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12th October 2017, 10:37 #1
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Improved current mirror
I do not understand "The circuit on the right is a feedback amplifier with loop gain T. Since all time constants in that loop are of the same order of magnitude, they create a system with several poles"
Could anyone point me to some maths reference / equations ? Razavi book does not mention about the circuit on the right though.

12th October 2017, 10:37

12th October 2017, 17:06 #2
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Re: Improved current mirror
How do I derive the feedback loop gain T = (gm1)*(Rin) ?

12th October 2017, 21:28 #3
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Re: Improved current mirror
The improved Wilson current mirror is discussed in Gray/Hurst/Lewis/Meyer Analysis and Design.

13th October 2017, 05:30 #4
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13th October 2017, 05:30

13th October 2017, 10:56 #5
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Re: Improved current mirror
if you use a simple bipolar current mirror, the output current is smaller than the input current because the base currents are subtracted from the input current to generate input bias currents for the 2 transistors. The sys gain error is 1Iout/Iin, and if you can reduce the base currents the gain error will decrease. You can reduce the base currents by increasing the BF, but for bipolar devices it is finite. And gain error will also occur by the Early effect. At this circuit the explanation is similar, but more complicated because of the feedback.

13th October 2017, 11:46 #6
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Re: Improved current mirror
The sys gain error is 1Iout/Iin

13th October 2017, 12:15 #7
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Re: Improved current mirror
Gain Error = Ideal Gain (1)  Real Gain (Iout/Iin)
For 1:1 current mirror. For 1:M current mirror it is MIout/Iin.
And about Wilson current mirror you can read more here: https://wiki.analog.com/university/c...ext/chapter11

13th October 2017, 12:15

14th October 2017, 04:48 #8
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Re: Improved current mirror
For the loop gain T, you can image that from the drain node of M3, the impedance is 1/gm3 +
Rin=ro1+ 1/gm3.
For M1, from its gate to its drain, the gain is gm1*ro1, which is aproximately gm1*Rin. Also, from the drain of M1 to the gate of M3, to the source of M4, the gain is 1, so the total loop gain T is around gm1*Rin.

Yesterday, 02:30 #9
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Re: Improved current mirror
a) Why is the gain from the drain of M1 to the drain of M3 = 1 ?
b) In most CMOS applications, the use of resistive source degeneration to reduce noise and improve matching is restricted due to headroom constraints. Could anyone elaborate on this statement extracted from "Ultra HighCompliance CMOS Current Mirrors for Low Voltage Charge Pumps and References" ?
Ultra HighCompliance CMOS Current Mirrors for Low Voltage Charge Pumps and References.pdfLast edited by promach; Yesterday at 02:41.

Yesterday, 03:44 #10
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Re: Improved current mirror
Source follower, the gain is around 1.
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