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Bandgap Start up (Read 7028 times)
circuit_cook
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Bandgap Start up
Mar 14th, 2013, 4:05pm
 
Hi, all:

    I designed a bandgap, which is the Kunjit type circuit.  I have doubt about the startup. I know there is positive and negative feedback in the bandgap. And at start up, I need to gurantee that negative feedback is larger than the postive feedback, so the overall is negative feedback, but I don't know how in this case.  
   I ran two power supply ramping up simulations with a ramping time of 10u and 1s, the output is attached. va is the node of negative feedback, and vb is the node of positive feedback. The waveform shows sometimes the vb is larger than va, but it doesn't show any oscillation, the circuit seems to start up fine. Does the simulation shows the real scenario, or could there be problems, how can I gurantee the corret startup.
   The startup circuit is an inverter type that compares a reference current to generated IPTAT, and pulls the PMOS. Also the OTA biasing is a current mirror to MP1, so self biased, does this make it harder to startup?

Thanks.
Judy

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circuit_cook
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Re: Bandgap Start up
Reply #1 - Mar 14th, 2013, 4:07pm
 
The startup simulation graph
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test1.jpg
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circuit_cook
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Re: Bandgap Start up
Reply #2 - Mar 14th, 2013, 4:07pm
 
second graph
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rfidea
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Re: Bandgap Start up
Reply #3 - Mar 15th, 2013, 2:50am
 
I would start by analysing your circuit. The risk for not starting is if your va and vb nodes are at 0V, there is no current in your bgr core. If also the opamp is designed so do not pull down the gate of the pmos on top va and vb will still be at 0 V and the circuit does not start. So, have a look how the opamp is design. Is there any risk for this behavoural if there is mismatch in it?

But you have a startup circuitry. Is that "stronger" than the opamp?

My recommendation is that you understand how the circuit starts. Then you run simulations. Doing a slow and a fast rise is good, as you have done. But I would add simulations in all process corners and in temperatures as well. Maybe you should add monte-carlo to get some offsets as well.

You can newer simulate those things to little. If the bgr does not start you can not evaluate your chip.
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circuit_cook
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Re: Bandgap Start up
Reply #4 - Mar 17th, 2013, 7:56pm
 
rfidea:

    Thanks for your comments.  I had ran these slow and fast ramp with Monte Carlo simulations, which include both process and mismatch, and I ran at three different temperatures. The simulation all shows the BG starts up fine.  
   I still worry if the simulations show the real case, yes, if it does not start up right, I can't evaluate it.  
   How do I make sure the startup is stronger than the ota?

Thanks.
Judy
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rfidea
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Re: Bandgap Start up
Reply #5 - Mar 18th, 2013, 12:20pm
 
I would also run real process corner simulations as a complement to the MC ones. MC does not run the worse case, it only pick some random process corners. This give you a very good understanding of the distribution of the bandgap voltage for example. But it does not say anything about if the circuit malfunctions in some corner.

About how strong the startup circuit is. Maybe you can force the circuit in an unvanted state with a short circuit and measure the short current in the startup and in the bgr. The largest will win.
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circuit_cook
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Re: Bandgap Start up
Reply #6 - Mar 18th, 2013, 6:42pm
 
rfidea:

     Thanks a lot for your suggestions. I will try that.

Judy
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yvkrishna
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Re: Bandgap Start up
Reply #7 - Apr 27th, 2013, 10:42am
 
Judy,

"The waveform shows sometimes the vb is larger than va, but it doesn't show any oscillation"

This situation is probably OK as node 'vb' should go high for sometime or else the amp out will try to go towards supply and not to its final steady state(vdd-vgs of pmos)  Remember amp out has to start from supply during the startup process and the settle to its final value.

As already suggested use .IC / .nodeset to force the circuit to see wrong OP and test the strength of your startup circuit.

Regards,
yvkrishna
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circuit_cook
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Re: Bandgap Start up
Reply #8 - May 1st, 2013, 7:27pm
 
yvkrishna:

    Thanks, that explanation is enlightening.

Judy
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