BJT hybrid pi model,why "always" ro resistor is shown?

[Alim Khan]

In BJT hybrid pi small signal model, we have a resistor "ro" between Collector and Emitter.

This resistor is to include the change in collector current when there is a small signal voltage change between collector and emitter. (due to early effect)

I am finding that all BJT hybrid small signal models keep this "ro" even when there is no
small signal voltage "change" between collector and emitter.

Assume that there is no load resistor between collector and ground. That means always VCC will be maintained between collector and ground.
If Emitter is grounded, then this means that there cannot be any change in voltage between collector and emitter.
"Delta VCE" = 0, means , there cannot be any change in collector current.
As there cannot be any change in collector current, we can avoid this "ro" resistor which
essentially is ( "ro = "Delta VCE"/ "Delta IC" )
In this case why do we include this "ro" resister?Also why don't we maintain the voltage across this ro as " delta VCE" ?
Please recollect that for " r pi" resistor between Base and Emitter we always maintain the
voltage as "delta VBE"

 

[analogdesign]

ro is part of hybrid-pi model that is used if the BJT is biased in the forward active region. In your scenario vce = 0 so the device is cut off and not operating as a transistor anymore.

If we don't have ro how do we express the non-ideal behavior of the BJT? Remember that ideally ic is constant over vce but ro expresses the fact that this is not exactly true.

 

[Alim Khan]

hi in my case VCE is not zero.

VCC is maintained between collector and ground . And Emitter is grounded.
This means VCE is equal to VCC ( dc battery)
VCE =VCC, but ΔVCE = 0 ( since we are not applying any small signal at VCE, ΔVCE= 0)

please assume BJT is biased in the forward active region itself.

Ic = Is {exp(VBE/VT) -1} (1 +VCE/VA)

ΔIc = Is {exp(VBE/VT) -1 } ΔVCE / VA

ro = ΔVCE / ΔIc

if we see hybrid-pi model there is no ΔVCE as part of the model.
that is my doubt. if ΔVCE is not shown, that means ΔVCE = 0.

if ΔVCE = 0 , then ro also must be zero. Am i right?

 

[LvW]

if ΔVCE = 0 , then ro also must be zero. Am i right?

No - ro is always present.
However, if there is no change in VCE (ΔVCE = 0) it has no influence at all (it is a dynamic resistance only).
Note that, in general, we often have some specific operational conditions in which some parts have no effect.

 

[Alim Khan]

yes,
if there is no change in VCE (ΔVCE = 0) it has no influence at all.but in a case where emitter is collected to the ground, and in the absence of a load resistance at the collector,
the small signal output taken from collector junction will be equal to ( -gm * Vπ * r0 ).

Is this correct behaviour?

 

[Jony130]

How can you have any output voltage if you tied the collector to Vcc (AC ground) ??

 

[analogdesign]

please assume BJT is biased in the forward active region itself.

How can we assume something that is wrong? To be in forward active region the BE junction must be forward biased and the CE junction must be reverse biased. If vc = ve = 0 this is impossible.

 

[Alim Khan]

How can we assume something that is wrong? To be in forward active region the BE junction must be forward biased and the CE junction must be reverse biased. If vc = ve = 0 this is impossible.

as I told before, VC = VCC ( VCC = collector supply line voltage in a common NPN circuit , it is a finite value, say VCC =2 V )

 

[Alim Khan]

How can you have any output voltage if you tied the collector to Vcc (AC ground) ??

ofcourse, that is my doubt. The output is tied to Vcc.
But the presence of ro in the model, accounts for the "Intrinsic Gain" in BJT.
Gain becomes -gm*ro.

Where does this gain (Intrinsic Gain) show up? As output is tied to Vcc, the output cannot change. So where does this gain go?

 

[analogdesign]

ofcourse, that is my doubt. The output is tied to Vcc.
But the presence of ro in the model, accounts for the "Intrinsic Gain" in BJT.
Gain becomes -gm*ro.

Where does this gain (Intrinsic Gain) show up? As output is tied to Vcc, the output cannot change. So where does this gain go?

OK I finally understand your question!

You have connected the output to VCC which looks like a small-signal ground. So in your small-signal model you have gm driving ro in parallel with ground (or zero ohms). So, you have gain = gm*zero = zero. That is where the gain goes.

 

[Alim Khan]

ofcourse, that is my doubt. The output is tied to Vcc.
But the presence of ro in the model, accounts for the "Intrinsic Gain" in BJT.
Gain becomes -gm*ro.

Where does this gain (Intrinsic Gain) show up? As output is tied to Vcc, the output cannot change. So where does this gain go?

OK I finally understand your question!

You have connected the output to VCC which looks like a small-signal ground. So in your small-signal model you have gm driving ro in parallel with ground (or zero ohms). So, you have gain = gm*zero = zero. That is where the gain goes.

I disagree because gain is not zero.

r0 = VA/ Ic
Va not equal to zero.
Ic is the dc bias current, which is also not zero.
Hence r0 is not zero.

For the resistance ro, which is connected from collector to ground( as emitter is grounded), the current passing through resistor is ( -gm Vπ).
So Voltage is ( -gm Vπ) * ro.

Gain is ( -gm Vπ) * ro / Vπ = -gm*ro.
This particular gain is called intrinsic gain of the transistor.
This is different from normal gain.


Can anyone please tell me what is the purpose of this intrinsic gain?

 

[analogdesign]

I disagree because gain is not zero.

r0 = VA/ Ic
Va not equal to zero.
Ic is the dc bias current, which is also not zero.
Hence r0 is not zero.

For the resistance ro, which is connected from collector to ground( as emitter is grounded), the current passing through resistor is ( -gm Vπ).
So Voltage is ( -gm Vπ) * ro.

Gain is ( -gm Vπ) * ro / Vπ = -gm*ro.
This particular gain is called intrinsic gain of the transistor.
This is different from normal gain.

Call it whatever you want. You are mixing up concepts. If you connect the collector to VCC you are shorting out your small-signal gain. You might have an "intrinsic gain" but you can't measure it because you've connecting the device in parallel with zero ohms.

You asked where the gain went and I told you. Try hooking up a BJT as you've described and measure its gain. It won't be gm*ro.

 

[Alim Khan]

Call it whatever you want. You are mixing up concepts. If you connect the collector to VCC you are shorting out your small-signal gain. You might have an "intrinsic gain" but you can't measure it because you've connecting the device in parallel with zero ohms.

You asked where the gain went and I told you. Try hooking up a BJT as you've described and measure its gain. It won't be gm*ro.

Hmm. I think i

Call it whatever you want. You are mixing up concepts. If you connect the collector to VCC you are shorting out your small-signal gain. You might have an "intrinsic gain" but you can't measure it because you've connecting the device in parallel with zero ohms.

You asked where the gain went and I told you. Try hooking up a BJT as you've described and measure its gain. It won't be gm*ro.

I think you are very right.
I didnt consider the zero resistance path which is parallel with ro.

Now this makes very much sense.
Thank you.

 

[analogdesign]

I think you are very right.
I didnt consider the zero resistance path which is parallel with ro.

Now this makes very much sense.
Thank you.

You are welcome. Small-signal analysis is a confusing concept and it takes a long time to get the hang of it. Remember that models are called "models" for a reason. They are not the real thing. They can be very useful in certain situations but don't fall into the trap of thinking because the model says something then it must be true in reality.