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Design >> Analog Design >> Diode Connected BJT: Current Eqn:: a fundamental doubt
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Message started by ipsc on May 26^th, 2011, 11:24am

Title: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by ipsc on May 26^th, 2011, 11:24am

The current for a diode connected BJT (base & collector shorted) is given by
I = Is * exp(qVBE/kT)

Now, my doubt is: what is ‘I’ here – IE or IC?

My intuition is that, it has to be ‘IE’ because above eqn is similar to a normal diode eqn and the diode here is VBE. And moreover, IC is a secondary form of this diode current.

It would be great, if someone can please clarify whether it is 'IE' or 'IC' and why (if possible with a reference)?

Title: Re: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by loose-electron on May 28^th, 2011, 6:14pm

Its a scalar for the current that is proprotional to the diode geometry - Dig out a copy of Grey and Meyer (or newer variant has more authors these days) and its all explained in there.

Treat it as Id and don't think like its a transistor anymore.

Or treat it like its a bipolar, and that Vbase = Vcollector and you are done.

Both are valid approaches

Considering how high beta is, is it really important anyhow?

Title: Re: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by raja.cedt on May 29^th, 2011, 6:49pm

hi,
it is Ic only, just derive from device physics basics. I didn't understand your intuition. Again i will go with loose-electron, when you have higher beta (typically that is the case), more or less it doesn't matter.

Thanks.

Title: Re: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by ipsc on May 30^th, 2011, 3:41am

Thanks loose-electron & raja.cedt for your replies.

@loose-electron

Quote:

Treat it as Id and don't think like its a transistor anymore.

Or treat it like its a bipolar, and that Vbase = Vcollector and you are d

If I treat this way i.e. if I reduce BJT to two terminals:

Termial1 = Emittor
Termial2 = Collector + Base

Hence the diode current 'Id' becomes
Id = IE = IC+IB = Is * exp(qVBE/kT)

please correct me if I am wrong.

@ raja.cedt

Quote:

it is Ic only, just derive from device physics basics

sir, i am bit out of touch with the device physics. Could you please explain it to me conceptually/intuitively or at least methodology of derivation in brief?

Quote:

Considering how high beta is, is it really important anyhow?

But for CMOS technologies 'beta' is only in single digits, right?

Regards
ipsc

Title: Re: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by boe on May 30^th, 2011, 8:04am

ipsc wrote on May 30^th, 2011, 3:41am:

But for CMOS technologies 'beta' is only in single digits, right?

For the parasitic pnp in modern technology nodes, yes.
B O E

Title: Re: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by raja.cedt on May 30^th, 2011, 9:16am

agree, for parasitic pnp diode beta is lesser in that case you should use IC rather than IE.
Thanks.

Title: Re: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by loose-electron on May 30^th, 2011, 6:07pm

If it is a good quality bipolar, Beta is on the order of 100 (YMMV) (a horizontal NPN with a thin base well structure)

If it is the collector tied to substrate PNP associated with a lot of CMOS then its Beta is around 6 to 20 (YMMV) this is a lateral PNP

You asked about bipolars and all of a sudden the question turns to parasitic PNP's on CMOS?

Title: Re: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by ipsc on May 31^st, 2011, 10:55pm

oops... Considering the generic nature of the Question, I didn't expect that not mentioning about parasitic pnp will cause so much confusion.

Let me also put in my main objective for asking this question. I am just trying to understand effects of BJT beta on a CMOS Bandgap. Hope everything is clear now !!!

Now coming back to my original question,

if I treat diode connected (B & C shorted) BJT as a diode and reduce it to two terminals, I will have:

Termial1 = Emittor
Termial2 = Collector + Base

Hence the diode current 'Id' becomes:
Id = IE = IC+IB = Is * exp(qVBE/kT)

i.e. in the diode connected BJT current eqn,
I = Is * exp(qVBE/kT)
I is represented by IE not IC.

But as per raja.cedt it should be IC, if we derive the eqn from device physics basics.

Now, the question is where am I going wrong in my analysis?

Regards
ipsc

Title: Re: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by boe on Jun 1^st, 2011, 12:51am

Ipsc,

Since IB = IC/β, Id = IE = IC + IB = (1+1/β)*IC. Therefore, IE = Is1*exp(qVBE/kT), with Is1 = (1+1/β)*Is.

B O E

Title: Re: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by Alexandar on Jun 1^st, 2011, 1:48am

loose-electron wrote on May 28^th, 2011, 6:14pm:

[..]

Considering how high beta is, is it really important anyhow?

Well, you are less effected by the variations on Rb, which is nice.

Title: Re: Diode Connected BJT: Current Eqn:: a fundamental doubt
Post by loose-electron on Jun 4^th, 2011, 1:51pm

The equations you are going back to are "academic simplifications" - good for illustration of concepts.

However, I would not design a circuit based upon them.
Take the foundry provided models and work with those. If you want bias curves, take the foundry models and do a DC sweep curve set.