Bandgap reference BJT question

Hi
I've a question about bgr's BJT.
Attatched is my bgr circuit. and it operates well.

But I've known that BJT used in figure is used as diode.
so, using diode's temperature characteristic of forward drop voltage.
Isn't it??

Now I have a question about that.
Why that circuit using BJT instead of diode?
I used CMOS process and it gives me models of both diode and bjt.

Someone helps me
Thannks in advance/

Diode P+ - NWELL is really substrate P - N - P transistor.

Thanks gevy
my question is exactly that!
you said "Diode P+ - NWELL is same as substrate P - N - P transistor".
Then why not using diode??(only p-n)(not pnp)????

Ljy4468,

If you use a Nwell to substrate diode, how are you going to make the diode conduct any current if your subs = gnd (which is the case for a p-sub process) unless you take your Nwell potential to negative?You need an isolation; and that is why you have an Nwell. That way you can use the "diode" (better known as the base emitter junction) formed between the p+ and Nwell. That way neither the base nor the emitter is committed to gnd.

Hi
Realy if You have diode p+ nwell You have pnp-p+-nwell-psub

I believe that what you are getting at is that there is a specific structure of P+/Nwell/Psub that is used for the "bipolar", so you are asking why use that structure rather than simply a P+/Nwell "diode". Here is my take on this:

1- The "bipolar" will simulate more accurately than the "diode", since it will include the substrate current that is probably not modeled for the "diode".
2- There usually is a specific structure for the "bipolar" that has characterization data available. When building a bandgap structure, the good characterization is needed in order to properly determine the tempco of the Base-emitter voltage.
3- The additional structure of the bipolar should help prevent current injection into other substrate tied devices.

There is, of course, nothing preventing the use of a P+/Nwell diode in your application.

The Vbe of BJT is more accurately than the diode.

Thanks all!
I understood well, good luck!