voltage gain in decibels

Thread Starter

djstar

Joined Jan 26, 2008
39
hi, ive just started a gcse in electronics and im still quite new to the subject. ive been given the question

Q2) A circuit consists of a 1.5v battery and connected to it, from the positive terminal, is a 15 Ω resistor, followed by 2 resistors in parallel, whose junction back to the battery negative. Find the voltage gain or loss in decibels across the parallel network.

im not overlly to sure on what to do, could some one point me in the right direction, but please put it in basic layman terms, many thanks , liam
 

Papabravo

Joined Feb 24, 2006
13,931
You need to start with a ratio. That is a fancy word for one number divided by another. In our case there are two voltages, and you need to form a ratio by dividing one number by the other.

If both numbers are the same the ratio will be one.
If one of the numbers is twice as large as the other then the ratio will be either 2 or 1/2 depending on which number is in the numerator, and which is in the denominator.

Once you have a ratio, you need to take the logarithm of the ratio. If you have a calculator you can get this number with minimal mental effort. It is just a matter of punching a key.

To wrap things up you multiply the result by the number 20. That's it, that's all there is, and there isn't any more. Why do we need this you may well ask. The answer is that is helpful when comparing things that may be different by several orders of magnitude.

Some examples:
Rich (BB code):
20 * log(2/1)  = 6.02 ≈ 6 dB
20 * log(10/1) = 20 dB
20 * log(1,000,000/1) = 120 dB
 
20 * log(1/2) = -6.02 ≈ -6 dB
20 * log(1/10) = -20 dB
20 * log(1/1,000,000) = -120 dB

	
	
	
	



	



	










Rich (BB code):
Get the idea?
 

mik3

Joined Feb 4, 2008
4,843
Find the voltage (Vo) across the parallel resistors and then divide the battery voltage by Vo to get the gain of the circuit, which in this case is less than 1 and thus attenuates the signal rather than amplify it.

Gain=1.5/Vo

then to convert it to dB use this formula

gain in dB=20log(Gain)
 

hobbyist

Joined Aug 10, 2008
887
followed by 2 resistors in parallel, whose junction back to the battery negative. Find the voltage gain or loss in decibels across the parallel network.
Do we assume the parrallell resistors are 15 ohms also, if not what is the Vout?

Would the Av. be [1.5 x Rp / 1.5 x (15+Rp)] where Rp is parrallel resistors?
 
Last edited:

KL7AJ

Joined Nov 4, 2008
2,225
Whenever figuring out DB using voltage ratios, it's CRUCIAL that the impedances are the same. If using voltage ratios at two different impedances (for example the input and output voltages of a step-up transformer) you can show that the transformer has POWER GAIN....obviously an impossible situation!

Whenever humanly possible, always figure dB using power ratios....which also forces you to know the impedances you're working with.


Eric
 

Papabravo

Joined Feb 24, 2006
13,931
Are you saying that it is inappropriate to measue the gain of an inverting opamp circuit because the input and output impeadances are different? I don't think I would subscribe to that notion.
 

KL7AJ

Joined Nov 4, 2008
2,225
Are you saying that it is inappropriate to measue the gain of an inverting opamp circuit because the input and output impeadances are different? I don't think I would subscribe to that notion.
It's appropriate to measure voltage gain, as long as it's SPECIFIED as voltage gain. Once you start converting to DB, you MUST either match impedances, or use the power formula.

Eric
 

Papabravo

Joined Feb 24, 2006
13,931
So if you know the reference units then the ambiguity is removed?

For example dBmV is a voltage relative to 1 mV regardless of impedance. Right?

Also the use of 20 as a multiplying factor means you are comparing the ratio of the voltage squared, which with a constant impedance is proportional to the power.
 
Last edited:

KL7AJ

Joined Nov 4, 2008
2,225
So if you know the reference units then the ambiguity is removed?

For example dBmV is a voltage relative to 1 mV regardless of impedance. Right?

Also the use of 20 as a multiplying factor means you are comparing the ratio of the voltage squared, which with a constant impedance is proportional to the power.
That would be a correct usage.

Just as another example of why one needs to be specific...quite the opposite of the transformer example above....let's take the case of a voltage follower. A voltage follower has no voltage gain at all, but it has (in theory)INFINITE power gain. Assuming the input impedance is infinite (real world cases come pretty close!), it draws NO input power, but puts out a finite amount of power into its load resistance. Using the voltage formula for dB without consideration of the impedances, one would come up with the conclusion that the voltage follower has 0 db of gain...or perhaps less, when in reality the power gain is tremendous.

Hope this helps!

Eric
 

Thread Starter

djstar

Joined Jan 26, 2008
39
i think ive manged to work it out thanks to your replies . i got the loss as 0.333 which is - 9.5 decibels, does that sound any where near ?
 
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