To part 4OSCILLOSCOPES - BASIC USE AND MEASUREMENTS
PART 3 - LET'S MAKE SOME MEASUREMENTS Amplifier gain and insertion loss ---------------------------------------------------------------------- NOTE: This is a text version of an article appearing in the Summer 1997 issue of "QRPp." The article contains numerous illustrations and photos of oscilloscopes displays, which unfortunately can not be included in a text file. AMPLIFIER GAIN The gain of an amplifier can be measured in terms of VOLTAGE GAIN, which is simply Av = Vout/Vin. For example, if the input to an amplifier is 1Vpp, and the output is 4Vpp, then the amplifier has a VOLTAGE gain of 4. GAIN IN DB is often more useful and is how the gains of amplifiers are usually expressed. With dB (decibels), everytime you DOUBLE the AC voltage, you ADD 6dB of gain. It is the RATIO of output to the input, and this RATIO is easy to measure on a scope, even for signals that exceed the cited bandwidth of your scope to some extent. Say you just built a single transistor amplifier to boost the audio signal before the final audio amplifier (usually an LM386). It is often easier to start with the OUTPUT for measuring amplifier gain. Inject an audio signal into the amplifier (transistor base). Place the scope lead on the transistor COLLECTOR, and set the scope so the output waveform is exactly 4 divisions peak-to-peak. Do not disturb oscope settings. Now move the scope leads to the amplifier INPUT, the transistor's base. You will of course have a much smaller signal, and the ratio of the input to the output will be the gain in dB. For example, say the input signal is 2 divisions peak-to-peak. This would be 6dB of gain, since you are DOUBLING the signal in the amplifier. Everytime you DOUBLE the voltage, it is 6dB of VOLTAGE gain. If the input signal is 1 division peak-to-peak, then the amplifier gain is 12dB. (With the output still at 4Vpp or 4 divisions). Going from 1 division to 2 division is 6dB gain; going from 2 divisions to 4 divisions is 6dB gain. Therefore going from 1 division to 4 divisions is 12dB (6dB + 6dB). This illustration shows how 0dB---|------|------|------|------| to measure Amplifier GAIN | | | | | on an oscope. FIRST, adjust | | | | | the scope so the amplifier 3dB---|----- | -----|------|------| OUTPUT is 4 DIVISIONS peak- | | | | | to-peak. THEN, switch to | | | | | the INPUT, and position the 6dB---|--**--|------|--**--|------| waveform so the NEGATIVE | * * | | * * | | peaks are on the bottom |* *| |* *| | division, as shown. Where 12dB--*----- * -----*------*------* the POSITIVE peaks of the 18dB__| |* *| |* *| input occur will be the GAIN | | * * | | * * | in dB by reading the scale ---|------|--**--|------|--**--| on the left. You may want Amplifier INPUT displayed to make such a scale and attach it to the side of your CRT screen for making quick measure- ments in dB's. IF YOU WANT TO DO IT MATHEMATICALLY ... Vout = 4Vpp Vin = 1Vpp Therefore, voltage gain Av = Vout/Vin = 4v/1v = 4 and gain in dB is: dB = 20log(Av) = 20log(4) = 20(0.602) = 12dB OR AS SHOWN DIRECTLY ON THE O-SCOPE AS DESCRIBED ABOVE. Since this is a relative measurement (a RATIO), the absolute value of Vin or Vout does not need to be determined. INSERTION LOSS. In some circuits, such as filters or attenuators, the LOSS in dB needs to be determined, and this is called the INSERTION LOSS. It is determined the same way as amplifier gain, except the INPUT will be GREATER than the OUTPUT since there is a LOSS involved. For example, with a signal generator connected to your QRP rig antenna input, you want to measure the insertion loss of your IF crystal filter. At the filter input, you can just barely squeek out 2 divisions of input signal on your scope at its most sensitive setting. (Perhaps due to exceeding the scope's bandwidth). The output from the crystal filter is 1 division, or a 50% reduction. The insertion loss would be 6dB (since if the power were HALF, or 50%, it would be a 6dB LOSS). OR MATHEMATICALLY USING SCOPE DIVISIONS: Insertion loss (dB) = 20log(Vin/Vout) = 20log(2 div./1 div.) = 20log(2) = 20(.30) = 6 dB If the output were 1.5 divisions, Insertion loss (dB) = 20log(2 div/1.5 div) = 2.5 dB Again, you are determining the insertion loss of a circuit element from the RATIO of input to output. You do not need to make absolute measurements. So if the frequency is beyond the bandwidth of your scope, as long as you can get enough vertical deflection to measure it's magnitude in some terms of divisions, and able to see the signal either get smaller or larger, you can estimate the gain or loss in dB fairly accurately. If the signal DOUBLES, it is 6dB; if it is about half of doubling, it is about 3dB; if the change is barely noticeable, it is around 1dB. This is usually sufficient for determining if circuit elements are working properly. For example, using the insertion loss of an IF crystal filter as above, if you determine the loss to be a few dB, the crystal filter loss is acceptable. If it is much more than around 6dB, you may have a problem. And if you can't see any output, you have a real problem. (Loss is about 1 to 1.5 dB per crystal in filter). Same with checking the gain of amplifiers. It's not important if the gain is 6.2 dB or 6.5 dB, but whether the gain is ABOUT what you'd expect. If the output of an amplifier is ABOUT DOUBLE the input, you have 6dB of gain. If the output is just a bit larger than the input (or the same) ... then you've got a problem (no or little gain occuring). With a little practice on your o-scope, you will learn to recognize approximate gains and losses in dB's very quickly from the oscope display. You might want to go through your QRP kit with the circuit and measure the gains through different stages. If your rig has an MC1350 IF amplifier, what is it's gain with a strong signal vs. a small signal to see if the AGC is working properly. What is the gain of the LM386 audio output amplifier? With larger bandwidth scopes, check the gains of the RF driver transistor and output PA transistor. Knowing what these gains are could help troubleshoot the circuit later should a problem develop. END OF PART 3 72, Paul Harden, NA5N NA5N@Rt66.com by Paul Harden, NA5N
Many thanks to Paul for allowing me to use his work
Frank G3YCC