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Design >> Analog Design >> Startup problem in fully differential amplifiers
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Message started by rajeee1000 on Feb 5^th, 2009, 8:27am

Title: Startup problem in fully differential amplifiers
Post by rajeee1000 on Feb 5^th, 2009, 8:27am

Hello,

I am seeing a start-up problem in a two-stage fully differential amplifier in a TIA configuration. Does anybody know of how to solve this problem at architectural level without using an explicit start-up circuit?

Regards
Rajesh

Title: Re: Startup problem in fully differential amplifiers
Post by aaron_do on Feb 6^th, 2009, 1:13am

Hi Rajesh,

a schematic would help. However, I've run into a similar problem before and i guess you're talking about the same thing. Start-up can be achieved using just a pair of start-up transistors (pretty easy to design), or you can just add a high impedance load (current source) to the stage which is driving the TIA.

cheers,
Aaron

Title: Re: Startup problem in fully differential amplifiers
Post by rajeee1000 on Feb 6^th, 2009, 4:03am

Hi Aaron,

Please see the schematic here. One solution seems to be to use current sources across the first diff pair which always provide some loop gain for the common-mode feedback to operate. (Mistake in the schematic: there are PMOS level shifters on both the inputs of the CMFB error amplifier).

Your second solution to use a current source at the input of the TIA might be problematic when feedback resistors are larger (can result in input common-mode voltage for the TIA out of range)

Could you please elaborate on the your first solution, the start-up transistors ?

Thanks!

Regards
Rajesh

Title: Re: Startup problem in fully differential amplifiers
Post by aaron_do on Feb 12^th, 2009, 10:10pm

Hi,

your output current sources are pulling the output CM low at start-up. This is fed back to the input diff pair so that it cannot turn on. I'm not sure how well it would work, but you could try this:

Use a pair of PMOS with their drain source accross the gate source of the input diff pair, and the PMOS gates connected to the outputs. So when the output is low, the PMOS will turn on and pull the gates of the input diff pairs up. This will pull the output up and turn off the PMOS start-up transistors.

Like i said, i'm not so sure how well it would work, but as far as i know, when you have a self-biased circuit like that, you need some kind of start-up...

Sorry i took so long to reply,
Aaron