Help understanding voltage db vs power db.

[amdx]

The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.

I'm confused after using this calculator.
http://www.sengpielaudio.com/calculatorVoltagePower.htm

The first calculator labeled,
“ Find decibel voltage gain and ratio out/in by entering input and
output voltage:”
If I put 1 volt on the input line and 2 volts on the output line, the
calculator gives a 6db gain.

The second calculator labeled,
“Find decibel power gain and ratio out/in by entering before and after
power“
In order to enter power in Watts, I'll assume a 50 ohm system and
calculate power using
V^2/R=P. As above 1volt^2/50 =0.02 Watts for input power and 2^2/50=0.08
Watts for output power.
When I enter 0.02 and 0.08 into the second calculator, I get 6db power gain.

Back to the fact that I know, a 6db voltage gain is a 3db power gain.

So, what am I doing wrong that I don't get 3db for the power?

Thanks for your time, Mikek

 

The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.

As the wise man said, "It ain't what you don't know that hurts you--it's what you do know that ain't so."

Decibels are a convenient means for expressing *power ratios*. Keep that in mind and your troubles will disappear. (These ones, anyway.)

dB = 10 log(P1/P2),

where P1 and P2 are in units of power.

That formula with a 20 in it is okay if the circuit impedance is the same for both P1 and P2, e.g. if they're measured at the same point in the circuit or in different places in a well-matched 50-ohm system.

There are places. such as op amp outputs, where the impedance level isn't too well defined, so the formula with 20 in it is good for expressing SNR or distortion.

But don't just toss that 20 formula around, or you'll wind up thinking that a step-up transformer has gain.

Cheers

Phil Hobbs

 

On Friday, April 17, 2020 at 8:02:18 AM UTC-4, amdx wrote:

The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.

I'm confused after using this calculator.
http://www.sengpielaudio.com/calculatorVoltagePower.htm

The first calculator labeled,
“ Find decibel voltage gain and ratio out/in by entering input and
output voltage:”
If I put 1 volt on the input line and 2 volts on the output line, the
calculator gives a 6db gain.

The second calculator labeled,
“Find decibel power gain and ratio out/in by entering before and after
power“
In order to enter power in Watts, I'll assume a 50 ohm system and
calculate power using
V^2/R=P. As above 1volt^2/50 =0.02 Watts for input power and 2^2/50=0.08
Watts for output power.
When I enter 0.02 and 0.08 into the second calculator, I get 6db power gain.

Back to the fact that I know, a 6db voltage gain is a 3db power gain.

So, what am I doing wrong that I don't get 3db for the power?

Thanks for your time, Mikek

dB is a way to relate one POWER (always power) to another POWER. Frequently it will relate one measured power to another reference power.

A reference power will be something like 0dBm. That is an absolute reference power of 1 mW. Now you can say that a signal is 10dB lower than another power. In the case of 10dB lower than 0dBm you get 10dBm - 10 dB = -10 dBm.

You have to remember that dB's relate one power to another and it only works if you know which one is the reference power.

Now regarding voltage and power.......

remember dBs are all about power...power ......power......

10 dB multiplies power by a factor of 10. 3dB doubles the power. 3 dB always doubles the power. 6 dB multiplies the power by 4. remember that.

Back to voltage.....

In most cases we cannot measure power so we have to measure voltage instead.. We want to relate things in terms of power but we can tend to have voltage measurements available to us. Now if you have 1V across 1 ohm that is 1 Watt

Lets call that 0dBw ( that is 0 dB relative to 1Watt which equals.....1Watt)

Now if you double the voltage on that resistor you get 2 volts but the POWER (remember dbs are about POWER) goes up by 4 ... which is 6 dB.

You now have 6dB above the reference of 0dBW...or 6dB above 1 W ==== 4 Watts

so when taking voltage measurements you get
20log ( voltage measured/ voltage on reference).


You have to figure out when something is an absolute....dBm, dBW

and when it is relative....dB.

 

On Friday, April 17, 2020 at 9:10:13 AM UTC-4, amdx wrote:

On 4/17/2020 7:49 AM, Pimpom wrote:
On 4/17/2020 5:32 PM, amdx wrote:
   The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.


That's where you got it wrong. The same ratio - e.g. 3:1 - expressed in
dB for voltage is twice that for the same power ratio. This convention
arises from the assumption that the two voltages are applied to the same
resistance/impedance, which is not always the case and can create
confusion.

Mathematically, for voltage the dB value is 20log(V1/V2) whereas for
power it's 10log(P1/P2)

Take an amplifier with a gain of 5x and an input voltage of 1V. By
convention, the voltage gain is 20log(5/1) = 14dB.

If the input resistance of the amplifier and the output load resistance
are both 100 ohms, the input power is 1²/100 = 0.01W; output power is
5²/100 = 0.25W. The power ratio is 10log(0.25/0.01) = 14dB.

Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB

If the input and output resistances are *not* the same, the gain in dB
will NOT be the same for voltage and power.


I am assuming the same impedance for all calculations.
RE: Voltage gain = 5 or 14dB
Is the voltage gain of 5 expressed as a 14db power gain?

Meaning, when you calculate gain using a voltage ratio, is the answer
in units as power gain?

That would solve my confusion.

Thanks, Mikek

We are trying to teach you to fish and I think you want us to toss you a fish

 

[amdx]

On 4/17/2020 7:49 AM, Pimpom wrote:

On 4/17/2020 5:32 PM, amdx wrote:
   The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.


That's where you got it wrong. The same ratio - e.g. 3:1 - expressed in
dB for voltage is twice that for the same power ratio. This convention
arises from the assumption that the two voltages are applied to the same
resistance/impedance, which is not always the case and can create
confusion.

Mathematically, for voltage the dB value is 20log(V1/V2) whereas for
power it's 10log(P1/P2)

Take an amplifier with a gain of 5x and an input voltage of 1V. By
convention, the voltage gain is 20log(5/1) = 14dB.

If the input resistance of the amplifier and the output load resistance
are both 100 ohms, the input power is 1²/100 = 0.01W; output power is
5²/100 = 0.25W. The power ratio is 10log(0.25/0.01) = 14dB.

Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB

If the input and output resistances are *not* the same, the gain in dB
will NOT be the same for voltage and power.

I am assuming the same impedance for all calculations.
RE: Voltage gain = 5 or 14dB
Is the voltage gain of 5 expressed as a 14db power gain?

Meaning, when you calculate gain using a voltage ratio, is the answer
in units as power gain?

That would solve my confusion.

Thanks, Mikek

 

[Arie de Muynck]

On 2020-04-17 14:49, Arie de Muynck wrote:

On 2020-04-17 14:02, amdx wrote:
  The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.

Yes, that is wrong, your memory exchanged voltage and power.

A 3 dB voltage gain (x1.413) is a 6 dB power gain (x1.995), assuming the
same impedance.
See e.g. the table in <https://en.wikipedia.org/wiki/Decibel>.

Arie

Oops, those numbers are also wrong. Sorry, I was up all night...

Arie

 

[Pimpom]

On 4/17/2020 5:32 PM, amdx wrote:

The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.

That's where you got it wrong. The same ratio - e.g. 3:1 -
expressed in dB for voltage is twice that for the same power
ratio. This convention arises from the assumption that the two
voltages are applied to the same resistance/impedance, which is
not always the case and can create confusion.

Mathematically, for voltage the dB value is 20log(V1/V2) whereas
for power it's 10log(P1/P2)

Take an amplifier with a gain of 5x and an input voltage of 1V.
By convention, the voltage gain is 20log(5/1) = 14dB.

If the input resistance of the amplifier and the output load
resistance are both 100 ohms, the input power is 1²/100 = 0.01W;
output power is 5²/100 = 0.25W. The power ratio is
10log(0.25/0.01) = 14dB.

Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB

If the input and output resistances are *not* the same, the gain
in dB will NOT be the same for voltage and power.

 

[Arie de Muynck]

On 2020-04-17 14:02, amdx wrote:

 The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.

Yes, that is wrong, your memory exchanged voltage and power.

A 3 dB voltage gain (x1.413) is a 6 dB power gain (x1.995), assuming the
same impedance.
See e.g. the table in <https://en.wikipedia.org/wiki/Decibel>.

Arie

 

[amdx]

On 4/17/2020 8:15 AM, blocher@columbus.rr.com wrote:

On Friday, April 17, 2020 at 9:10:13 AM UTC-4, amdx wrote:
On 4/17/2020 7:49 AM, Pimpom wrote:
On 4/17/2020 5:32 PM, amdx wrote:
   The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.


That's where you got it wrong. The same ratio - e.g. 3:1 - expressed in
dB for voltage is twice that for the same power ratio. This convention
arises from the assumption that the two voltages are applied to the same
resistance/impedance, which is not always the case and can create
confusion.

Mathematically, for voltage the dB value is 20log(V1/V2) whereas for
power it's 10log(P1/P2)

Take an amplifier with a gain of 5x and an input voltage of 1V. By
convention, the voltage gain is 20log(5/1) = 14dB.

If the input resistance of the amplifier and the output load resistance
are both 100 ohms, the input power is 1²/100 = 0.01W; output power is
5²/100 = 0.25W. The power ratio is 10log(0.25/0.01) = 14dB.

Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB

If the input and output resistances are *not* the same, the gain in dB
will NOT be the same for voltage and power.


I am assuming the same impedance for all calculations.
RE: Voltage gain = 5 or 14dB
Is the voltage gain of 5 expressed as a 14db power gain?

Meaning, when you calculate gain using a voltage ratio, is the answer
in units as power gain?

That would solve my confusion.

Thanks, Mikek

We are trying to teach you to fish and I think you want us to toss you a fish

I do know the 10log and 20log formula, but a simple yes/no to the
20log(V2/V2) Is, the answer expressed in power.

At this point, I think it is, otherwise, why the 20log.

Sorry to be dense, but if I didn't need help, I wouldn't be asking.

Mikek

 

[John S]

On 4/17/2020 8:41 AM, amdx wrote:

On 4/17/2020 8:15 AM, blocher@columbus.rr.com wrote:
On Friday, April 17, 2020 at 9:10:13 AM UTC-4, amdx wrote:
On 4/17/2020 7:49 AM, Pimpom wrote:
On 4/17/2020 5:32 PM, amdx wrote:
    The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.


That's where you got it wrong. The same ratio - e.g. 3:1 - expressed in
dB for voltage is twice that for the same power ratio. This convention
arises from the assumption that the two voltages are applied to the
same
resistance/impedance, which is not always the case and can create
confusion.

Mathematically, for voltage the dB value is 20log(V1/V2) whereas for
power it's 10log(P1/P2)

Take an amplifier with a gain of 5x and an input voltage of 1V. By
convention, the voltage gain is 20log(5/1) = 14dB.

If the input resistance of the amplifier and the output load resistance
are both 100 ohms, the input power is 1²/100 = 0.01W; output power is
5²/100 = 0.25W. The power ratio is 10log(0.25/0.01) = 14dB.

Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB

If the input and output resistances are *not* the same, the gain in dB
will NOT be the same for voltage and power.


   I am assuming the same impedance for all calculations.
RE: Voltage gain = 5 or 14dB
   Is the voltage gain of 5 expressed as a 14db power gain?

Meaning, when you calculate gain using a voltage ratio, is the answer
in units as power gain?

   That would solve my confusion.

                          Thanks, Mikek

We are trying to teach you to fish and I think you want us to toss you
a fish


I do know the 10log and 20log formula, but a simple yes/no to the
20log(V2/V2) Is, the answer expressed in power.

At this point, I think it is, otherwise, why the 20log.

Sorry to be dense, but if I didn't need help, I wouldn't be asking.

                                                   Mikek

Actually, that should be V2/V1. I made that same mistake in my reply. Sorry.

 

[John S]

On 4/17/2020 8:41 AM, amdx wrote:

On 4/17/2020 8:15 AM, blocher@columbus.rr.com wrote:
On Friday, April 17, 2020 at 9:10:13 AM UTC-4, amdx wrote:
On 4/17/2020 7:49 AM, Pimpom wrote:
On 4/17/2020 5:32 PM, amdx wrote:
    The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.


That's where you got it wrong. The same ratio - e.g. 3:1 - expressed in
dB for voltage is twice that for the same power ratio. This convention
arises from the assumption that the two voltages are applied to the
same
resistance/impedance, which is not always the case and can create
confusion.

Mathematically, for voltage the dB value is 20log(V1/V2) whereas for
power it's 10log(P1/P2)

Take an amplifier with a gain of 5x and an input voltage of 1V. By
convention, the voltage gain is 20log(5/1) = 14dB.

If the input resistance of the amplifier and the output load resistance
are both 100 ohms, the input power is 1²/100 = 0.01W; output power is
5²/100 = 0.25W. The power ratio is 10log(0.25/0.01) = 14dB.

Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB

If the input and output resistances are *not* the same, the gain in dB
will NOT be the same for voltage and power.


   I am assuming the same impedance for all calculations.
RE: Voltage gain = 5 or 14dB
   Is the voltage gain of 5 expressed as a 14db power gain?

Meaning, when you calculate gain using a voltage ratio, is the answer
in units as power gain?

   That would solve my confusion.

                          Thanks, Mikek

We are trying to teach you to fish and I think you want us to toss you
a fish


I do know the 10log and 20log formula, but a simple yes/no to the
20log(V2/V2) Is, the answer expressed in power.

Yes, with the constraint that the input & output resistances are the
identical.

The reason that 20log(V2/V2) is used is to describe voltage gain without
regard to input & output resistances.

At this point, I think it is, otherwise, why the 20log.

Sorry to be dense, but if I didn't need help, I wouldn't be asking.

                                                   Mikek

 

[Pimpom]

On 4/17/2020 6:40 PM, amdx wrote:

On 4/17/2020 7:49 AM, Pimpom wrote:
On 4/17/2020 5:32 PM, amdx wrote:
   The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.


That's where you got it wrong. The same ratio - e.g. 3:1 - expressed in
dB for voltage is twice that for the same power ratio. This convention
arises from the assumption that the two voltages are applied to the same
resistance/impedance, which is not always the case and can create
confusion.

Mathematically, for voltage the dB value is 20log(V1/V2) whereas for
power it's 10log(P1/P2)

Take an amplifier with a gain of 5x and an input voltage of 1V. By
convention, the voltage gain is 20log(5/1) = 14dB.

If the input resistance of the amplifier and the output load resistance
are both 100 ohms, the input power is 1²/100 = 0.01W; output power is
5²/100 = 0.25W. The power ratio is 10log(0.25/0.01) = 14dB.

Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB

If the input and output resistances are *not* the same, the gain in dB
will NOT be the same for voltage and power.


I am assuming the same impedance for all calculations.
RE: Voltage gain = 5 or 14dB
Is the voltage gain of 5 expressed as a 14db power gain?

Meaning, when you calculate gain using a voltage ratio, is the answer
in units as power gain?

That would solve my confusion.

Thanks, Mikek

The dB unit is based on power ratios. dB for voltage ratio is
derived from that and its use regardless of impedance is just
convention.

 

[Adrian Tuddenham]

<blocher@columbus.rr.com> wrote:

On Friday, April 17, 2020 at 8:02:18 AM UTC-4, amdx wrote: > The fact that
I know: a 6db voltage gain is a 3db power gain.

dB is a way to relate one POWER (always power) to another POWER.
Frequently it will relate one measured power to another reference
power.

This is what you need to have burned into your brain. Anything else is
going to confuse you and lead to the wrong answer eventually.

Decibels are all about POWER, there are no voltage decibels.


--
~ Adrian Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk

 

[amdx]

On 4/17/2020 8:49 AM, Pimpom wrote:

On 4/17/2020 6:40 PM, amdx wrote:
On 4/17/2020 7:49 AM, Pimpom wrote:
On 4/17/2020 5:32 PM, amdx wrote:
    The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.


That's where you got it wrong. The same ratio - e.g. 3:1 - expressed in
dB for voltage is twice that for the same power ratio. This convention
arises from the assumption that the two voltages are applied to the same
resistance/impedance, which is not always the case and can create
confusion.

Mathematically, for voltage the dB value is 20log(V1/V2) whereas for
power it's 10log(P1/P2)

Take an amplifier with a gain of 5x and an input voltage of 1V. By
convention, the voltage gain is 20log(5/1) = 14dB.

If the input resistance of the amplifier and the output load resistance
are both 100 ohms, the input power is 1²/100 = 0.01W; output power is
5²/100 = 0.25W. The power ratio is 10log(0.25/0.01) = 14dB.

Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB

If the input and output resistances are *not* the same, the gain in dB
will NOT be the same for voltage and power.


   I am assuming the same impedance for all calculations.
RE: Voltage gain = 5 or 14dB
   Is the voltage gain of 5 expressed as a 14db power gain?

Meaning, when you calculate gain using a voltage ratio, is the answer
in units as power gain?

   That would solve my confusion.

                          Thanks, Mikek


The dB unit is based on power ratios. dB for voltage ratio is derived
from that and its use regardless of impedance is just convention.

Ok guys, now it makes sense.
Although, now, I'll work it a bit more.
Thank, Mikek

 

On Friday, April 17, 2020 at 8:02:18 AM UTC-4, amdx wrote:

The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.

I'm confused after using this calculator.
http://www.sengpielaudio.com/calculatorVoltagePower.htm

The first calculator labeled,
“ Find decibel voltage gain and ratio out/in by entering input and
output voltage:”
If I put 1 volt on the input line and 2 volts on the output line, the
calculator gives a 6db gain.

The second calculator labeled,
“Find decibel power gain and ratio out/in by entering before and after
power“
In order to enter power in Watts, I'll assume a 50 ohm system and
calculate power using
V^2/R=P. As above 1volt^2/50 =0.02 Watts for input power and 2^2/50=0.08
Watts for output power.
When I enter 0.02 and 0.08 into the second calculator, I get 6db power gain.

Back to the fact that I know, a 6db voltage gain is a 3db power gain.

So, what am I doing wrong that I don't get 3db for the power?

Thanks for your time, Mikek

Power gains and voltage gains are numerically equal when expressed in dB. Try working out the formula. Forget about the obvious common impedance jazz.

 

[Ricky C]

On Friday, April 17, 2020 at 8:02:18 AM UTC-4, amdx wrote:

The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.

Ok, that is wrong. 3 dB is a doubling of the power. 6 dB is a doubling of the voltage. 6 dB is a quadrupling of the power. 3 dB is not the same as 6 dB no matter what. dB is always indicating power ratios no matter how it is measured.

This comes from power being proportional to the square of the voltage.

I'm confused after using this calculator.
http://www.sengpielaudio.com/calculatorVoltagePower.htm

The first calculator labeled,
“ Find decibel voltage gain and ratio out/in by entering input and
output voltage:”
If I put 1 volt on the input line and 2 volts on the output line, the
calculator gives a 6db gain.

The second calculator labeled,
“Find decibel power gain and ratio out/in by entering before and after
power“
In order to enter power in Watts, I'll assume a 50 ohm system and
calculate power using
V^2/R=P. As above 1volt^2/50 =0.02 Watts for input power and 2^2/50=0.08
Watts for output power.
When I enter 0.02 and 0.08 into the second calculator, I get 6db power gain.

Back to the fact that I know, a 6db voltage gain is a 3db power gain.

So, what am I doing wrong that I don't get 3db for the power?

6 dB is not 3 dB... That's what's wrong. You are thinking in terms of doubling the power and doubling the voltage as somehow being the same thing. It's not. See my explanation above.

dB is always expressing a power ratio. Power is the square of voltage. That's why the formulas for calculating dB use 10 and 20 multipliers. It doesn't make 3 dB the same as 6 dB ever. Adding the power or voltage label is confusing you.

--

Rick C.

- Get 1,000 miles of free Supercharging
- Tesla referral code - https://ts.la/richard11209

 

[Ricky C]

On Friday, April 17, 2020 at 9:10:13 AM UTC-4, amdx wrote:

On 4/17/2020 7:49 AM, Pimpom wrote:
On 4/17/2020 5:32 PM, amdx wrote:
   The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.


That's where you got it wrong. The same ratio - e.g. 3:1 - expressed in
dB for voltage is twice that for the same power ratio. This convention
arises from the assumption that the two voltages are applied to the same
resistance/impedance, which is not always the case and can create
confusion.

Mathematically, for voltage the dB value is 20log(V1/V2) whereas for
power it's 10log(P1/P2)

Take an amplifier with a gain of 5x and an input voltage of 1V. By
convention, the voltage gain is 20log(5/1) = 14dB.

If the input resistance of the amplifier and the output load resistance
are both 100 ohms, the input power is 1²/100 = 0.01W; output power is
5²/100 = 0.25W. The power ratio is 10log(0.25/0.01) = 14dB.

Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB

If the input and output resistances are *not* the same, the gain in dB
will NOT be the same for voltage and power.


I am assuming the same impedance for all calculations.
RE: Voltage gain = 5 or 14dB
Is the voltage gain of 5 expressed as a 14db power gain?

Meaning, when you calculate gain using a voltage ratio, is the answer
in units as power gain?

That would solve my confusion.

Thanks, Mikek

Yes, exactly.

So long, and thanks for all the fish.

--

Rick C.

+ Get 1,000 miles of free Supercharging
+ Tesla referral code - https://ts.la/richard11209

 

On Friday, April 17, 2020 at 8:02:18 AM UTC-4, amdx wrote:

The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.

I'm confused after using this calculator.
http://www.sengpielaudio.com/calculatorVoltagePower.htm

The first calculator labeled,
“ Find decibel voltage gain and ratio out/in by entering input and
output voltage:”
If I put 1 volt on the input line and 2 volts on the output line, the
calculator gives a 6db gain.

The second calculator labeled,
“Find decibel power gain and ratio out/in by entering before and after
power“
In order to enter power in Watts, I'll assume a 50 ohm system and
calculate power using
V^2/R=P. As above 1volt^2/50 =0.02 Watts for input power and 2^2/50=0.08
Watts for output power.
When I enter 0.02 and 0.08 into the second calculator, I get 6db power gain.

Back to the fact that I know, a 6db voltage gain is a 3db power gain.

So, what am I doing wrong that I don't get 3db for the power?

Thanks for your time, Mikek

"The decibel scales differ by a factor of two so that the related power and field levels change by the same number of decibels with linear loads."

A decibel is one tenth of a bel. So why is decibel 10 x Log( pwr ratio) ?

https://en.wikipedia.org/wiki/Decibel.

 

[George Herold]

On Friday, April 17, 2020 at 8:36:47 AM UTC-4, pcdh...@gmail.com wrote:

The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.

As the wise man said, "It ain't what you don't know that hurts you--it's what you do know that ain't so."

Decibels are a convenient means for expressing *power ratios*. Keep that in mind and your troubles will disappear. (These ones, anyway.)

dB = 10 log(P1/P2),

where P1 and P2 are in units of power.

Just as an aside, I find the factor of ten (10) to be 'an issue'*
in my brain. I'd much rather have a straight LOG(P1/P2).

George H.
*makes me have to stop and think a bit.

That formula with a 20 in it is okay if the circuit impedance is the same for both P1 and P2, e.g. if they're measured at the same point in the circuit or in different places in a well-matched 50-ohm system.

There are places. such as op amp outputs, where the impedance level isn't too well defined, so the formula with 20 in it is good for expressing SNR or distortion.

But don't just toss that 20 formula around, or you'll wind up thinking that a step-up transformer has gain.

Cheers

Phil Hobbs

 

[George Herold]

On Friday, April 17, 2020 at 9:10:13 AM UTC-4, amdx wrote:

On 4/17/2020 7:49 AM, Pimpom wrote:
On 4/17/2020 5:32 PM, amdx wrote:
   The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.


That's where you got it wrong. The same ratio - e.g. 3:1 - expressed in
dB for voltage is twice that for the same power ratio. This convention
arises from the assumption that the two voltages are applied to the same
resistance/impedance, which is not always the case and can create
confusion.

Mathematically, for voltage the dB value is 20log(V1/V2) whereas for
power it's 10log(P1/P2)

Take an amplifier with a gain of 5x and an input voltage of 1V. By
convention, the voltage gain is 20log(5/1) = 14dB.

If the input resistance of the amplifier and the output load resistance
are both 100 ohms, the input power is 1²/100 = 0.01W; output power is
5²/100 = 0.25W. The power ratio is 10log(0.25/0.01) = 14dB.

Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB

If the input and output resistances are *not* the same, the gain in dB
will NOT be the same for voltage and power.


I am assuming the same impedance for all calculations.
RE: Voltage gain = 5 or 14dB
Is the voltage gain of 5 expressed as a 14db power gain?

Meaning, when you calculate gain using a voltage ratio, is the answer
in units as power gain?

That would solve my confusion.

Thanks, Mikek

Power ~ V^2, (also I^2 and V*I) so when you do the log,
there is a factor of 2 you have to take care of.
Log(Pow) = ~ Log(V^2) = 2*Log(V)

When working in dB I just think of power and convert to voltage
at the end if I have to.

GH