# Notes on Opamp Bandwidth

20 Mar 2015Opamp datasheets usually provide certain parameters to specify the frequency response of the opamp. An often misunderstood spec is on the `Gain Bandwidth Product`

. I would try to elucidate the specification today.

The transfer function of a typical opamp has a real dominant pole at a fairly low frequency (a few Hertz). The opamp's transfer function can be approximated by:

$$\frac{A_0}{(1 + s/\omega_0)}\tag{where}$$

\(A_0\) is the DC gain and \(\omega_0\) is the dominant pole frequency (which is also referred as the \(-3dB\) bandwidth of the opamp). The gain of the opamp at frequency \(\omega\) can then be calculated as:

$$A = \frac{A_0}{\sqrt{1 + (\omega / \omega_0)^2}}$$

$$\text{For}\quad \omega >> \omega_0 \text{,}\quad A \approx \frac{A_0}{\omega / \omega_0} => A \omega = A_0 \omega_0 = \text{GBWP}$$

** Gain Bandwidth Product is usually defined for dominant pole compensated opamps**. For frequencies beyond \(-3dB\) frequency (\(\omega_0\)), the product of gain and frequency is constant.

*Unity Gain Bandwidth*and

*Gain Bandwidth Product*have the same value for a dominant pole compensated opamp. ie, \(GBWP = A_0 \omega_0\). Shown below is the Bode plot of the frequency response of the opamp discussed above.

If we were to plot the Gain vs Frequency graph on a linear scale (instead of the log-log scale as in the Bode Plot), the curve would approximate a hyperbola (a bit beyond \(\omega_0\)).

## GBWP of non-inverting amplifier

There is yet another way to interpret the GBWP of opamps. Here is the classic non-inverting amplifier configuration.

We usually come across statements which say that the \(-3dB\) bandwidth of this amplifier is GBWP divided by Gain. However, this statement is not quite obvious. Let us formally prove that this is indeed the case. At a block diagram level, this circuit can be equivalently represented as follows:

where \(\beta = \frac{R_2}{R_1 + R_2}\). The closed loop transfer function of this new system is

$$\frac{A_1}{(1 + s/\omega_1)}\tag{where}$$

$$A_1 = \frac{A_0}{(1 + \beta \omega_0)} \text{ and } \omega_1 = \omega_0(1 + \beta A_0)$$

Observe that \(A_0 \omega_0 = A_1 \omega_1\). The graph below shows the effect of the \(\beta\) on the Gain Bode Plot of the amplifier.

## Bandwidth of Cascaded Opamps

Consider \(n\) identical cascaded opamps, where each of the opamp has a resistive feedback loop with a certain \(\beta\). If the GBWP of the opamp is known, then can we somehow predict the GBWP of the cascaded system?

Each of the amplifier has a GBWP as \(A_0 \omega_0\), and since the gain of each amplifier stage is determined by \(R_1\) and \(R_2\), (gain = \(K = A_0 \frac{R_1 + R_2}{R2}\)) => the bandwidth of each amplifier stage is \(\omega_0 \frac{R_2}{R_1 + R_2}\). On first look, it appears that the bandwidth of the cascaded system should also be \(\omega_0 \frac{R_2}{R_1 + R_2}\), but that is not exactly the case. The straight line approximation in bode plot is not good enough in this case! Let us find out the true \(-3dB\) bandwidth.

$$\big(\frac{A_1}{\sqrt{1 + (\omega_{3db} / \omega_1)^2)}}\big)^n = \frac{1}{\sqrt{2}}$$

$$\omega_{3dB} = A_1 \omega_1 \sqrt{2^{-1/n} - 1} = GBWP \sqrt{2^{-1/n} - 1}$$

Say we have a specific opamp with a \(1 MHz\) GBWP. If we were to use this opamp to get a gain of \(1000\), the \(-3dB\) bandwidth would drop to around \(1 kHz\). Instead, if we use \(3\) cascaded opamps (each with a gain of \(10\), and hence, an individual bandwidth of \(100 kHz\)), the cumulative bandwidth of the cascaded system would be

$$100 \sqrt{2^{-1/3} - 1} : kHz \approx 51 : kHz$$

That's it for today! I traditionally had a pretty poor understanding of this subject matter; hope this post helps clear things up for those in similar shoes!

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