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I was making a non inverting amplifier,in the first iteration a provided a gain 1000 and at that time I can see the line voltage frequency 50Hz is coming at the output.

I reduced the gain and made it 51 and I can see that there is no 50Hz signal at the output.

May I know how the high gain circuit is capturing the 5Ohz line frequency.What is the mechanism behind it

Is it a radiated noise or conducted noise

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    \$\begingroup\$ Please provide a schematic. Also a layout picture would help. \$\endgroup\$ – Math Keeps Me Busy May 1 at 13:32
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May I know how the high gain circuit is capturing the 50 Hz line frequency.

All around you are electricity wires in your walls and in your equipment. These produce an alternating electric field in the room where you are testing your amplifier. You don't feel it because the connection with your body is via capacitance from the building wiring and, that capacitance is very low (usually much less than 10 pF typically) but, nevertheless it is there.

So, if your amplifier input has a high input impedance (unlike the human body), an equivalent single digit pF of capacitance between the amplifier input and the room wiring is usually enough to generate a few microvolts of AC voltage at the amplifier input.

That gets amplified by 1000 and you see several millivolts of signal on the output.

You don't necessarily see that voltage on your o-scope when the gain is only 51 because the 50 Hz amplitude signal is hidden in the noise caused by your o-scope input.

Is it a radiated noise or conducted noise

Well, first of all, it's not noise because it is something you can see as a sine wave on your o-scope. However, it can be regarded as being interference. Probably that's a better phrase to use.

Secondly, it is loosely termed "radiated" in that it is not conductive but, the preferred mechanism is "capacitively coupled".

Of course, if your input wires are long and unbalanced, you could be picking up magnetic induced signals due to the currents flowing in your room wires but, more likely, it's an electric/capacitive coupling.

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    \$\begingroup\$ put a cell phone nearby and call it. Your circuit will most likely pick up the RF from the phone as well. \$\endgroup\$ – Kartman May 1 at 11:48

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