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I was watching a video online (https://youtu.be/YlP2u1mqw7A?t=1380). Minute 23:00.

He mentioned that the time constant for an op-amp is (for a first-order exponential system)

$$\tau = \frac{1}{2\pi f_{3dB}}$$

I'm a bit confused by that. I haven't seen that before. Usually, I associate time constant with some sort of capacitacne, but this is frequency? I realise it's due to the internal compensation of the op-amp but is there no capacitance?

Or is the capacitance value embedded within the f3db pole frequency?

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4 Answers 4

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The time constant comes in picture when you have a circuit whose response depends on frequency.

Take a simple RC circuit, it's time constant is just RC,

Take an op-amp. An op-amp by default is like a LPF in most of the cases. Hence, it will have a cut-off frequency where it's like lies. That pole is nothing but 1/time constant of the op-amp.

So basically the time constant of an op-amp is 1/it's corner freq/pole freq/cut-off frequency.

These frequencies are dependent on the transconductance of the devices and their dependibility on the frequency with the MOS capacitors.

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A parameter used to characterize an op amp is its GBW product (Gain Bandwidth Product) which is, if the op amp was designed for stability at unity gain, equal to the frequency at which the loop gain is 1 or 0 dB. The GBW is the product of the gain of any configuration (inverting amplifier, non inverting amplifier, etc) with the bandwidth of that configuration. At low frequencies, Vout/Vin is defined by the gain of your configuration. At high frequencies, Vout/Vin is defined by the gain of the op amp itself, since you can't get more gain in your circuit than the gain of the op amp itself. The bandwidth of your closed loop circuit is the frequency at which Vout/Vin starts to be defined by the gain of the op amp, and this frequency is GBW/Av, being Av the DC gain of your circuit. The gain of your closed loop circuit starts dropping at 20 dB/dec at this frequency. Thus, if you look at the transfer function of your closed loop circuit, it is very similar (at least until the GBW product frequency) to a first order RC low pass filter. For that reason you can think of the closed loop circuit as a first order system with one pole and calculate the time constant associated with the pole.

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there is a capacitor involved in setting the slewrate and the single-pole rolloff so valuable in bandwidth-thinking.

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but is there no capacitance?

\$\tau\$ is C multiplied by R i.e. the time constant components in the basic op-amp compensation circuit.

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