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Why is Vbe less when emitter area is more?

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cheenu_2002

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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?
 

dick_freebird

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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

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Cheenu,

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

Vbe = Vt ln(Ic/Is).

Is increases with emitter area. Is it clear now?
 

cheenu_2002

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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

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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

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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

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cheenu_2002 said:
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 said:
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

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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

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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?
 

ssankurathri

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hi cheenu,
have u got solution for this? Given the same area, for increase in current, increases vbe. coz for same area, resistance remains same.
for given resistance as current increases, voltage drop increases.

correct me if I am wrong.

Regards
 

kashyap89

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Think about it this way....With a wider emitter area, the charge/unit area has reduced...remember, we are only increasing the emitter area , say width and not the actual number of electrons , the doping still is the same.
By increasing the emitter area, the electrons moving from the n-region (in the emitter) through the base experience lesser recombination (as the charges are more spatially spread-out) and reach the collector. This means electrons are not lost in the base due to recombination and the depletion potential is less...so lesser vbe

Prasad
 

Ibkay

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Some background physical electronics first:

Basically, Vbe is the fundamental quantity that determines the carrier CONCENTRATION injected into the base from the emitter (i.e. majority carriers in the emitter which become minority carriers on crossing over into the base). This is the main constituent of Ie.

This injected minority carrier CONCENTRATION at the edge of the emitter-base junction space charge layer diffuses across the base to the base-collector space charge layer edge because of the CONCENTRATION GRADIENT created by the reverse bias existing at the base-collector junction which creates an almost zero value of minority carriers there (remember that diffusion rather than drift is the principal cause of charge transport during BJT action, and diffusion depends on a concentration gradient and not the intensity of the electric field). These minority carriers then get swept into the collector because the reverse potential existing there favours the minority carriers and therefore the strong electric field pulls them into the collector (therefore creating Ic).

Some minority carriers don't make it across due to recombination with some of the base majority carriers, but this is typically a small fraction because the base is usually very narrow and much lightly doped compared to the emitter. This is partly responsible for Ib, the other part being the reverse injection into the emitter.

To answer your question, all the device physics charge conservation equations are usually developed starting with carrier CONCENTRATIONS (or current density), only at the end converting to total current by multiplying by the cross sectional area A. If Vbe causes a certain CONCENTRATION of carriers to get injected into the base, then the ACTUAL QUANTITY of charge and hence the CURRENT is proportional to the PRODUCT of Vbe and the Area. Therefore, to maintain a constant CURRENT, you increase EITHER Vbe or Area. In other words, to maintain a constant current, if you increase Vbe, then you must decrease area, or vice versa.

Hope this is useful! A very good book that elegantly covers semiconductor and transistor physics without overwhelming you with equations is Electronic Principles by P Gray and C Searle (very old book but a masterpiece in my opinion).
 
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vysakhk

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Hi Cheenu
Let me put it in this way. When you increase the emitter size its like keeping so many pnp transistors in parallel. So when you pump a current, lesser current will be flowing through a single trasistor, which decreases the vbe of that transistor. As these are in parallel, the vbe of one will be equal to vbe of all.

Regards
Vysakh
 

SunnySkyguy

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40 years ago , we just accepted "size matters" and the result was lower bulk resistance at the expense of junction capacitance and bandwidth. gold doping was used to increase current gain and a few other tricks.

But these days Diodes Inc (nee Zetex) hold a huge number of patents for "extremely low" low junction voltages for both Vbe and Vce and they would be my first choice for any power saturated switch. They also have a brand of diodes called Click to view >>
SBR® (SUPER BARRIER RECTIFIERS)

e.g. ZXTD6717E6 dual complementary SOT23-6
NPN: Vceo =15V; Vce.sat = 0.10 V; I = 1.5A ; Rce=135 mΩ
PNP: Vceo =-12V; Vce.sat =-0.175V; I = -1.25A; Rce=150 mΩ

or SOT23 diodes.com/datasheets/ZXT13N50DE6.pdf Rce=36 mΩ

Putting devices in parallel "can" work ok, if they are matched from the same batch or binned as I get them from my supplier, but thermal runaway can occur when current hogging causes lower diode drops with NTC effect. This can be alleviated by ensuring low thermal resistance substrate heatsink and adding enough ESR in wiring resistance to cover variations in voltage, when used at max current. So essentially the external resistance is on the same order of magnitude as the ESR so that net IR drop is small and "Shockley effect" sensitivity is reduced. I have a rule of thumb for which I have my own design equation that standardizes the required additional resistance for a given ESR , power and thermal resistance for stability against thermal runaway that doesn't require a 2 volt drop with a Constant current source.

Bulk resistance can be modelled as "ESR" in a linear fashion over a useful range of rated current for most diodes including LEDs. There is a direct inverse relationship between ESR and the power handling capability of diodes which depends on both the electrical & thermal resistance and area of the junction.
Luxeon.jpg

Then there are another class of diodes they specialize in called Avalanche diodes which are suitable for Laser , LIDAR with absolute max current of 60A in 10ns, all in a tiny SOT23, but have high Vcesat at low currents.
 
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