2
Use a PWL or PULSE source. In addition to what ThePhoton said, you can change the graph to an x-y source by clicking on the time axis and changing the time axis to a signal.
2
Don't make the source a DC source.
With the DC source, LTSpice is assuming the voltage has been 10 V since \$t=-\infty\$ and all transients have settled before the simulation starts.
Make it a transient source that turns from 0 V to 10 V a few ms after the simulation starts.
2
Add all of the traces you want manually to the plot window, then click the top bar of the plot window to activate it. Go to "File->Save Plot Settings As" and you can save the current plotting configuration. Next time you open your simulation, run it, then click the top bar of the plot window to activate it. Go to "Plot Settings->Open Plot Settings File", ...
2
LTspice always uses db(1V) for the y-axis of the ac simulation results. So, since you actually used 0.01V as the input signal amplitude the actual gain of the circuit is 100 times greater than the output dB value. If you converted -16.4dB to a gain of 0.15 and you expected a gain of 15 then your simulation is actually spot on.
1
All parameters need curly braces around them, this lets the compiler know that it's a parameter. Another tip is to check the netlist by going to view->spice netlist to see what spice code the graphical interface is 'compiling'
The .param directive allows the creation of user-defined variables.
This is useful for associating a name with a value for the ...
1
Here's an envelope detector:-
What you have is a diode on the input of an op amp which has bias current of 10fA (0.00000000000001A). At this current the diode is practically a short circuit.
To work properly the envelope detector must have a load. The capacitor needs to be much larger than the diode's parasitic capacitance. The resistor needs to draw much ...
1
The period of your square wave is 100 us, but its duty cycle is 50%, so the input is at +5 V for only 50 us, not 100, before switching again.
This means it's only at +5 V for about 2.3 time constants rather than 4.6, and it's not able to come as close to fully charging the capacitor as it would have if kept at that voltage longer.
As another answer points ...
1
Let's take a simple voltage divider AND you are only interested in resistor tolerances, not supply tolerances (assume psu is ideal)
It is trivial to calculate the nominal, minimum and maximum
max: \$10\cdot \frac{10k*1.01}{(10k*1.01) + (10k*0.99)}\$ = 5.05V
nom: \$10\cdot \frac{10k}{10k + 10k}\$ = 5V
min:\$10\cdot \frac{10k*0.99}{(10k*1.01) + (10k*0.99)}\...
1
A brute force solution can be implemented using the following number system.
This described method will cover the minimum and maximum result.
For example, using a decimal number system, you can specify the tolerance by 10 values.
The number of digits is the number of components you want to vary.
ABCD
0000 all components have their minimum value
0001
0002
...
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