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on my avatar i show a self biased current mirror, i had been studing this circuit, but i have still one question, how, exactly, does it comes on estability.
when i turn on a self biased current mirror, how the current become stable?
i had been reading papers about this, but is still my question, betwen turned on and current stable, what hapens.
the current mirror Q3-Q4 make the crrent in left branch is equal to the current in right. and the Q1-Q2-R make the current in two branched have another function. You can know the two different functions decide the output current have a stable value, which is the normal output current of this struction.
But you should know that this self-bias structure also have a zero-current stable state. So when the power supply is on, if you don't have a startup circuit the circuit will work under zero-current stable state.
'analysis and design of analog integrated circuits---P.R.Gray' has a good explaination.
I think it can be called bootstrap bias, right? the Q3 Q4 is Current mirror, Q1 Q2 is current source. they built a positive feedback loop. there are two operating point (Iin=Iout), the zero point (Iin=Iout=0) is undesired. but don't worry about starup, because this loop can drive it away from zero by itself.
I have seen this kind of circuit in DC-DC chips. Without start-up circuits, it still works well.
the current is depent on the value of R, Bangap circuits work better than it.
without the startup circuit it can work well, but usually it would need more time to get to stable. i think you can choose whether to design a startup circuit in your project acorrding to your demand.
Let's assume that node 1 is at zero potential. The moment you power up your device, i.e. VDD for your bottom rail and Gnd for your upper rail, both Q3 and Q4 will be ON.
Node 1 then will be pulled up to a higher potential. This situation will lead to positive voltage developed at the gate of Q1. Hence, once it reaches a value that is greater than the threshold voltage both transistors Q1 and Q2 will be ON and drive node 1 back to zero eventually.
I think maybe it cannot be started up if the node 1 is at high level. Let's assume there is some charge in node 1. So it is at high level. At the best, you can add a start up circuit. The start up circuit is also rather simple. There is only several Tr.
I think maybe it cannot be started up if the node 1 is at high level. Let's assume there is some charge in node 1. So it is at high level. At the best, you can add a start up circuit. The start up circuit is also rather simple. There is only several Tr.
When you turn on ANY electrical circuit, after a certain initial transient, the voltages and currents get very near values that can be obtained by a DC analysis - OPERATING POINT. This is a STABLE point, where the circuit will remain if nothing is changed.
The problem with your circuit is that it has two stable operating points: one with all currents being zero and another one defined in the way leebluer described.
The startup circuit is used to avoid having the operating point with all currents being zero. It basically detects if there is no current flowing in the circuit and, in that case, it forces a certain current thought the transistors. In this way the situation with all currents being zero is no longer a stable operating point.
The circuit will then approach the desired operating point as any other electrical circuit does.
The startup circuit's job is to inject current when your circuit is zero current and NOT to inject ANY current whent the circuit works.
So there you must decide how do u sense "NO current"
a.you can use voltage difference between circuit works or zero to judge,
watch all node voltage different by simulation and choose one with max difference voltage as startup input
b.you can use current comparator to compare a current u designed
when the circuit starts...the startup circuit inject current
because the current in main circuit is near zero,it's gm is small
so the startup circuit domain...after some threshold,the circuit comes domain,and the startup is bring to off
so the startup can't too stong or too weak...
some main circuit seems no startup circuit,maybe they use leakage current to start ,but this is not a good design,very process depands,
and careful simulate is need to port to other process..
I think this circuit cant work without a startup, it may be stable at zero.
Actually this loop is a positive feedback loop if circuit is work porperly, but the loop gain is smaller than 1, so it can also be stable. Startup circuit pushes this circuit to a state that the positive loop functions, then the circuit will go stable with positive loopgain < 1
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