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# BJT operation

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#### PG1995

##### Active Member
Hi,

Question 1:
It says, "Saturation is the state of a BJT in which the collector current has reached a maximum and is independent of the base current."

The word 'saturation' suggests a situation or point beyond which no further increase is possible.

You can see that in the saturation region, Ic increases for increasing value of Vce until point 'B' where saturation occurs for the given value of Ib; and from point 'B' onward there is no further increase in Ic for increasing value of Vce.

In my opinion, the phrase 'saturation region' is not a good descriptive word because Ic does increase between the point 'A' and 'B'. Yes, 'B' is a point where saturation occurs and hence should be called ''saturation point'. Further, on the contrary, the region between points 'B' and 'C' is more of a saturation region than a active or linear region because Ic doesn't increase. What's your opinion on this? Perhaps, the choice of this terminology has to do something with amplification using BJT.

Question 2:
As it says that saturation is the state of a BJT in which the collector current has reached a maximum and is independent of the base current. I don't see how it's independent of base current. You can see that all three points 'X', 'Y', and 'Z' lie in the saturation region but still value of Ic is different for each point.

Question 3:
In the cutoff region, Ic is not entirely zero. Is this leakage current?

Thanks a lot for your help.

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"Saturation is the state of a BJT in which the collector current has reached a maximum and is independent of the base current."
I see saturation different that some people. In class we plot Ib and Ic whenRc-0hms.

I see saturation as a point where Vcc is mostly across Rc and very little voltage is across C-E. The collector current will not increase with more base current is because Vc-e is very low.

Low voltage transistors have Vsat of 0.2 volts. 1200 volt transistors are more like 1 volt.

Will post more in a moment.

Here is data from a transistor I know very well. "HD1760JL"
Here the saturation voltage is described as the C-E voltage (Ic=18A, Ib=4.5A)
s
Next graph: current gain is forced to 4. Measure Vce, load is a resistor,
Note that with a Ic of 10A the Vce=2V.

A 50V transistor will have a Vce(sat) in the 0.2 range.

As it says that saturation is the state of a BJT in which the collector current has reached a maximum and is independent of the base current.
I agree that is a confusing definition.
The definition I prefer is, when saturated, the collector-emitter voltage does not change significantly with an increase in base current.

Hi,

Question 1:
It says, "Saturation is the state of a BJT in which the collector current has reached a maximum and is independent of the base current."

The word 'saturation' suggests a situation or point beyond which no further increase is possible.

You can see that in the saturation region, Ic increases for increasing value of Vce until point 'B' where saturation occurs for the given value of Ib; and from point 'B' onward there is no further increase in Ic for increasing value of Vce.

In my opinion, the phrase 'saturation region' is not a good descriptive word because Ic does increase between the point 'A' and 'B'. Yes, 'B' is a point where saturation occurs and hence should be called ''saturation point'. Further, on the contrary, the region between points 'B' and 'C' is more of a saturation region than a active or linear region because Ic doesn't increase. What's your opinion on this? Perhaps, the choice of this terminology has to do something with amplification using BJT.

Question 2:
As it says that saturation is the state of a BJT in which the collector current has reached a maximum and is independent of the base current. I don't see how it's independent of base current. You can see that all three points 'X', 'Y', and 'Z' lie in the saturation region but still value of Ic is different for each point.

Question 3:
In the cutoff region, Ic is not entirely zero. Is this leakage current?

Thanks a lot for your help.

The active region is defined as the e-b terminals being forward biased and the b-c terminals being reversed biased. Supposedly, there needs to be some reverse voltage across the b-c terminals to receive the current to/from the emitter, past the base, and onward/inward to/from the collector. The charge flow direction depends on PNP or NPN. Most literature says that when no voltage exists between the b-c terminals, the transistor does not operate in the active region anymore. However, the link below says otherwise. You can pontificate all you want about how the saturation mode works or is supposed to work, but for sure it works differently than in the active region.

This link shows that strange and hard to understand things happen near the boundary of active and saturation regions. You might also search the web for part-1 of that same topic. It's out there waiting for you.
https://www.electronicdesign.com/power/whats-all-vsubbesub-stuff-anyhow-part-2

Ratch

When a transistor is driven on hard (saturated) it gets slow and there is a storage delay time.

Thank you!

Question 1:
Please have a look on the attachment, bjt_amp1.

I don't understand how Vc=Ic*Rc. In my opinion, Vc should be Ic*r'_e and Vcc=Ic*Rc + Ic*r'_e.

Question 2:
Please have a look on the attachment, bjt_datasheet1.
What would be the corresponding base current, I_b, for this maximum collector current?

Question 3:
This question is also about bjt_datasheet1 attachment.
What do these values, V_CE(sat) and V_BE(sat), suggest?

Question 4:
At most places only internal ac emitter resistance, r'_e, is given. Does there exist internal ac collector resistance, r'_c?

Question 5:
Does there exist a minimum value of base current which is required to get the transistor started?

Thank you!

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• bjt_amp1.jpg
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• bjt_datasheet1.jpg
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Thank you!

Question 1:
Please have a look on the attachment, bjt_amp1.

I don't understand how Vc=Ic*Rc. In my opinion, Vc should be Ic*r'_e and Vcc=Ic*Rc + Ic*r'_e.

Why make it complicated? Obviously Vc is (Vcc - Ic * Rc). Can't you understand that there is a voltage drop across resistor Rc., and the drop value is Ic * Rc?

Question 2:
Please have a look on the attachment, bjt_datasheet1.
What would be the corresponding base current, I_b, for this maximum collector current?

Depends on the BETA value at that operating point. BETA varies widely even among the same transistor types and batch lots.

Question 3:
This question is also about bjt_datasheet1 attachment.
What do these values, V_CE(sat) and V_BE(sat), suggest?

They allow you to calculate the power dissipated in the collector and emitter circuits. Allowing too much current in those circuits will burn up the transistor.

Question 4:
At most places only internal ac emitter resistance, r'_e, is given. Does there exist internal ac collector resistance, r'_c?

Resistance is the same no matter which direction the charge flows. However, AC (Alternating Cycle) voltage produces reactance if capacitance or inductance is present, which will reduce the current and change its phase.

Question 5:
Does there exist a minimum value of base current which is required to get the transistor started?

Thank you!
No, a transistor is a transconductance device (voltage controls current) . Some literature shows a transistor with base and collector resistors and try to tell you that since Ic is dependent on Ib, then the transistor is a current driven device and a current amplifier. But that is not true. A transistor is a voltage driven device (transconductance). What they are showing you is a current amplifying circuit, using a transconductance BJT. The physics of the transistor operation prove it.

Ratch

Edit: OK, I see your confusion now. The book is talking about the AC voltage at the collector. In that case, ignore the DC (Defined Constant) supply voltage and the calculation becomes the AC collector current times Rc.

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No, a transistor is a transconductance device (voltage controls current)
The physics of the BJT transistors says that it is a voltage operated device (with the input looking like a forward-biased diode with a low, non-linear input impedance which generates a base current from the base-emitter voltage).
And using its transconductance value for the small-signal AC gain calculations works well.

But for DC bias and large-signal switching applications, treating the transistor as a black-box, current operated device, and using the current gain of the transistor (Beta or hfe from the data sheet) for the design calculations is way easier.
Trying to use transconductance for those calculations (which no design engineer ever does) is generally an exercise in pedantic masochism.

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Thank you!

Why make it complicated? Obviously Vc is (Vcc - Ic * Rc). Can't you understand that there is a voltage drop across resistor Rc., and the drop value is Ic * Rc?

Possibly, I'm missing something obvious. In the text, on the contrary, it is written Vc=Ic*Rc. In my last post I also said "Vcc=Ic*Rc + Ic*r'_e" where Vc=Ic*r'_e. Thanks.

The physics of the BJT transistors says that it is a voltage operated device (with the input looking like a forward-biased diode with a low, non-linear input impedance which generates a base current from the base-emitter voltage).
And using its transconductance value for the small-signal AC gain calculations works well.

I think we are coming to an agreement. Unlike a FET or tube, a BJT works by diffusion. Electrons from the N slap diffuse into the P slab and holes from the P slab diffuse into the N slab. This is the diffusion current. The loss of electrons from the N-P boundary of the N slap, and the loss of holes from the N-P boundary of the P slab sets up a reverse counter voltage that gradually inhibits further diffusion current. Applying an external forward voltage to the N-P slab lowers the counter voltage and allows more diffusion current to occur. This diffusion operation explains why the voltage-current plot of a diode is non-linear and exponential instead of a straight line like a FET or tube.

But for DC bias and large-signal switching applications, treating the transistor as a black-box, current operated device, and using the current gain of the transistor (Beta or hfe from the data sheet) for the design calculations is way easier.
Trying to use transconductance for those calculations (which no design engineer ever does) is generally an exercise in pedantic masochism.
For biasing, the waste current (base current Ib) is a factor. For operation in the active region, most of the emitter current passes through the base and onward to the collector when the base voltage lowers the reverse counter voltage caused by the emitter-base diffusion. However, some current gets diverted from the collector current into the base terminal where it does no good. This waste base current has to be taken into consideration when calculating bias, so the beta of the transistor is useful for that purpose. For switching circuits, the transistor is operating from the cut-off to the saturation regions, with as little time possible spent in the active mode. Obviously, transconductance is meaningless when no current exists at cut-off, and also meaningless when base voltage does not control the collector current in the saturation region.

Ratch

Hi,

This is an edit to my previous post.

Possibly, I'm missing something obvious. In the text, on the contrary, it is written Vc=Ic*Rc. In my last post I also said "Vcc=Ic*Rc + Ic*r'_e" where Vc=Ic*r'_e and it is the voltage drop across the transistor. How come the voltage drops across the transistor and resistor are equal as the text assumes Thanks.

Hi,

This is an edit to my previous post.

Possibly, I'm missing something obvious. In the text, on the contrary, it is written Vc=Ic*Rc. In my last post I also said "Vcc=Ic*Rc + Ic*r'_e" where Vc=Ic*r'_e and it is the voltage drop across the transistor. How come the voltage drops across the transistor and resistor are equal as the text assumes Thanks.
Hi,

This is an edit to my previous post.

Possibly, I'm missing something obvious. In the text, on the contrary, it is written Vc=Ic*Rc. In my last post I also said "Vcc=Ic*Rc + Ic*r'_e" where Vc=Ic*r'_e and it is the voltage drop across the transistor. How come the voltage drops across the transistor and resistor are equal as the text assumes Thanks.

Because the text is talking about AC voltaqe, and the AC voltage across the collector voltage source is zero. That makes the transistor and resistor virtually in parallel with each other, so they have the same voltage value.

Ratch

Thank you.

Because the text is talking about AC voltaqe, and the AC voltage across the collector voltage source is zero. That makes the transistor and resistor virtually in parallel with each other, so they have the same voltage value.

Ratch

I'm sorry but I still don't understand how Vc and Vr are equal. Could you please give it another try? Thanks.

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Thank you.

I'm sorry but I still don't understand how Vc and Vr are equal. Could you please give it another try? Thanks.

First all, we are talking about AC voltage. Therefore, the AC voltage across Vcc is zero. That makes the south end of RC at the same potential as the ground bus. Then, since the collector and the north end of Rc share the same terminal, the AC voltage of each of the components must be equal to the AC voltage Vc.

Ratch

Thank you!

First all, we are talking about AC voltage. Therefore, the AC voltage across Vcc is zero. That makes the south end of Rc at the same potential as the ground bus. Then, since the collector and the north end of Rc share the same terminal, the AC voltage of each of the components must be equal to the AC voltage Vc.

Ratch

If the south of Rc is at the potential as the ground for AC voltage then why is the potential of north of Rc greater than the ground bus? Is the north also at the same potential as ground which would make Vc equal to ground? Thanks for the help and patience.

Thank you!

If the south of Rc is at the potential as the ground for AC voltage then why is the potential of north of Rc greater than the ground bus? Is the north also at the same potential as ground which would make Vc equal to ground? Thanks for the help and patience.

For the north of Rc, the voltage is positive for half a cycle, negative for the other half cycle, and zero for two instants per cycle.

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Thank you!

Please have a look on this attachment, bjt_amp1a.

Question 1:
Shouldn't it be Vc = - (Ic*R_C)?

Question 2:
Shouldn't it be Vin instead of Vs?

#### Attachments

• bjt_amp1a.jpg
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Thank you!

Please have a look on this attachment, bjt_amp1a.

Question 1:
Shouldn't it be Vc = - (Ic*R_C)?

Question 2:
Shouldn't it be Vin instead of Vs?

The book is correct. Keep in mind that it is referring to AC voltage. What is "R_" to which you keep referring?

Ratch

Thanks.

It's collector resistor; the resistor connected to collector terminal of transistor.

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