Timeline for How to solve this. Multiple op amps circuits in series
Current License: CC BY-SA 3.0
16 events
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| Jan 11, 2018 at 18:30 | history | edited | Andy aka | CC BY-SA 3.0 |
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| Jan 11, 2018 at 18:07 | vote | accept | J. Doe | ||
| Jan 11, 2018 at 18:07 | comment | added | J. Doe | Okay, this question was from last years exams. I am studying robotics engineering and the teacher probably wanted the students to think that way then. I do not know. But at least I know now how to get the correct answer. The textbook would have solved that question the way I did but I remember that the teacher said the textbook was too nice to the students. Maybe it is something like that because this one was made by the teacher | |
| Jan 11, 2018 at 18:02 | comment | added | Andy aka | It still contradicts the answer - power amplification literally means watts out divided by watts in. Clearly there can be watts in because of R1 hence you apply an input voltage and current is taken and power in = volts_in x current_in. But, with no load resistor there can be no output current and therefore no power out. Having said all that, it's a common mistake to make and most electrical engineers would look at the question, see the shortfall and assume that an attempt at an answer based on voltage amplification can be made. | |
| Jan 11, 2018 at 17:57 | comment | added | J. Doe | A better translation would be: The two cascaded amplifiers below shall give a power amplification of 30 dB. What resistance must you choose for R1 for this to become true? That is the best translation and most accurate I could do | |
| Jan 11, 2018 at 17:57 | comment | added | Andy aka | Well, if you translated it correctly and the recognized correct answer is 3.2 kohm then the words in the question are incorrectly phrased for that answer. | |
| Jan 11, 2018 at 17:54 | comment | added | J. Doe | The picture is all I got. The question is written in swedish so I believe there is no good reason to post it here so I did a somewhat rough translation but missed out on the power part but as concluded it did not matter. | |
| Jan 11, 2018 at 17:54 | history | edited | Andy aka | CC BY-SA 3.0 |
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| Jan 11, 2018 at 17:52 | comment | added | Andy aka | That is correct so, either your question is badly phrased when it refers to power gain (because there can be none without an output load) or you omitted the load resistor from where you copied the question. Maybe you need to photograph the original question and post it. | |
| Jan 11, 2018 at 17:49 | comment | added | J. Doe | Okay but our textbook states that: power gain = 10^(power gain (dB)/10) and voltage gain = 10^(power gain (dB)/20) so you mean that because there is no load attached I should calculate as voltage gain despite that the text tell me that the power amplification is 30dB? | |
| Jan 11, 2018 at 17:47 | comment | added | Andy aka | Apart from anything else, unless you show an output load resistor there is no power amplification because the load is open circuit and therefore takes zero power. | |
| Jan 11, 2018 at 17:43 | comment | added | J. Doe | Second circuit got amp of 10 times. 20*log(10)=20dB voltage amplification. First circuit must have 10dB amp which is 3.16 times. I get the correct answer when I calculate for voltage gain but not for power | |
| Jan 11, 2018 at 17:43 | comment | added | Andy aka | 30 dB of power amplification is still an equivalent voltage amplification (assuming same value input and output resistances) of 31.62. Do the math. | |
| Jan 11, 2018 at 17:40 | comment | added | J. Doe | I forgot to mention but it was 30dB power amplification and not voltage | |
| Jan 11, 2018 at 17:30 | history | edited | Andy aka | CC BY-SA 3.0 |
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| Jan 11, 2018 at 17:15 | history | answered | Andy aka | CC BY-SA 3.0 |