0
\$\begingroup\$

I have simulated the frequency response of an amplifier circuit. Frequency range is from 10 Hz to 2.5 MHz. It gave me a plot with a y-axis range going from -50 to 50. I assume these are dBs but I am not sure.

I need to find the bandwidth of the amplifier. To do this, I know that I need to find the lower 3b and upper 3b point. But I am not sure how to find them. I know I have to multiply something by 0.707 but I am not sure what. Is it the dB value at the peak of the plot that I need to multiply by 0.707, and the result would be the corresponding dB value for the upper and lower frequencies?

Thank you.

\$\endgroup\$
1
  • \$\begingroup\$ If you're not sure about what you see on the graph - show your code. The answer by Olin is good, but if you ran a wrong Spice simulation you'll still get a wrong value. \$\endgroup\$
    – Vasiliy
    Aug 4, 2013 at 13:53

4 Answers 4

3
\$\begingroup\$

First, a graph without labels is no graph at all. Ditch whatever irresponsible software created it and move on.

Second, if the Y axis is in dB and the X axis in frequency or log frequency, then it is easy to find the "bandwidth" of the amplifier by inspection. In this case we are assuming you have defined bandwidth as the frequency range over which the gain ranges by no more than 3 dB. This is merely the high frequency limit minus the low frequency limit. This figure is of little use, with the frequency range (like 20 Hz to 20 kHz for a good audio amp) generally being more useful. Saying such a audio amp has 19.98 kHz bandwidth is rather less meaningful.

To find the upper and lower frequency limit of the range that is all within your 3 dB spec, find the highest point on the graph, draw a line at 3 dB below that, then look at where that line intersects the plot. Those intersection points are the upper and lower frequencies of your passband. To get bandwidth, subtract the two.

\$\endgroup\$
0
\$\begingroup\$

In Spice and most of it's derivatives, you need to specify what you are plotting. The tool assumes that you have a basic understanding of what you are looking for. In your case you need to "add a trace" of the form of 20*log10*(Vout/Vin). I don't have a copy of Pspice to give you a precise answer.

The tool is very configurable and often comes un-configured, so it's not very friendly to beginners.

Older version of PSpice had a graphical interface that allowed you to select all the parameters, so I bet this information is already in the menus that you used (default values). It's probably a case of you just looking it up.

\$\endgroup\$
0
\$\begingroup\$

dB values are never multiplied by \$0.707 \approx \sqrt{\frac{1}{2}}\$. Rather, a -3 dB drop occurs, relative to some maximum, when the signal voltage drop to 0.707 of that maximum. In other words, 70.7% voltage is -3dB.

What is special about -3dB is that it represents half the power. If the voltage drops to \$\sqrt{\frac{1}{2}}\$, then since power is proportional to the square of voltage, power has dropped to \$\frac{1}{2}\$.

Actually, half power is not exactly -3dB; this is an approximation. The true value is \$10\log_{10}\frac{1}{2} = -3.010299...\$.

The multiplier \$10\$ becomes \$20\$ when we calculate decibels using voltage, because of the squaring. We want the decibels corresponding to \$\sqrt{\frac{1}{2}}\$ voltage to be the same as those for \$\frac{1}{2}\$ power, thus:

$$20\log_{10}\sqrt{\frac{1}{2}} = 20\log_{10}\left(\frac{1}{2}\right)^{1/2} = \frac{1}{2}20\log_{10}\frac{1}{2} = 10\log_{10}\frac{1}{2}$$

If your graph shows dB values, then you can find the -3 dB points directly, relative to the amplitude across the amplifier's mid-band. If the graph shows linear signal level as a voltage, then you use 0.707. If the graph shows power, then use 0.5. You must know what units are on the dependent axis of the graph, of course.

\$\endgroup\$
0
\$\begingroup\$

I was under the impression that you typically plot the amplifier's gain on the Y-axis (in dB), relative to frequency (in Hz) on the X-axis. Now, simply inspect the plot to find the frequency where the gain has reduced by 3 dB. I have no idea why you'd need to multiply anything by \$ \sqrt{\frac{1}{2}} \approx 0.707\$ as long as your plots are in dB.

I assume this is for a class. Make sure you understand not only what you're looking for, but also why. Good amplifiers have flat frequency response over the entire operating range, but due to parasitic capacitances, amplifiers lose their ability to (stably) amplify a signal when it reaches high frequencies. That's why an audio amplifier with a gain of 10 is much cheaper to design than an RF amplifier with a gain of 1000.

Most EEs don't design amps for a living -- but knowing this material will help you learn to spec an amplifier for a larger system. Good stuff.

\$\endgroup\$
1
  • \$\begingroup\$ thanks for your answer. I will think about it and come back later. \$\endgroup\$ Aug 4, 2013 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.