Help with Gain

[lewisj91]

Hi all,

I'm doing a level 3 electronics course and stuck on the following
Input = 0.95v
output = 3.7v
Noise = 23mV

calculate:

Gain In Volts
Gain in Decibels
Signal to noise ratio in decibels

 

[lewisj91]

I have the current values

Gain in volts = 3.89
Gain in Decibels = 11.809

I don't know if this is right and also I can figure out how to work out SNR, can someone help please ?

 

[davenn]

hi there
welcome to EP

show us the formula you used and working you did to get your answers

Dave

 

[pebe]

The decibel is a measure of power ratios, so you will need to know the input and output impedances.

 

[KrisBlueNZ]

The decibel is a measure of power ratios, so you will need to know the input and output impedances.

The bel, and the decibel, are measures of ratios, not necessarily power ratios. dB can be used to express current ratios and voltage ratios; the formula is dB = 20 log₁₀ (V_OUT / V_IN).

 

[Arouse1973]

I have the current values

Gain in volts = 3.89
Gain in Decibels = 11.809

Yes I agree with Kris. If you only have voltages and nothing else then express this in dBV, using the formula as above #5.
Adam

 

[KrisBlueNZ]

Yes I agree with Kris. If you only have voltages and nothing else then express this in dBV, using the formula as above #5.

dBV is dB relative to a reference level of 1V RMS. It's used when you're specifying an absolute signal level. When you're specifying a gain or attenuation, i.e. the ratio between two voltages, you just use dB.

 

[Arouse1973]

dBV is dB relative to a reference level of 1V RMS. It's used when you're specifying an absolute signal level. When you're specifying a gain or attenuation, i.e. the ratio between two voltages, you just use dB.

Yes your right Kris. I am so used to working with definite levels in EMC I missed that one.
Thanks
Adam

 

[pebe]

The bel, and the decibel, are measures of ratios, not necessarily power ratios. dB can be used to express current ratios and voltage ratios; the formula is dB = 20 log₁₀ (V_OUT / V_IN).

When I was studying electronics almost 50years ago, I was censured by my tutor for not realizing that the result of the formula dB = 20 log10 (VOUT / VIN) could only be correct if the two voltages were measured across a common impedance. This is also implied in the second para of the section ‘Field Quantities’ in the Wiki link below.

http://en.wikipedia.org/wiki/Decibel#Standards

Looking at it a bit deeper, the gain of an amplifier with an input of 1V across 20ohms and an output of 2V across 80ohms calculated using dB = 20 log10 (VOUT / VIN), would give a gain of 6dB. But using the formula 6dB = 10 log10 (POUT / PIN) to get the two powers shows a power ratio of 4, whereas in fact the amplifier has only unity power gain.

The term dBV relates to absolute voltages where 0dB = 1V and in cable television dBmV relates to an absolute level where 0dB = 1mV across 75ohms. Other suffixes relate to other absolute values. But I have always understood that a figure with only a dB suffix relates to a power gain.

However, in the modern world where ‘positive’ and ‘negative’ have come to mean ‘good’ and ‘bad’, and I have seen a man walking down the street referred to on television as ‘self-ambulating’, nothing surprises me any more.

 

[KrisBlueNZ]

I agree that using dB for a pure voltage ratio, with no associated impedance, takes the idea somewhat out of context, but it's normal. The dBV unit is widely used in audio and has no associated impedance; it's simply a convenient way to represent an absolute AC voltage level because amplifications add and attenuations subtract, rather than multiplying. I guess units, like words, get used in different ways over the years.