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I know the importance of phase margin to ensure stability in closed loop systems, but what about in open loop systems?

I am designing a regulated common based LNA IC that operates in open loop.

Can we just ignore the phase response and just focus on the gain and bandwidth?

I ensure all the poles are on the left half plane of the s-plane but two poles are located close to each other so I got the phase is -260deg when the gain is 1V/V.

The phase is that low because the DC gain is high, 9000 V/V.

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3 Answers 3

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Can we just ignore the phase response and just focus on the gain and bandwidth?

You can but, if the spectrum of the signal you are trying to amplify is wide AND the time-domain waveform shape is important to maintain, then you probably can't.

What is the significance of Phase Margin in open loop amplifier?

None.

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  • phase margin only applies to a closed loop , where the feedback voltage can't keep up with the current but with high feedback gain and low forward gain, it can be compensated. Thus current feedback and common base amplifiers tend to have greater bandwidth.

  • beware that open-loop amplifiers can also have internal feedback that may cause phase margin loss in the RF band.

LNA design is all about Unconditional stability with any load present to the input or output of the device, the circuit will not become unstable –will not oscillate.

Instabilities are primarily caused by three phenomena: internal feedback of the transistor, external feedback around the transistor caused by an external circuit, or excess gain at frequencies outside of the band of operation.

  • stability analysis: numerical (Kand μfactors)and graphical(stability circles on Smith Chart).
  • An option of numerical analysis consists of calculating a term called Rollett Stability Factor (or K-factor). K-factor is given by:$$K= \dfrac{1-|S_{11}|^2-|S_{22}|^2+ |S_{11}*S_{22}-S_{12}*S_{21}|^2}{2*|S_{12}*S_{21}|}$$
  • When K>1, the circuit will be unconditionally stable for any combinations of source and load impedance.
  • When K<1, the circuit is potentially unstable and oscillation may occur for a certain combination of source and/or load impedance present to the transistor.
  • However, the two necessary and sufficient conditions for unconditional stability are that the stability factor Kis greater than unity (K>1) and the stability measure bis positive(b>0).
  • Where b is given by:$$b = 1+ |S_{11}|^2-|S_{22}|^2-|S_{11}*S_{22}-S_{12}*S_{21}|^2$$
  • The K-factor represents a quick check for stability at a given biasing condition. A sweep of the K-factor over frequency for a given biasing point should be performed to ensure unconditional stability outside of the band of operation.
  • When working with K-factor, it is important that it is greaterthan1. However, the actual value is not telling anything. You cannot say that an amplifier with K=3 is "more stable" than an amplifier with K=1.1.
  • As noted, if K>1, says the amplifier is stable but gives no hint at how far outside the Smith Chart the stability circles are. Similarly, if you get K<1 it doesn't give you much insight as to what to do, or where to do it credit
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  • \$\begingroup\$ Thanks for your answer. my circuit is not in the RF range. the BW is 5 MHz :) Just to make sure, this sentence..."LNA design is all about Unconditional stability". Does is it mean that the LNA must be stable regardless the external impedance at the input and output? I cannot think that it is possible. Could you please explain about it a little bit more? \$\endgroup\$ Jan 5, 2021 at 20:02
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You can if indeed the amplifier is 'open loop'. There are always paths that you may have ignored-- transistor input-output capacitance for example. Cgs in Mos, Cbe in bjt. Also there may be a path through the power supply lines, ground planes, layout lines crossing ...

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