How to design the dead point for bandgap circuit?

[lhlbluesky]

as we know, bandgap circuit has two stable operating points,so a start up circuit is needed;but i find that when i disable my start up circuit,and make vdd a fixed dc voltage, for example 1.8v for smic 0.18mm process, the output vref can still start up
after a longer time about 200ms;while i enable my circuit, the start up time is only 3ms or so;i guess the former appear because there is nonzero static leakage current in the circuit, so even if there is no strat up circuit, it can start up also after a long enough time,is that right?if not right , what is the correct case？

the other question, someone says the bandgap circuit needs a "dead point",that is the static stable operating point which appears for no start up circuit,i add nodeset for nmos bias transistor(gate=0) and pmos bias transistor(gate=vdd) to realize the "dead point"when i do my simulation, is that right? if not, how to design the "dead point" correctly?

pls give me some detailed advice,thanks all.

 

[darkk]

bandgap problem,also

Show the schematic!

 

[Fom]

bandgap problem,also

Introduce transistor mismatch into simulation circuit.

 

[edge_tv]

bandgap problem,also

So I thinke your right, my Prof. was once telling me that leakage current will charge the input Caps of the transistors on till the starting to operat normal. So if you want to power down somthing you have to pull down (or pull up) every transistor wiche could be dangures

 

[fnlee]

Re: bandgap problem,also

I think you also consider the other case,vdd power up(0 to vdd), not fixed vdd,if this circuit (no start up circuit)can start up, your simulation result will be right,and also your said will be right.

 

[eagleliujun]

bandgap problem,also

one: if you test your startup circuit ,you should set the vdd from 0 to 1.8 ,not fixing 1.8 .
the startup is accelerate the start up and disable the zero current condition .
two : i don't understand it ,pls discribe it clearly!

 

[engronger]

Re: bandgap problem,also

when simulation, the bandgap is not always die. why we need a startup circuit ,that is because there is a mismatch.

the dead point, you must set the point which will make your circuit un-work