Video Lectures created by Tim Feiegenbaum at North Seattle Community College.
We're continuing in 10.1 and we're looking at amplifier basics. Characteristics of amplifiers are common across devices including integrated circuits. Gain is a term to indicate the increase in signal given by an amplifier. Gain is expressed as gain = output/input and is commonly represented as A. Amplifiers may be designated to amplify current, voltage or power.
Gain or loss can also be expressed as decibels as in dB. Decibels are based upon logarithms and provide a convenient way to express very large ratios. Conversion formulas for voltage, current, and power are…
Here we have the conversion to decibels for voltage, current, and power. Now, you'll notice here. For voltage, we're going to take 20log of the output over the input. Let's just do some examples here. Let's pretend that our output voltage--and we'll just use very round numbers for ease. The output voltage is 100. The input voltage is 1. If we were evaluating gain, we would say the amplification was the input or the input is 100 times larger than the input.
Now, to look at the gain in decibels, we will need to bring out our calculator. We're going to say 100 and then we'll take the log of 100 which is 2, and then we need to multiply it by 20 so times 20 equals, and the value is 40. We could say that we have and what we commonly do say is that we see 40 decibels of gain in this amplifier, and this would be voltage gain, 40 dB of voltage.
Now, keep in mind that when we use the term decibel, this is a unitless value. By that, I mean this simply represents the relationship between input and put since we have 1 volt of input, 100 volts of output. We're taking 20 times the log of 100. Let's look at current and we'll amplify this side.
Let's say we had the same amplifier and this wouldn't necessarily be always the case, but let's just pretend that in the same amp, we also had a current gain. This time, the current gain would be 100 milliamps and then we had 1 milliamp of current. Like I said, this would not necessarily be the case, but just for ease of numbers again, we'll do the same thing.
Now, we could say the current gain is 100. To simply say, 100 times the input. Now, if we were taking the log of that, we took 20log times--again, it would be just 100. We get the same thing. Again, we would say we get 40 dB. Now, the fact is that, again, this is a much smaller unit. This is in milliamps, but the ration remains the same so the output is the same. Again, we're looking at--the decibels is unitless. It simply represents the relationship between input and output. When we're looking at voltage in current, the relationship is going to be 20 times the log of that value.
Now, when we look at power gain, we're going to find that power gain is not 20log. It's going to be 10 times the log. Now, recall that when we calculate power, it is the function of voltage and current. Now, amplifiers are usually optimized for either voltage current or power, but when it comes down to the final analysis and you have to calculate what is the power, you simply look at what is the voltage and what is the current in the circuit and you multiply them out. What we could do, if we pretend that this was the same amplifier, we could pretend that this is the same--we want to evaluate the power for this amp. When we say, well, we had 100 volts of input, we had 100 milliamps, so that would represent 100 times. The 100 milliamps would represent the total power. The 100 times the 100 is 10,000 to keep our units right. I remember this is watts. Since this is milliamps times 100, then this would be milliwatts.
Our input, we would have 1-volt times 1--again, this is in milliamps so this would be 1 milliwatts. Now, we're looking at 10,000 milliwatts of power. Now, there's quite a difference between 100 over 1 and 10,000, but we're looking at the same amps. Let's again look at our calculator and calculate what is the power out here. We start out with 10,000 and we take the log of that. Now, remember this time it's times 10 so times 10, and notice that we still get 40. Interestingly enough, remember this is actually what we would expect since this is the same amplifier, and this brings into play why we take 10 times the log by 20 with voltage and current.
Okay, let's take a little closer look. With our voltage gain, we had 100/1. Remember that when we do decibels, we're looking at the log. Remember, when we do logs, we're thinking about the log represents the exponent. When we take the log of 100, we're looking at what exponent we would raise 10 to get this value. In this case, the log would be 2 and so the log of 100 is 2. Remember the 10 times 10. That's where we get that value.
Since we're doing the decibels, remember to calculate the decibels, we would take 20 times that exponent and times 20, and that's where we got the value of 40 dB. With power, it was much the same. The value was 10,000/1. Again, the log represents the exponent if we had 10 and we wanted to find out what is the exponent that we would raise 10 to get 10,000. Well, that value would be 4. If we just took the log of 10,000, we would get 4, but to convert it to decibels, we would take that times 10. Hence, we get our value which we previously calculated which was 40 dB.
Okay, this addresses the first item that we are looking at in this section which has to do with amplifier basics. We're looking at the term of gain so we looked at the value A which represents the value of amplification. Then we looked at converting that gain into decibels which are generally used to represent very large values.
Video Lectures created by Tim Fiegenbaum at North Seattle Community College.
In Partnership with Future Electronics
by Jake Hertz
by Jake Hertz
by Robert Keim