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| converting Volt into dB |
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| bob91343:
Strictly speaking, dB has to be power gain. Output power divided by input power gives the power gain ratio. Take the logarithm of that (base 10) and multiply by ten to get the result in dB. However, the situation is more complex, as the definition has been perverted. One way it's been perverted is to ignore the impedance, thus destroying the meaning. Another way is to assume a reference that, for some reason, isn't stated. Example might be sound levels, which are usually expressed in dB but no reference given. There is an assumption but it's seldom stated. Standard references are 1 milliwatt, 6 milliwatt, and I forget the sound pressure level, something like 0.0002 dynes per sq cm but don't quote me. Sometimes 1 Watt is the reference. For radio communications, the reference is 1 milliwatt. It's indicated by, rather than dB, dBm. In a 50 Ohm system, -73 dBm is 50 microvolts. (Also referred to as signal strength S9) So try to keep the air clear and be precise in what you mean. |
| radiolistener:
--- Quote from: Mepakos on August 06, 2019, 09:55:30 pm ---why -20V is wrong. Ua=20Vss and Uce=Ua. UCE is the voltage gain of the transistor. So why its wrong. --- End quote --- because input voltage is Vin = 1 mV AC and output voltage is Vout = 20 mV AC. So, voltage gain is: Vgain = Vout / Vin = 20 / 1 = 20 times --- Quote from: Mepakos on August 06, 2019, 09:46:58 pm ---The question is what is the voltage gain of the transistor in dB? --- End quote --- There is 1 mV AC on the input and 20 mV AC on the output. So we have voltage gain: Vgain = Vout/Vin = 20/1 = 20 times You can convert it into dB in the following way: Vgain_dB = 20 * log10( voltage_gain ) = 20 * log10( 20 ) = 26 dB This equation 20*log10(x) is actual for voltage and current gain. But when you needs dB for power gain, then you're needs to replace 20 with 10: Pgain_dB = 10 * log10( power_gain ) For example, if you have 5W on the input and 30W on the output: Pgain = Pout/Pin = 30/5 = 6 times Pgain_dB = 10 * log10( 6 ) = 7.7815 dB |
| magic:
--- Quote from: bob91343 on August 07, 2019, 12:26:06 am ---Strictly speaking, dB has to be power gain. Output power divided by input power gives the power gain ratio. Take the logarithm of that (base 10) and multiply by ten to get the result in dB. However, the situation is more complex, as the definition has been perverted. One way it's been perverted is to ignore the impedance, thus destroying the meaning. --- End quote --- All true. And one way it was perverted, and the way OP probably is interested in, is in using dB to refer to voltage gain, without taking into account load resistance and therefore actual power output of the amplifier. For this purpose, you simply assume that increasing RMS voltage of the signal 20 times increases power 20·20=400 times. So that's 52dB gain. Everybody reports voltage gain that way and virtually no one will say that the gain of your circuit is 26dB. I think this perversion originates from pro audio, where load impedance used to be assumed 600Ω for some reason (could it be the typical impedance of input transformers in old gear which employed them? no idea). If load impedance is constant and if the amplifier is able to drive it then of course 10x voltage gain produces 100x power gain so this "perverted" rule makes sense. Also, note that power is always positive. So it doesn't matter if the amplifier is inverting or not. In particular, -52dB gain means that output is 1/400 of the input. --- Quote from: bob91343 on August 07, 2019, 12:26:06 am ---For radio communications, the reference is 1 milliwatt. It's indicated by, rather than dB, dBm. In a 50 Ohm system, -73 dBm is 50 microvolts. (Also referred to as signal strength S9) --- End quote --- Indeed. If you want to accurately report power output of this amplifier, you multiply output RMS voltage times output RMS current into whatever load will be connected. That's your power which you express in dB referred to watt (dBW) or milliwatt (dBm) or whatever. For example, 20V RMS into 50Ω is 8W so probably about 8dBW=38dBm or so. If you want to report power gain, you divide output power as above by input power going into the amp and get a unitless (just dB) gain ratio. |
| NaxFM:
You can't have a negative gain, the gain is always positive, less than one if it's an attenuation, more if it's a gain. Also, the gain in this case is a unitless number, because it's a ratio between output and input voltage, so the unit cancels out. Doing the ratio, the result is negative, in this case for the gain you just take the absolute value and say that the output is inverted Inviato dal mio Moto G (5S) Plus utilizzando Tapatalk |
| bob91343:
The broadcasters' use of 600 Ohm stems from telephone engineering. The telephone systems are traditionally 600 Ohms, most likely derived from practical considerations of building the transmission lines. They were balanced lines. Most broadcasters had some link to the telephone system, primarily for long distance relay of programs. So they had to match the line to avoid reflections (echo). So the terms dbW and dBm and so on referred to the base, or 0 figure. (Note that the Bel was named after Alexander Graham Bell, and so the B needs to be capitalized. The leading d means one-tenth and so is lower case. The m suffix means milliwatt. The W suffix means Watt, after James Watt. Capitalization is important and often neglected.) |
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