converting Volt into dB

[Mepakos]

How can i convert -20V into dB?

[magic]

You can't, logarithm is only defined on positive numbers. Decibel scale is usually used to refer to signal power and power is always positive too.

It's you the Decibel Guy again, I see. You would get more sensible responses if you explained what you are trying to calculate and why instead of asking one-liner questions that don't even have a valid answer.

[Mepakos]

The question is what is the voltage gain of the transistor in dB?    its -20V! I need to convert that to dB

[Kalin]

No it is not -20V. People are not going to do your homework for you. If you want to know how something works or where you went wrong in a calculation many people will help you, but noone is going to just give you answers. Gain is a ratio of input to output and doesn't have a unit.

Sent from my SM-G965W using Tapatalk

[Mepakos]

I dont ask for answers just for good advice on how to solve the task.Instead of saying u are not doing my homework which i dont want, u can tell me why -20V is wrong. Ua=20Vss and Uce=Ua.   UCE is the voltage gain of the transistor. So why its wrong.

[Kalin]

https://en.m.wikipedia.org/wiki/Gain_(electronics)

Read this and report back what you think your gain is and how you think you would convert that to dB

Sent from my SM-G965W using Tapatalk

[Mepakos]

10log(20/1)=13,01dB.

[Bassman59]

How can i convert -20V into dB?

What is the reference to which the -20 V is compared?

[chrisl]

10log(20/1)=13,01dB.

That is a power gain in dB. What you are looking for is a voltage gain in dB.

[Audioguru again]

He is wrongly subtracting the gain number because the transistor inverts the signal. The voltage gain is still 20 times whether it is inverted or not.
A voltage gain of 2 times is +6dB.
A voltage gain of 10 times is 20dB.
Then isn't a voltage gain of 20 times 6dB added to 20dB= +26dB?

[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]

why -20V is wrong. Ua=20Vss and Uce=Ua.   UCE is the voltage gain of the transistor. So why its wrong.

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

The question is what is the voltage gain of the transistor in dB?

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]

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.

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.

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)

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.)

[soldar]

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.

I don't think that's right.  https://en.wikipedia.org/wiki/Decibel shows negative numbers. 0 dB is gain of unity. Negative gains get negative numbers in dB. WIFI ranges in the -60 dB range. Audio too.

[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.

I don't think that's right.  https://en.wikipedia.org/wiki/Decibel shows negative numbers. 0 dB is gain of unity. Negative gains get negative numbers in dB. WIFI ranges in the -60 dB range. Audio too.

Well, yes, speaking in terms of deciBels you can have negative gain. I didn't say that, but I was talking about gain expressed as a pure number.
If the ratio is less than one, then the gain in deciBels is negative

Inviato dal mio Moto G (5S) Plus utilizzando Tapatalk

[radiolistener]

Negative gains get negative numbers in dB.

Negative gain in dB is not a gain, this is attenuation

[soldar]

Negative gain in dB is not a gain, this is attenuation

Attenuation is the same as "negative gain" and both are correct. Just like a negative profit is a loss, negative growth is shrinking, etc.

A program designed to calculate gain can output the result as "Gain: -20 dB" and it is not incorrect.

[mikerj]

Negative gain in dB is not a gain, this is attenuation

Attenuation is the same as "negative gain" and both are correct. Just like a negative profit is a loss, negative growth is shrinking, etc.

A program designed to calculate gain can output the result as "Gain: -20 dB" and it is not incorrect.

Exactly, the sign changes depending on whether you refer to a value as a gain or an attenuation.

[mikerj]

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.

I don't understand this, dB is simply a log scale, it can be used with many different units.  Are you saying that using it for e.g. electric field strength or bandwidth is wrong?

[radiolistener]

you can use it for different units. For example there is popular unit dBuV (decibel micro-Volt)

[Audioguru again]

The inverted output of the transistor has no load so its voltage gain is 20 times which is +26dB. It has some output impedance so its output level and voltage gain will drop when it has a load. Then you can calculate its power gain if you want.
edit: Because it inverts, its output phase is negative, not its gain.

[pepelevamp]

ya know i figure you don't need to be talking about power or what not to use decibels. its simply a way to talk about scales easily. but the power db versus voltage db - that can be confusing.

[bob91343]

I reiterate that decibels are a measure of power ratio.  Any other use is perversion.

Having said that, it's popular culture to use decibels for other purposes, although it's clearly wrong.

As for using it for field strength, etc., it's still a valid power ratio.

We don't say that a 2 Ohm resistor is 6 dB higher than a 1 Ohm resistor.  Decibels are properly used for power ratio.

Saying an amplifier has such and such decibel gain is improper.  But common practice.

The bottom line is that you can be precise or not but if your audience understands your meaning you can get away with however you say it.

[schmitt trigger]

Indeed, one can use decibels for many purposes. Many prepostperous.

In my previous job, since it was a public company, the CEO's compensation was available.

As a prank, we started calculating how many decibel dollars (dB$) our pay compared to the CEO's compensation.

Because it was purchasing POWER, it was calculated as: 10log(my$/$CEO$)

Of course, the results were always negative 

[soldar]

I reiterate that decibels are a measure of power ratio.  Any other use is perversion.

Can we have a cite for that? Because Wikipedia disagrees and cites ISO 80000-3:2006.

https://en.wikipedia.org/wiki/Decibel
In the International System of Quantities, the decibel is defined as a unit of measurement for quantities of type level or level difference, which are defined as the logarithm of the ratio of power- or field-type quantities.

[Audioguru again]

The voltage gain of opamps is measured in decibels.
The voltage gain of audio power amplifiers is measured in decibels.
The voltage cutoff frequency of a filter is measured at -3dB.
The voltage reduction of a logarithmic volume control is measured in decibels.

[soldar]

Another cite:

https://en.wikipedia.org/wiki/Gain_(electronics)

The term gain alone is ambiguous, and can refer to the ratio of output to input voltage (voltage gain), current (current gain) or electric power (power gain).[4] In the field of audio and general purpose amplifiers, especially operational amplifiers, the term usually refers to voltage gain,[2] but in radio frequency amplifiers it usually refers to power gain.

Another: http://www.sengpielaudio.com/calculatorVoltagePower.htm

[mikerj]

I reiterate that decibels are a measure of power ratio.  Any other use is perversion.

You can reiterate it as many times as you want, but unless you back it up with some actual evidence then I'm just going to ignore it.

[magic]

In the field of audio and general purpose amplifiers, especially operational amplifiers, the term usually refers to voltage gain,[2] but in radio frequency amplifiers it usually refers to power gain.

Worth noting that when it refers to voltage gain, it's still reported as the equivalent power gain under assumption of equal load impedance. That is, a 10x voltage gain is reported as 20dB because it corresponds to 100x power gain, not as 10dB because it's a 10x gain.
Just Wikipedia being a misleading pile fo turd as usual

[soldar]

I reiterate that decibels are a measure of power ratio.  Any other use is perversion.

You can reiterate it as many times as you want, but unless you back it up with some actual evidence then I'm just going to ignore it.

You pervert, you!