Why we can't measure Vbe?

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Osawa_Odessa

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In a PN junction, electrons migrate from the N-material to the P-material, leaving a net positive charge on the N side of the junction, and net negative charge on the P side. This continues until the resultant E field it too much to overcome, and the device is in a state of equilibrium. You cannot measure the voltage that results from this charge migration, because the device will simpley continue to adjust itself to maintain equilibrium.

Can you help me to explain the bold part?
 

The measured voltage is zero because no electric field is created. The particles arrange themselves and create a 'potential barrier' in a sort of way that they need a certain electric filed to start conducting some current.
 

Osawa_Odessa said:
You cannot measure the voltage that results from this charge migration, because the device will simple continue to adjust itself to maintain equilibrium.
Click to expand...
The contact potential appearing across junction is a built-in potential barrier, in that it is necessary to the maintenance of equilibrium at the junction; it does not imply any external potential. Indeed,the contact potential cannot be measured by placing a voltmeter across the devices, because new contact potentials are formed at each probe, just cancelling the equilibrium potential difference.
 
Indeed,the contact potential cannot be measured by placing a voltmeter across the devices, because new contact potentials are formed at each probe, just cancelling the equilibrium potential difference.
Click to expand...
How about if two probes are made from different materal (metal)?
And one more question.
In transistor Vbe = 0.7 V
Is this Vbe = Vbuilt-in - Vbarrier?
 

Osawa_Odessa said:
How about if two probes are made from different materal (metal)
Vbe = Vbuilt-in - Vbarrier?
Click to expand...
It doesn't matter which type of material you use in making the contact,you can't measure it.
What do you mean by Vbarrier??? and the Vbe = 0.7 V,are you applying 0.7 V ??
 
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rahdirs said:
It doesn't matter which type of contact you use,you can't measure it.
Click to expand...

I meant two probes are differental then the readings is not zero.

What do you mean by Vbarrier??? and the Vbe = 0.7 V,are you applying 0.7 V ??
Click to expand...

I want to know where does this voltage come from?
 

Osawa_Odessa said:
View attachment 91954
I want to know where does this voltage come from?
Click to expand...

Oh,your talking about voltage drop between the p-n junction in a transistor.
Some digital multimeter manufacturers equip their meters with a special “diode check” function which displays the actual forward voltage drop of the diode in volts. These meters work by forcing a small current through the diode and measuring the voltage dropped between the two test leads.

Were you asking where does this voltage come from or are you trying to say that when we can measure this voltage,why can't we measure built-in potential?

If you're trying to say why can't we measure that voltage if we can measure this voltage,in this case you're measuring voltage drop between two ohmic-contacts,even if there is a contact-potential between the probe & ohmic contact,it would get cancelled on the both probes to give only the drop in voltage.
 
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Thanks,
but I am not quite understand that.
In PN junction, there is a built-in potential Vbi
.When you use the meter a above to measure, the meter forces a small current through the junction and the built-in potential vanish?
Then according to Ohm's law: V= I*R = 0 if we assume that the junction has no resistance.
 

No,the built-in potential doesn't vanish.
When the two p-n junctions are brought together,as you already said in post#1 due to diffusion the built in-potential develops(say,V0).
If i apply a forward voltage Vf,the potential is decreased by Vf to V0-Vf,it doesn't become zero,the barrier still exists.
 
Thanks, that is a bit clearer.
I still don't know where does Vbe =0.7 from. Is Vbe = V0 - Vf?
But V0 is built in voltage at equilibrium and it is a constant. Vf is external voltage source that can change its value. But Vbe = 0.7 V is relatively constant.
Then it seems that Vbe = V0 - Vf is wrong.
And if I apply Omm's law for the second configuration (forward bias) it seems wrong:
Vf + ε = 0 <=> Vf + V0 - Vf =0 <=> V0 = 0:-?
 

Osawa_Odessa said:
And if I apply Omm's law for the second configuration (forward bias) it seems wrong:
Vf + ε = 0 <=> Vf + V0 - Vf =0 <=> V0 = 0:-?
Click to expand...
No,you're again thinking it wrong in your application of KVL.You are forgetting Contact Potential.

If i attach(by attach,i mean making electrical contact,not physical contact) two wires to a semiconductor,one to the p-side & other n-side & i short them.Does a current flow ?
No,common sense that current cannot flow by shorting two ends of p-n semiconductor.
why doesn't current flow even though there is a barrier potential V0.Where does this potential difference go?
This potential difference is cancelled by the contact potential between the wire & semiconductor.
Applying KVL,
Combined Contact-potential at both ends + V0 = 0 (this is when no external potential was applied)
=> Comb.Contact Potential = -V0.

Now,on applying Potential difference Vf using the same wires,apply loop equation,
Vf + V0-Vf + Combined Contact Potetial = 0 (KVL Satisfied & V0 not equal to 0)
 
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Thanks a lot! I am starting to understand it.
Now,on applying Potential difference Vf using the same wires,apply loop equation,
Vf + V0-Vf + Combined Contact Potetial = 0 (KVL Satisfied & V0 not equal to 0)
Click to expand...
How about if two wries are different in material?
Then the contact potential of each wire with semiconductor are different.
And if possible could you give me a link about contact potential? I need to look it.
 

Osawa_Odessa said:
How about if two wries are different in material?Then the contact potential of each wire with semiconductor are different.
And if possible could you give me a link about contact potential? I need to look it.
Click to expand...

If you're using two wires say one is Aluminium & other is Iron (i don't know if u even use Aluminium & Iron as connecting wires,take it just for example).
Then there unlike my previous post,in this case the Contact potential of Aluminium & Silicon is different from Iron & Silicon.But in this case,there is a contact potential between Iron & Aluminium also.

Thus if i short those two wires under equilibrium,& apply KVL
Contact-Potential(Al-Si)+Contact-Potential(Fe-Si)+Contact-Potential(Al-Fe)+V0 = 0
=> Contact-Potential(Al-Si)+Contact-Potential(Fe-Si)+Contact-Potential(Al-Fe) = -V0.
unlike in earlier post if i used both wires of Aluminium,
=> Contact-Potential(Al-Si)+Contact-Potential(Al-Si)+V0 = 0

Contact-Potential is the potential developed between two materials when-ever you electrically connect the two materials.It is defined as the work function of the first material tip, φ1, minus that of the second material, φ2.It can be positive or negative.
I think there should be a table listing contact potentials between two materials listing all materials.I read it from here & there.
If you are confused,just try applying known situations like current doesn't flow when you short a semiconductor.
 
Hi rahdirs,
Now I am interested in this too.
in this case you're measuring voltage drop between two ohmic-contacts
Click to expand...
Do you mean "Combined Contact-potential at both ends" here?
I am confused why the meter can measure the drop voltage in diode (or BE junction of trans).
In this case, the potential difference between two probes is:
Vmeasure =Combined Contact-potential at both ends + V0 but this is equal to zero. How the meter show 0.65V? I only think a possible that V0 has to be zero but don't know if it is right or not.
 

Osawa_Odessa said:
I am confused why the meter can measure the drop voltage in diode (or BE junction of trans).
Click to expand...

Actually, the voltage is not direct measured. The diode check scale in multimeters works as a current source. It applies a small current (like 1 mA) and show the necessary voltage to conduct this current. In silicon semiconductor junctions, the voltage needed to conduct 1 mA is between 0.55 V and 0.7 V. You see, this is the principle of operation for semicondutors devices. They will not conduct electric current unless you apply a minimum amount of energy across their terminals.
 
Can you help me find contact potential ψKL in this image?
I got stuck at voltage source part. Assuming that both leads of the voltage source are made of the same material.


Assume that the material next + lead of voltage source is Mk and the one next to -lead is Mk+1. Also let's call contact potential of + and - leads is ψ+ and ψ- respectively.
ΨKL = ΨM1 - ΨM2 + ΨM2 - ψM3 +...+ ψMk-1 - ΨMk + ψMk - (ψ+) + Vsource + ( ψ-) - ψMk+1 +...+ ψM(n-1)-ψM
ψKL = ΨM1 + Vsource -ψM
Is this right?

I am confused if there are contact potentials between each + and - lead with voltage source?
 
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Osawa_Odessa said:
Can you help me find contact potential ψKL in this image?
I am confused if there are contact potentials between each + and - lead with voltage source?
Click to expand...
I don't understand where did ΨKL come??Is it the potential on joining M1 & Mn-1.
There is a contact potential between the positive terminal & Mk,there is contact potential between negative terminal & Mk+1.
Are you asking if there is contact potential between terminal of battery & electrolyte in the battery ???
 

I don't understand where did ΨKL come??
Click to expand...
It is only a symbol to indicate the total contact potential between M1 and Mn.
Are you asking if there is contact potential between terminal of battery & electrolyte in the battery ???
Click to expand...
yes, that is confusing me.

- - - Updated - - -


Let's consider forward bias case.
And call ψ1, ψ2, ψ3 are contact potentials of wire (or leads of voltage source), P, N respectively.
Now apply Krichoff's law:
Vf = ψ1 - ψ2 + ψ2 -ψ3 + ψ3 - ψ1 = 0???
Vf is external source and it is nonzero.
Can you tell me where I am wrong?
 

Vf + Contact.Potential between the wire connecting + lead & pregion + V0-Vf + Contact.Potential between the wire & n-region =0
Vf + ψ1- ψ2 + V0-Vf + ψ1- ψ2 =0
Click to expand...
I am a bit confused now.
I think "Contact.Potential between the wire connecting + lead & pregion" and "Contact.Potential between the wire & n-region" are equal in magnitude but opposite sign.
Let's rename contact potential( all potential is in reference to vacuum level):
ψ1: wire contact potential
ψ2: Si contact potential
Then: Vf + ψ2- ψ1 + V0 - Vf + ψ1 - ψ2 =0 => Vf = V0 - Vf???
 

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It seems logical that both potentials should be equal & of opposite sign.But if i say write the KVL loop equation when p-n junction is shorted then,
ψ2- ψ1 + V0 + ψ1 - ψ2 =0 => V0 = 0.
But we know that V0 ≠ 0.I think we are forgetting a very-important point.The built in voltage was developed due to space-charge regions that were developed at the junction.This built-in potential was developed to stop the diffusion process.I think we are missing some voltage term -V0 that has caused for this built-in potential to appear..Let's wait for some-one to continue this.
 
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