Voltage Gain Question

[EddieB]

Hi Guys,

I'm slowly working through the Book teach yourself electronics by Dr. Malcolm Plant.. I've been dabbling with electronics for years but just started to jump into the theory behind things.

The Chapter I am currently studying is Amplifiers.. and doing the questions at the end of the chapter I have ran into a problem I can't solve.. Was wondering if someone could give me a hand here to help me understand the question and work it out correctly..

The question word for word: 'The voltage gain of an Amplifier falls by 3dB from a value of 1000. What is the new voltage.'

To me the question seems vague and I've tried various methods to work it out, but no luck.. the answer in the back is 500v.

Additional: I've been using the formula for voltage gain: 20 log10 vout/vIn.

Regards,
Eddie

[alm]

Voltage gain is the ratio between two voltages (Vout/Vin), so it's not measured in volts, but a dimensionless ratio (eg. 1000). You can either convert the current gain to dB, subtract 3dB and convert back to straight ratio, or convert the 3dB to a ratio and divide 1000 by that number. The book may also have taught to memorize the easy ones like 3dB, 6dB and 20dB.

The formula you've been using is fine.

[EddieB]

Hi Alm thanks for the quick reply

I understand that it is a ration between the voltages and is dimensionless and previously calculated the gain of 1000 to = 60dB, subtracting 3dB to arrive at 57dB but haven't had much luck from that.

My main confusion is how the answer in the back is in Volts, perhaps a typo?

Thanks again for the help
Eddie

[alm]

I missed that 500V was the supposed correct answer. I assume the volts is an error, and they apparently consider the -3dB a power ratio, not a voltage ratio (hence the 10*log10(Vin/Vout)), which I consider misleading unless stated somehow in the question.

[EddieB]

Hmmm I think I'll give this one a miss and find other questions like it ( hopefully correct ) on the internet.

Thanks for all your help

Cheers

[Zero999]

A 3dB fall in the power gain is simply half the original gain.

A 3dB fall in the voltage gain is the original gain multiplied by 1/?2.

[jahonen]

For me, there is just one kind of decibels, power decibels. The "voltage decibel" is just power expressed via voltage to same reference inpedance. 10*log10(P1/P2) = 10*log10((U1^2/R)/(U2^2/R)) = 10*log10((U1/U2)^2) =  2*10*log10(U1/U2) = 20*log10(U1/U2).

Halving the voltage results power reduction to 1/4 of original value, thus -6 dB ratio in decibels.

Regards,
Janne

[EddieB]

Ahhh all makes sense considering it's ratios.. 

Thanks for clearing it up guys

[.o:0|O|0:o.]

Post corrected for power (thanks tecman):

For illustration purposes, if you ever have a non-trivial value, you can manipulate the power gain formula in the following way to obtain the result.

10log(Po/Pin)=-3;
log(Po/Pin)=-0.3;
Po/Pin=antilog(-0.3);
Po=Pin*antilog(-0.3);

For Pin=1000;
Po=1000*0.501 =~ 500;

.o:0|O|0:o.

P.s.
10log(500/1000) =~ -3;

[tecman]

For illustration purposes, if you ever have a non-trivial value, you can manipulate the voltage gain formula in the following way to obtain the result.

10log(Vo/Vin)=-3;
log(Vo/Vin)=-0.3;
Vo/Vin=antilog(-0.3);
Vo=Vin*antilog(-0.3);

For Vin=1000;
Vo=1000*0.501 =~ 500;

.o:0|O|0:o.

P.s.
10log(500/1000) =~ -3;

For voltage ratios, it is 20log(..

For power it is 10log(...

paul

[.o:0|O|0:o.]

20log(Vo/Vin)=-3;
log(Vo/Vin)=-0.15;
Vo/Vin=antilog(-0.15);
Vo=Vin*antilog(-0.15);

For Vin=1000;
Vo=1000*0.708 =~ 708;

.o:0|O|0:o.

P.s.
20log(708/1000) =~ -3;