02 volts.

Have a look at your equations again using these values. I could see some wrong assumptions there.

Why is it that I have no response?
Am I missing something?
Is my question not in the right place?

Either wrong or right, I should be informed at least something!

No, not at all. This is not an easy problem because the circuit is very unusual.
In the homework section, we have to be very careful not to do all the work for you and replying is entirely optional.

4. May 6, 2010

### I_am_learning

I'm sorry.
I understand that my assumption I(c) sat = 15 /(12 + 3) is wrong, because same current don't flow through both of the resistors.
Also, It seems that the graph (of load line and characteristic curve) that is mostly seen in books is also wrong! It shows that If we increase base current beyond saturation, the collector current remains pixed at the I(c) saturation, but actually the collector current decreases (as proved by your analysis, I(c) = 0.69mA , but I(c) sat = 1mA)

Now, given that I don't know I(c) also, I don't have the slightest idea of how do I proceed!
Any hint will be greatly appreciated!

Last edited: May 6, 2010
5. May 6, 2010

### I_am_learning

V(b) should have been, V(b) = 15 - 1.15mA * 5K
= 9.25 Volts!
And that would make the cb junction forward bias with 2.53 Volts!

6. May 6, 2010

### vk6kro

[PLAIN]http://dl.dropbox.com/u/4222062/Sat.PNG [Broken]

The circuit can be redrawn as above because the Vce voltage across the transistor is negligible. The diode is the base-emitter junction.

As a first approximation, you could even ignore the diode.
Work out the voltage across the 3 K resistor.
Then work out a new current through the 5 K resistor allowing for the diode (0.8 volts) and adjust the emitter voltage accordingly.

Then you can work out the collector current.

This will give a result that is quite close to the real values.

EDIT: sorry that was a typo. Base current = 1.5131 mA

Last edited by a moderator: May 4, 2017
7. May 6, 2010

### I_am_learning

How could you know it in the first place? T
he problem is not this particular question. (infact this question is made by myself so as to develop insight into solving transistor problems, i.e. this question isn't my homework or assignment)

What is the general procedure to solving? How do I proceed in the case when base current is just little more than the Saturation current; At that condition V(ce) won't be negligible I think!

(I already did the calculation based on your model and I got, I(b) = 1.51 and I(c) = 1.70, it is consistent with your previous post:|| I am just trying to understand it further)

Last edited by a moderator: May 4, 2017
8. May 6, 2010

### vk6kro

If you look at a circuit like that and see the base resistor is not about 100 times the collector resistor, then it should ring alarm bells and tell you this is a saturated transistor.

If it is not saturated, you have to examine the circuit and the transistor characteristics in more detail and apply Kirchoff's Laws to solve for the different currents and voltages.

Fortunately, there are some short cuts you should be aware of if you just want to set up a transistor amplifier to give undistorted output. Suppose you wanted to keep your collector and emitter resistors but find a new base resistor that will work better.

If the transistor has reasonable gain (=/>100), you can ignore the base current in the emitter resistor. Transistor gains vary by +/- 50% so a 1 % error here doesn't matter much.

Also, for a linear amplifier, there should be about half of the supply voltage across the transistor. So, in this case, the transistor would have 7.5 volts across it and the collector and emitter resistor will have the remaining 7.5 volts across them in the ratio of their resistances.

So, you can work out the collector current and then work out the base current from the known gain, and hence the base resistor. If a base resistor is given, you can then estimate whether it is likely to bias the transistor in the middle of its range or turn it off or saturate it.

In your circuit, suppose you wanted to make it linear. You would have 6 volts across the 12 K resistor and 1.5 volts across the 3 K emitter resistor. So, there must be a collector current of 0.5 mA flowing. The transistor has a gain of 100 so the base current would be 5 uA.

The base resistor has 15 volts at one end of it and (1.5 + 0.6) volts at the other end of it.
So, it has 12.9 volts across it. So the resistor should be 2.58 MegOhms. (12.9volts/.000005Amps).

So, you can see why your 5 K resistor would not be a good choice.

9. May 6, 2010

### I_am_learning

Ok, I got that. Thank you very much. Bye.