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Design >> Analog Design >> current source design
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Message started by filipe on Jul 20^th, 2007, 2:21pm

Title: current source design
Post by filipe on Jul 20^th, 2007, 2:21pm

I'm desingning an operational amplifier, and I'm implementing a current source with a MOS transistor M1.
All the projects that I have seen have a small length to M1, probably due to the matching, I think.
But if I have a big L to M1, I have a greater impedance to the current source, and this is good. What
should I do??? Why dont use a big L???
Thanks

Title: Re: current source design
Post by aaron_do on Jul 21^st, 2007, 12:33am

Is M1 the input transistor (with all the gm)? If so then the small L gives higher Gm*BW since cgs is smaller for the same aspect ratio(W/L). Large L gives better matching, output resistance, and potentially better noise performance.

Title: Re: current source design
Post by filipe on Jul 23^rd, 2007, 9:59am

The M1 is the current source, as shown in the figure.
Why should I use a small L to it?? Why don’t use a big L to M1 and MB?

Title: Re: current source design
Post by joeb on Jul 24^th, 2007, 1:16am

Hi,

When u use small L values , you have the benefit of low Vdsat. This is low power constraints.
It's clear that u have trade offs with matching

Title: Re: current source design
Post by Sid on Jul 24^th, 2007, 3:46pm

Your M1 current source should have a large L for better output impedance. However, if you make L of M1 large you limit your input common-mode range (since now M1 has a large vdsat, your PMOS inputs will need to have a lower input common-mode for M1 to be in saturation). The high output impedance of M1 will improve your input common-mode rejection (CMRR). So, it is a trade-off between input common-mode compliance (i.e. range) and CMRR.

-Sid

Title: Re: current source design
Post by Monkeybad on Aug 2^nd, 2007, 7:04pm

Why not? The current source will become more ideal with large L.
Besides, I'm wondering that why the large L will affect the Vdsat and the input common mode voltage.
If you keep the Vgs of M1 the same, then you can still have the same Vdsat of the M1 for the saturation.
But now the current of M1 is lower when you increase the L, so you can increase the W to keep the
same current. Isn't it? (It's just my thought and correct me if I'm wrong.)

BEST REGARD

Title: Re: current source design
Post by doho on Aug 5^th, 2007, 4:58am

Monkeybad wrote on Aug 2^nd, 2007, 7:04pm:

Yes, you are absolutely right. The drain-source current is proportional to W/L and the output resistance to 1/L, so by increasing W and L with the same factor you can achieve a higher output resistance. But a larger transistor also has higher capacitances, so the effective output capacitance of your current source will increase. If you do not expect to have any voltage swing at the output node of the current source, then the extra capacitance will not matter. But if you expect large voltage swings at high frequenies, it might actually degrade the output impedance. A larger transistor also takes up more area which might be valuable. Summarizing you can say that, increasing W and L with the same factor for the output transistor in this current source results in higher output resistance, larger area and larger output capacitance.