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Question about cadance analoglib PORT

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visionbjp

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Im designing a current mode amp
and the port is giving me a headache

the problem is the amplitude 1 (dBm)
--> which power is this talking about

I tried a simple test using a port and a resistor connected

port_____1_____RL
l_____________l
l
gnd

here are the results
--------------------------------------------------------
Port resistor = 200 RL = 100 amplitude 1 = -30dBm
vpk at 1 = 13.298m ipk at 1 = 132.98u

nothing matches to have a power of -30dBm(1uW)
-------------------------------------------------------
Port resistor = 200 RL = 200 amplitude 1 = -30dBm
vpk at 1 = 19.9816m ipk at 1 = 99.908u

power at both port resistor and RL are 1uW
-------------------------------------------------------
Port resistor = 200 RL = 300 amplitude 1 = -30dBm
vpk at 1 = 23.9718m ipk at 1 = 79.906u

nothing matches to have a power of -30dBm(1uW)
------------------------------------------------------
HOW is this voltage or current at the output of the port decided when you give ampiltude 1 (dBm) ??.........
 

visionbjp said:
--> which power is this talking about
Click to expand...

Code: 
spectre -h port

"... However, the amplitude of the sine wave in the transient and PAC analyses can alternatively be specified as the power in dBm delivered by the port when terminated with the reference resistance. ..."

visionbjp said:
HOW is this voltage or current at the output of the port decided when you give ampiltude 1 (dBm) ??
Click to expand...

\[
V_{internal, amplitude} = \sqrt{8*R * 10^{\frac{P_{dBm}}{10}-3}}
\]

\[
V_{out, amplitude} = V_{internal, amplitude}*\frac{RL}{R+RL}
\]


Examples:

Amplitude = -30 dBm
Resistance = 200 Ohm

V(internal, amplitude) = sqrt(8*200*10^(-30/10-3)) = 40 mV

1) RL = 100

V(out, amplitude) = (40 mV)*100/(100+200) = 13.3 mV
I(out, amplitude) = (13.3 mV)/(100 Ohm) = 133 uA

2) RL = 200

V(out, amplitude) = (40 mV)*200/(200+200) = 20 mV
I(out, amplitude) = (20 mV)/(200 Ohm) = 100 uA

3) RL = 300

V(out, amplitude) = (40 mV)*300/(300+200) = 24 mV
I(out, amplitude) = (24 mV)/(300 Ohm) = 80 uA
 
Last edited by a moderator:
visionbjp said:
HOW is this voltage or current at the output of the port decided when you give ampiltude 1 (dBm) ??.........
Click to expand...
v = √(P*R) ; i = √(P/R) ; P=power-amplitude ; R = resistance-value
 

Where does the "8" come from, when you talk about Vinterner ( v^2/8R )

Added after 1 minutes:

dedalus said:
visionbjp said:
--> which power is this talking about
Click to expand...

Code: 
spectre -h port

"... However, the amplitude of the sine wave in the transient and PAC analyses can alternatively be specified as the power in dBm delivered by the port when terminated with the reference resistance. ..."

visionbjp said:
HOW is this voltage or current at the output of the port decided when you give ampiltude 1 (dBm) ??
Click to expand...

\[
V_{internal, amplitude} = \sqrt{8*R * 10^{\frac{P_{dBm}}{10}-3}}
\]

\[
V_{out, amplitude} = V_{internal, amplitude}*\frac{RL}{R+RL}
\]


Examples:

Amplitude = -30 dBm
Resistance = 200 Ohm

V(internal, amplitude) = sqrt(8*200*10^(-30/10-3)) = 40 mV

1) RL = 100

V(out, amplitude) = (40 mV)*100/(100+200) = 13.3 mV
I(out, amplitude) = (13.3 mV)/(100 Ohm) = 133 uA

2) RL = 200

V(out, amplitude) = (40 mV)*200/(200+200) = 20 mV
I(out, amplitude) = (20 mV)/(200 Ohm) = 100 uA

3) RL = 300

V(out, amplitude) = (40 mV)*300/(300+200) = 24 mV
I(out, amplitude) = (24 mV)/(300 Ohm) = 80 uA
Click to expand...

Where does the "8" come from, when you talk about Vinterner ( v^2/8R )
 
Last edited by a moderator:

visionbjp said:
Where does the "8" come from, when you talk about Vinterner ( v^2/8R )
Click to expand...

\[
\overline{P_{Out}} = \frac{V_{Out,RMS}^2}{RL} \Leftrightarrow V_{Out,RMS} = \sqrt{RL*\overline{P_{Out}}}
\]

\[
V_{Out,Amplitude} = \sqrt{2}*V_{Out,RMS} = \sqrt{2*RL*\overline{P_{Out}}}
\]

Internal resistance of port and load resistor form voltage divider:

\[
V_{Out} = V_{Internal}*\frac{RL}{R+RL}
\]

Output average power is specified for reference load (RL = R), thus

\[
V_{Internal,Amplitude} = 2*V_{Out,Amplitude} = \sqrt{4}*V_{Out,Amplitude} = \sqrt{8*R*\overline{P_{Out}}}
\]
 
Last edited by a moderator:
So here you are saying that Pout = PdBm(Port's amplitude(dBm))
I don't get it......
 

visionbjp said:
So here you are saying that Pout = PdBm(Port's amplitude(dBm))
I don't get it......
Click to expand...

No, \[\overline{P_{Out}}\] is power in Watts.

From definition of dBm:

\[P_W = 10^{\frac{PdBm}{10}}-3\]
 
Last edited by a moderator:
Hi,

your explanations helped me a lot to understand the PORT-device. But I still can't understand my simulation-results in the case of a complex RL.
For example: In the example calculations above, what would the V(out, amplitude) be if RL=200 and XL=-50 (200ohm resistor in series with a 135fF capacity)?
In which way do I have to factor the imaginary part in?

PS: Sorry for my bad English!
 

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