week_12

# week_12 - 113 + = p v j A j A 1 ) ( 14000 = v A in our case...

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Unformatted text preview: 113 + = p v j A j A 1 ) ( 14000 = v A in our case ? 8 . 1 2 14000 25 2 1 1 1 ) ( KHz f f A MHz A j A j A p p p v ta p v ta p v ta = = = = = + = = For ta p &lt;&lt; &lt;&lt; we can write: j j A ta = ) ( When you put the opamp in a feedback loop: x V in V _ ) ( j A out V + + 1 &lt; ) ( 1 ) ( ) ( 1 ) ( ) ( . j A j A V V V j A V V V j A V j A V V V V V in out in out out in out x out out in x + = = + = = = Use j j A ta = ) ( in the above equation + = + = = ta ta ta in out loop closed j j j V V A 1 1 . 1 1 Gain at low frequency = 1-3dB: ta dB = 3 . 114 = ta BW Gain always constant (gain bandwidth product) Linear settling time Setting time: time it takes for an OPAMP to reach a specified percentage of its final value when step input is applied. Linear due to finite ta of OPAMP (in small signal) Settling time Non-linear due to slew-rate of OPAMP under large signal input ta p 1 Without feedback ta dB = 3 So both gain and BW change with feedback for small step sizes in the output signal, OPAMP may not reach a slew-rate limit at all. 115 The time constant of OPAMP response in closed-loop system: ta dB 1 1 3 = = Since we are assuming a first-order system, transient response is: input = t step out e V t V 1 ) ( Time it takes for signal to reach its 99% final value is: ta settling t t e 6 . 4 6 . 4 01 . = = = For 90% is: ta settling t t e 3 . 2 3 . 2 1 . = = = Slope of the output: step t out V dt dV = = If slew-rate is larger than this value, there is no slew-rate limiting mechanism. 116 OPAMP compensation In order to analyze OPAMP stability under feedback, we have to consider OPAMP as a 2-pole system. + = eq ta j j j A 1 ) ( open-loop OPAMP transfer function 2 nd pole that takes into account all of OPAMPs poles and zeros. ta 1 6 or decade dB 2 1 12 or decade dB A + = = eq ta j j j A gain loop 1 ) ( Frequency at which loop gain becomes 1 = ta 2 2 2 2 1 1 1 1 1 ) ( eq t t ta eq t t ta eq t t ta t j j j A + = + = = + = Assuming that the second pole of T.F. ( eq ) is very high t ta When unity gain freq. is smaller than dominant pole without comp. dB ta 3 = 117...
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## week_12 - 113 + = p v j A j A 1 ) ( 14000 = v A in our case...

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