Thx Papabravo. I am seeing the values in the error log. Been messing around trying to get them to.meas TRAN acvolts RMS V(Vout)
.meas TRAN acamps RMS I(RL)
where TRAN tells which analysis you are interested in measuring, acvolts and acamps are "dummy names" for the results RMS is the function you want and V(Vout) and I(RL) are the things to be measured.
Other functions you could specify would be AVG, MAX, MIN, PP, INTEG. These measurements will be performed over the range of the abscissa unless further restricted by additional qualifiers.
View attachment 296306
Results of the .meas statement are in the SPICE Error Log. The "dummy name" is replicated in all lower case in the error log and was pasted back into the schematic (.asc) file.
You can place a behavioral voltage or current source on the schematic and write an expression for the thing you want to display. Generally speaking, voltage sources are easier to work with than current sources. You could also create functional blocks (a .subckt) that would compute RMS values for you.Thx Papabravo. I am seeing the values in the error log. Been messing around trying to get them to
display on the schematic but no luck so far.
Thx AG.Your 20 number for "tran" showed too many waves. I changed it to 92 to clearly see if there is distortion.
On the left display I show the collector DC output with peak input and output voltages, 5dB per graticule.
On the right display I show the frequency response to 10kHz at the output.
I have done a few hand calculations for the ac currents.Thx AG.
Correction. My irl agrees with Spice but my irc does not. Can you please tell me where I went wrong.I have done a few hand calculations for the ac currents.
irl = Vout/RL = 0.05297/10K = 5.30 μA
irc = (RL/RC) * irl = (10K/4.7K) * 5.3 μA = 11.27 μA
ic = irc + irl = 11.27 μA + 5.30 μA = 16.57 μA
ie = (ic/βac) * (βac+1)=(16.57/300)*(300+1)=16.63μA
ib = ie - ic = 16.63-16.57=6 μA
My irl agrees with Spice but my ic does not. Can you please tell me where I went wrong.
Maybe you could look at my calc and see where the problem is. The calc is based on the fact that there is an attenuation ofMaybe the error is because the input signal source has a series resistance of 1k ohms which reduces the actual input to 8mV peak instead of 10mV peak.
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