Why is Vbe less when emitter area is more?

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cheenu_2002

Joined: 27 Feb 2008
Posts: 9

15 Oct 2009 9:48 Why is Vbe less when emitter area is more?

Hi,
In BGR, the delta-Vbe is generated by maintaining a particular ratio of emitter area. Typically the transistor having higher emitter area has lesser Vbe for the same current. I would like to understand the physical reason behind the same. I have seen the literature, but they all prove this by means of equations.... there is no intuitive explanation. Can anyone help?

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dick_freebird

Joined: 04 Mar 2008
Posts: 312
Helped: 45
Location: USA

15 Oct 2009 14:48 Why is Vbe less when emitter area is more?

Because Is scales with area, more area is more current
for the same voltage. Or conversely, less voltage by
the diode equation for the same current.

Put Is1/A1, Is2/A2 into simultaneous diode equations
and see how area goes to voltage.

snafflekid

Joined: 09 May 2007
Posts: 120
Helped: 16
Location: USA

15 Oct 2009 23:09 Re: Why is Vbe less when emitter area is more?

Cheenu,

Ic=Is exp(Vbe/Vt) or reworked,

Vbe = Vt ln(Ic/Is).

Is increases with emitter area. Is it clear now?

cheenu_2002

Joined: 27 Feb 2008
Posts: 9

16 Oct 2009 14:07 Re: Why is Vbe less when emitter area is more?

Yeah. I am able to understand now from the equations. So, from the device physiscs point of view, more emitter area results in more minority carriers being swept across the junction and hence more Is? Is my understanding correct...

ssankurathri

Joined: 16 Mar 2006
Posts: 111
Helped: 6

20 Oct 2009 13:09 Re: Why is Vbe less when emitter area is more?

hi cheenu,

as far as i know, more area means more no.of recombinations and low resistance in the region. This low resistance leads to low voltage drop in the area. so, Vbe is low.

Regards,
skr

cheenu_2002

Joined: 27 Feb 2008
Posts: 9

04 Nov 2009 18:25 Re: Why is Vbe less when emitter area is more?

Hi Kr,
If I maintian the same emitter area for both the BJTs, and force more collector current through one of the BJTs, then what will be the change in Vbe?
I think the BJT carrying more current will have higher Vbe (as seen from the equations). Now, how does more collector current requirement increase the Vbe...
If I decrease the current, then will Vbe also decrease? How is that phenomenon happening? I am not able to understand the physics behind this.. pls help me.

snafflekid

Joined: 09 May 2007
Posts: 120
Helped: 16
Location: USA

04 Nov 2009 22:22 Re: Why is Vbe less when emitter area is more?

cheenu_2002 wrote:

Hi Kr,
If I maintian the same emitter area for both the BJTs, and force more collector current through one of the BJTs, then what will be the change in Vbe?

I understand your intention (a BJT in a negative feedback loop), but the BJT only has forward gain. Your suggestion of forcing Ic to increase Vbe implies reverse gain.

cheenu_2002 wrote:

I think the BJT carrying more current will have higher Vbe (as seen from the equations). Now, how does more collector current requirement increase the Vbe...
If I decrease the current, then will Vbe also decrease? How is that phenomenon happening? I am not able to understand the physics behind this.. pls help me.

Increasing Vbe exponentially increases the number of thermally injected carriers able to cross the base-emitter junction, just like how a diode works. Those carriers are the main component of collector current.

cheenu_2002

Joined: 27 Feb 2008
Posts: 9

05 Nov 2009 13:14 Re: Why is Vbe less when emitter area is more?

Actually why I want to understand this phenomena is that, in resistorless BGR design, 2 different currents are fed to BJTs having same area to generate delta-Vbe. So, I want to understand how the Vbe is increasing with increase in collector current

dick_freebird

Joined: 04 Mar 2008
Posts: 312
Helped: 45
Location: USA

05 Nov 2009 15:55 Why is Vbe less when emitter area is more?

You have to think in terms of current density, not plain
current.

A "transdiode" configured NPN takes most of its current
(as long as you are not internally saturating; respect
the Rc) through the collector and although you are "forcing"
Ic, you are really pushing Ib and Ic just follows in local
feedback.

I have never seen a resistorless bandgap. Where do you
get your positive TC term?