I'm learning more frequency domain analysis and LTspice tools, and I made this circuit and so far found some (jw) eq to get the correct branch currents.

But what foes it mean, for the current through the 1k resistor (green trace) to level off to -60dB , what does the -60dB mean, I can see the voltage at the node between 22n and 1k approaches the full 20V, and so it's amplitude in the bode plot (blue trace) , approaches 0dB. So why doesn't the current magnitude approach the branch current ratio of 2k||1k of 1:2, 10mA:20mA, so 2/3 of total, what's -60dB mean ?

the cursor box disappeared when taking the snap, so that's NYC wall paper in the pic too

Also in transient mode, how to I easily measure the phase and time between 2 traces ?

 

Decibels give you a relationship between your input and output. That relationship is often called a gain, which we can call A.

If you are using voltage the relationship is:

A(v) = Vout / Vin

Notice it will be unitless.

To convert your gain to decibels you use the following equation

A(dB) = 20 log (Vout / Vin)

To this we will attach the unit dB to indicate we are using the logarithmic scale.

If we are doing our calculations with power, in Watts, the following modification is needed.

A(p) = Pout / Pin
A(dB) = 10 log (Pout / Pin)

 

what does the -60dB mean

VAC = 1, so in 1kΩ the current tends towards 1mA. Power = 1 x 10^(-3) W.
dB = 20 x Log(power) = 20 x Log 10^-3 = 20 x -3 = -60dB.

Edit: '20' should read '10'. See post #5.

 

Yeah I was wondering that too, I tried setting 1 and 2 ohms and something else last night, trying move the 20mA closer to the number 20.

Just now I was about to calculate that in log terms, thanks.

I ordered a hardcover EE textbook, it should have been here a month ago. I guess i don't know the finer details of using dB for this

 

VAC = 1, so in 1kΩ the current tends towards 1mA. Power = 1 x 10^(-3) W.
dB = 20 x Log(power) = 20 x Log 10^-3 = 20 x -3 = -60dB.

With decibels we must do a comparison between two values, often comparing the input and output.

And for power calculations we must use 10 x log (power out/power in).

In the above example, since no powerIn is given, the denominator acts as a 1, which is 1 watt. 1 mW is then -30 dB from 1 W.

 

Yes. My bad. So I'm not sure how the -60dB figure is obtained.

 

If there is 1 volt on the input of the circuit, there must be 1 mV on the output.

A(dB) = 20 × log(0.001 mV / 1 V) = -60 dB

 

If there is 1 volt on the input of the circuit, there must be 1 mV on the output.

How is that possible with the post #1 circuit?

 

I am not quite sure exactly what all of the traces are in the graph since some elements of the graph appear to have been cut off, but the same thing happens with current as well.

Consider the output point to be measured across R for the voltage and through R for the current. Those voltages and currents are compared to the voltages and currents of the main source. Rf will not affect the voltage across R but will greatly affect the current through the source.

Therefore, the voltage across R will be able to achieve the same voltage as the source at high frequencies. achieving the same output as input is a 0 dB change.

Since Rf is in the picture the current through R will never achieve the same value as the source. In this circuit it can only achieve a magnitude that is 60 dB below that of the source, or -60 dB.

 

I'll try again.
The voltage across the 1k resistor is asymptotic towards 1V as the frequency rises. Power = V^2/R = 1/1k W = 10^-3 W
Power in dB = 20 x Log(power) = 20 x -3 = -60dB.

 

When you use power the multiplying factor is 10 instead of 20.

Let's look at the voltages. the input voltage is 1 Vrms. The output voltage tends toward 1 Vrms. If we calculate the dB where this is the case we do the following:

Av(dB) = 20 log (Vout / Vin) = 20 log (1 Vrms / 1 Vrms) = 20 log 1 = 20 × 0 = 0 dB

If you look at the blue curve it shows 0 dB for higher frequencies. 0 dB suggests there is no magnitude change from the input to the output.

Remember, dB is a comparison of two values..

 

When you use power the multiplying factor is 10 instead of 20.

You're right. Brain seems stuck on 20 .

 

@DarthVolta

Decibels are hard to think about until you realize the concept of the notation. Decibels are about energy transfer. This is why that figure is used in music, electronics, telephony, and anywhere that energy transfer is expressed in a log scale.

d (deci) is 1/10th of a Bell. In honor of A.G.Bell. dB is a ratio for power (1:10). Therefore:

0dB = 1 with zero 0's or 1
10dB = 1 with one 0 or 10
20dB = 1 with two 0's or 100
...
60dB = 1 with six 0's or 1,000,000.

so -60dB is very small.

Hope that helps