All About Circuits Forum Pspice graph with DB() function and without DB()
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#1
04-22-2008, 07:46 PM
 Jaywin New Member Join Date: Apr 2008 Posts: 9
Pspice graph with DB() function and without DB()

When put the DB() function the x-axis become log-scale is it?
And for the red circled region is it caused by the delay of capacitor?
Wondering why the voltage raise in the red-circled region...

And for the voltage gain for 2nd order low pass, why after certain frequency the gain raise again for the below graph(Left)?

Why the graph is different if we not using DB() function?
upper is using DB() lower is without using DB().

Thanks, if anyone could to explain to me...thz vry much.
#2
04-22-2008, 07:51 PM
 Caveman Senior Member Join Date: Apr 2008 Location: Austin, TX Posts: 471

The voltage is raised in the red region because that is what the circuit does. Look really close at the dB graph and you will see that it is just slightly over 0dB in the same area. That is because 1.1V is 0.87dB.
All dB scale is doing is converting the output to dB.

log_out = 20*log(linear_in)

Does that make sense?
#3
04-22-2008, 08:04 PM
 Jaywin New Member Join Date: Apr 2008 Posts: 9

The circuit is like this same as the previous circuit but just the voltage pointer shifted to the V(R1) node.
Ya, i got what u mean for the dB. Meaning putting dB the axis is change to the log-scale isn it? And observe that the dB graph for the Voltage gain is not started from 0db is this correct?
#4
04-22-2008, 09:08 PM
 Caveman Senior Member Join Date: Apr 2008 Location: Austin, TX Posts: 471

Yes, you have a dc gain of 1.58V/V which is approximately equal to 4dB. If you were to remove the 6k resistor and short the 3.5k resistor, you would have 0dB at dc.
#5
04-23-2008, 05:16 AM
 Jaywin New Member Join Date: Apr 2008 Posts: 9

Anyone able to explain for the voltage gain of 2nd order low pass, why the frequency increase until certain value the gain raise again for the below graph(Left)?

 Tags function, graph, pspice

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