Finding the magnitude of the voltage gain in dB

In summary, the conversation is discussing the effects of parasitic capacitances in a MOSFET amplifier and how to find the gain in dB. The person has derived the small-signal voltage gain including the capacitances and verified it to be correct. They are now stuck on finding the magnitude of the gain in dB and are trying to understand the solution provided, which involves using logarithms and substitutions. In particular, the conversation notes the use of the form log(A/B) = log(A) - log(B) and the substitution ω = 2πf. The solution also involves using the property log(C^n) = nlog(C) and simplifying the last term to [f(a+b)]^2.
  • #1
CoolDude420
198
8

Homework Statement


Hi,
so in a MOSFET amplifier question(a common-source amplifier in particular), we are examining the effects of the parasitic capacitances of the MOSFET when included in the small-signal model. In particular, the gain of the amplifier with these capacitances.
So I derived the small-signal voltage gain including the parasitic capacitances as follows,

80e590234d.png

I verified that this is correct.

Now, I'm stuck on how to find the actual magnitude of the this gain in dB. The answer just skips directly to this,

34c5cffa16.png


I understand the first part of that, i.e that's the magnitude of the gain in dB of the top of the fraction and the following that should be the magnitude of the gain of the bottom (subtracted of course). I have no idea where second part is coming from.

Homework Equations


e97d60e21f.png

dBgain = 20log10(gain)

The Attempt at a Solution

 

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  • #2
First of all, they are doing a quotient of the form ## log(\frac{A}{B})=log(A)-log(B) ##. They also do a substitution ## \omega=2 \pi f ##. In addition, they used ## 20 \, log(\sqrt{C})=20 \, log(C^{1/2})=10 \, log (C) ##. (Note ## log(C^n)=n \, log(C) ##). The rest you should be able to figure out. ## \\ ## And I don't know why they put the last term in the form: ## \Big(\frac{f}{\frac{1}{(a+b)}} \Big)^2 ##. They could simply have written it as ## [ f(a+b)]^2 ##.
 
Last edited:

1. What is the formula for finding the magnitude of voltage gain in dB?

The formula for finding the magnitude of voltage gain in dB is 20log10(Vout/Vin), where Vout is the output voltage and Vin is the input voltage.

2. Why is the magnitude of voltage gain expressed in decibels?

The magnitude of voltage gain is often expressed in decibels because it allows for a more convenient and compact representation of the gain. Decibels are also on a logarithmic scale, which makes it easier to compare gains of different magnitudes.

3. How is the magnitude of voltage gain related to the power gain?

The magnitude of voltage gain is related to the power gain through the formula 10log10(Pout/Pin), where Pout is the output power and Pin is the input power. This relationship is based on the fact that power is proportional to the square of the voltage.

4. Can the magnitude of voltage gain be negative?

Yes, the magnitude of voltage gain can be negative. This indicates that the output voltage is lower than the input voltage, resulting in an attenuation rather than an amplification.

5. How is the magnitude of voltage gain affected by frequency?

The magnitude of voltage gain can be affected by frequency, depending on the components and circuitry involved. Some components, such as resistors, have a constant gain regardless of frequency. However, other components, such as capacitors and inductors, have a varying gain based on frequency. This can result in a frequency-dependent magnitude of voltage gain for the overall circuit.

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