BJT

Thread Starter

arun9112

Joined Dec 3, 2011
6
hi i have attached an image file circuit, the whole question is in it. My question about this is how to know whether the transistor is saturated. Even though there is an explanation for question in the image itself but i don't understand. please help me elaborate on this thank you.
 

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Audioguru

Joined Dec 20, 2007
11,249
Don't you know that a transistor is saturated when it is turned on very hard so that its collector to emitter voltage is very low?

Then calculate the base current using simple Ohm's Law and multiply it by the beta to find the collector current.
Again use simple Ohm's Law to calculate the voltage across the collector resistor.

If the voltage across the collector resistor is near the supply voltage then the transistor is saturated. Simple!

Note: Beta is used only with a transistor that is linear and is not saturated.
 

Vahe

Joined Mar 3, 2011
75
For NPN BJT, there are two p-n junctions -- base-emitter junction (BEJ) and the base-collector junction (BCJ). There are four operating modes for the BJT depending on which junctions are forward and/or reverse biased.

Cutoff -- BEJ reverse-biased, BCJ reverse-biased
Saturation -- BEJ forward-biased, BCJ forward-biased
Forward Active -- BEJ forward-biased, BCJ reverse-biased
Reverse Active -- BEJ reverse-biased, BCJ forward-biased

Most common are the first three modes. Depending on the textbook that you use, you might see different values as to what constitutes a forward-biased junction. For example, the BEJ forward-bias voltage could be anywhere from 0.6-0.8V -- most books use 0.7V, which is right in the middle of that range. Also note that the BEJ and BCJ are not symmetric and therefore don't have the same forward-bias voltages.

The analysis of this type of problems requires that you assume a state of operation and verify your assumptions (after calculating the voltages and currents in the circuit). For example, if you assume that the circuit is in the forward active mode, you can assume that \(V_{BE}=0.7\text{V}\) and you can also use the fact \(I_C = \beta I_B\). After using these to analyze the circuit you must show that the BCJ is reverse biased; therefore, you must show that \(V_{BC} < 0.4\text{V}\). Note that this voltage is not 0.7V because the junctions are not identical.

If you want to assume that the transistor is in saturation, you can assume that \(V_{BE}=0.7\text{V}\) and \(V_{BC}=0.5\text{V}\). This results in \(V_{CE}=V_{CE,sat}=0.2\text{V}\). With these values, you can analyze the circuit to determine the voltages and currents. You should NOT use the \(\beta\) in this mode -- that is only for the forward-active mode. In the saturation mode, the ratio of the collector current and the base current should be less than \(\beta\). In other words, \(\beta_{sat} = I_{C,sat}/I_{B,sat} < \beta \).

In cutoff, both junctions are reverse biased; therefore all of the transistor currents are zero. In this mode, you have to show that \(V_{BE}<0.7\text{V}\) and \(V_{BC} < 0.4\text{V}\) for the cutoff assumption to be valid.

Does this make sense? It is actually not that difficult but you have to understand why you are doing certain things. Also, depending on the textbook, the forward-bias values, etc. might be slightly different.

Best regards,
Vahe
 
Last edited:

hobbyist

Joined Aug 10, 2008
887
Vahe,
Thankyou for taking the time to explain that in very good detail.

You have a very good way of explaining and teaching these things. It was very well written, and explained.

Keep up the great work..
 
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