5
\$\begingroup\$

Disclaimer: I am a Mechanical engineering student and I don't have a huge background in electrical engineering.

In on of my lab classes, we just dealt with the response of amplifiers to a variable DC voltage source. The lab consisted of DC power supply as the variable voltage source, an amplifier, amplifier power supply, and an oscilloscope. We varied the DC power supply from -12V to +12V in 2V increments and we had to record the output voltages and plot the final results. At 0V there was a offset of 2.54V caused by the amplifier being powered by the DC power supply.

After we plotted the results, the graph of input voltage vs output voltage looked as expected with lower saturation limits, then a linear increase in voltage and high saturation limits. Our results showed an amplification factor (the slope of the linear region) of less than 1, which caused the gain of the amplifier to be negative. My TA said the setup looked fine and could not explain any reasons why we got a negative gain.

Does anyone have any reason as to why the amplifier would produce a negative gain, thus defeating the purpose of the amplifier?

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Schematic? Pictures? Anything? \$\endgroup\$ – Matt Young Sep 7 '12 at 21:54
  • \$\begingroup\$ I could post a picture of the results, i tried to explain a schematic as best i could \$\endgroup\$ – Greg Harrington Sep 7 '12 at 21:55
  • \$\begingroup\$ @GregHarrington Can you draw the schematic? circuitlab.com works well. \$\endgroup\$ – Kortuk Sep 7 '12 at 22:22
  • \$\begingroup\$ The best picture I could get is the one from the lab manual. Most of the other amplifiers in the class were giving positive output gains \$\endgroup\$ – Greg Harrington Sep 7 '12 at 22:27
  • \$\begingroup\$ Do you think there could have been any procedural error that could have caused it to be wrong? I did the lab a week ago and can't quite remember if something was done wrong \$\endgroup\$ – Greg Harrington Sep 7 '12 at 22:44
7
\$\begingroup\$

The term "negative gain" is reserved for those cases where the line has a downward slope and so reverses the polarity of the signal (a 180° phase change). A op-amp in an inverting configuration is a prime example.

What you have is an attenuating amplifier; the signal out has a somewhat decreased amplitude from the signal coming in but not a different sign. (I'm ignoring the offset of the sloped line in order to keep it simple)

The relation of the voltage-over-voltage gain (V/V) to dB gain often confuses people since the dB measurement actually strips out the "inverting" property of negative-gain amplifiers (since you take the log of the absolute value of the gain). Let's set up fours scenarios for different line slopes (V/V gain):

  1. Gain: 2 V/V or 20*log(|2|) = 6 dB and 0° phase difference
  2. Gain: 0.5 V/V or 20*log(|0.5|) = -6 dB and 0° phase difference
  3. Gain: -0.5 V/V or 20*log(|-0.5|) = -6 dB and 180° phase difference
  4. Gain: -2 V/V or 20*log(|-2|) = 6 dB and 180° phase difference

Scenarios 1 and 2 have a positive slope/gain and thereby a 0° phase difference while scenarios 3 and 4 have negative gain (signal inverting) and thereby a 180° phase difference. Scenarios 2 and 3 have a gain who's absolute value is less than one and thereby are attenuating amplifiers, expressed by a negative dB gain, while scenarios 1 and 4 are "amplifying amplifiers".

The Wikipedia page on Gain might explain it better then I can.

The purpose of an amplifier isn't always to increase the voltage amplitude of a signal being passed through. It might for example be used to drive a current-hungry device when the source signal/device can't, like loudspeaker power amplifiers do.

\$\endgroup\$
  • \$\begingroup\$ Ok I think I am starting to understand the differences. I guess when I meant "negative gain", I meant that I was getting an amplification factor < 1 and then that gave a negative number when I calculated the gain in dB as G = 20*log(K). I always thought the gain of an amplifier should be > 0 and that's why I began to get confused. I didn't mean that the signal was being inverted \$\endgroup\$ – Greg Harrington Sep 9 '12 at 3:32
  • \$\begingroup\$ I thought it was a problem that i was calculating gain as < 0 but if it means the signal is being attenuated and that's fine with the amplifier then i understand that now \$\endgroup\$ – Greg Harrington Sep 9 '12 at 3:34
  • \$\begingroup\$ The last paragraph is a feasible explanation to my situation-it's what I was looking for. Found a device which has a volume output of -28 dB and I was wondering why the minus sign is present. \$\endgroup\$ – Daniel Tork Dec 15 '18 at 21:56
4
\$\begingroup\$

What you are saying is contradicting itself. You say the graph shows a gain a little less than 1, but suddenly you are saying the gain is negative (less than 0). The gain in the linear region does seem to be right around 1 according to the graph. I can't begin to guess how you got a negative gain out of that, especially since you apparently read the graph correctly.

At first approximation, the gain is 1.0. If you use the numbers from the table from input 0 to 6 V, the gain is 0.98. That is certainly not negative.

Added:

After reading some of your comments, I think you are confused by how you are expressing gain. Gain is simply the ratio of a output response size divided by the input size to get that response. In your case, over the range from 0 to 6 volts in, you get 2.54 to 8.41 volts out. Therefore gain is:

   Gain = dOut / dIn = (8.41V - 2.54V) / (6V - 0V) = 5.87V / 6.00V = 0.98

This is close to 1, meaning the amplifier actually attenuates slightly, but it is certainly not negative.

The problem seems to be that you then express this gain logarithmically, and the negative result confuses you. Voltage gain expressed in dB is 20*log10(gain), which is -.19 dB in this case. Note that logarithmic values, such as dB, will be negative when the linear value (the straight gain in this case) is less than one. In other words dB gives us a scale where 0 is unity gain, negative values represent attenuation, and positive values amplification.

While there is nothing wrong with gain less than 1, and therefore that gain expressed in dB being negative, this is not the same as negative gain. Negative gain means the output is inverted from the input. For the gain to be negative, the output would have to go down when the input goes up, which is clearly not happening in your case.

\$\endgroup\$
  • \$\begingroup\$ There is a different between gain and amplification factor. My teacher said the amplification factor for this should be the slope of the line between the upper and saturation limits. That slope comes out to be around .95. Gain (in dB) is 20*log(amplification factor). Since the factor is less than one, then the gain is negative \$\endgroup\$ – Greg Harrington Sep 7 '12 at 23:17
  • \$\begingroup\$ Could I say that the amplifier was defective? \$\endgroup\$ – Greg Harrington Sep 8 '12 at 0:43
  • 4
    \$\begingroup\$ Your gain is not negative. What you are calling the amplification factor is gain, expressed as a ratio. What I think is confusing you, is the fact that it is attenuating the signal, rather than amplifying it, which is a perfectly valid function of an amplifier. There may or may not be anything wrong with the amplifier, but we don't have a schematic. My money is on not. \$\endgroup\$ – Matt Young Sep 8 '12 at 1:23
  • \$\begingroup\$ the slope of the line is .95, making it the amplification factor (K). Gain, in dB is expressed as 20*log(K). Making my gain NEGATIVE. So why do you say it is not negative? \$\endgroup\$ – Greg Harrington Sep 8 '12 at 2:41
  • 1
    \$\begingroup\$ Because a negative logarithm is not the logarithm of a negative number. As Olin has stated, "In other words dB gives us a scale where 0 is unity gain, negative values represent attenuation, and positive values amplification." So the fact that the logarithm of K is negative does not mean that the gain (assuming you are referring to K as gain) is negative: it just means that K is less than one. \$\endgroup\$ – WhatRoughBeast Mar 15 at 23:13
0
\$\begingroup\$

"Negative gain" is an ambiguous term and should not be used.

"Negative Gain" could mean an inverting gain if the gain is specified as a ratio of output to input magnitudes or it could mean attenuation if the gain is specified in negative dBs.

Better to say that an amplifier has a gain or attenuation and also that this gain or attenuation is either inverting or non-inverting.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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