Proteus 8, two signal simulation and test.

Thread Starter

AlexLPD

Joined Aug 22, 2010
60
Hi to you all, Im working In this curious project;

The problem I want to resolve is Im using the software correctly, or if the software can simulate adequate, the signals?

I like to generate and inject a square wave on a live wire (24/120VAC).

I research the internet and found this configuration, to be use on the emitter /receiver;
-They are two filters for attenuate the signal of the 60Hz on the Main line:



They work like intended, This is the Scope output;

The Yellow channel is connected to the Mains AC: at 20Vpdv and 12divisons = 240VAC (the AC signal source was bad configured)
The blue channel is connected at the out of the firs filter: at 4.2Vpd and 20mV = 84mV
And the pink channel is connected to the out of the second filter; even at 20mV we get no signal.

An enlargement;


So there is no doub the filters woks for the AC sine wave of 60hz.

Know I tried to make the "complete" simulation, using a signal generator to "inject" the 10Khz PWM signal;
In theory, the square signal must pase the two filters and travel on the Main Line.


PS. I have to put the signal generator on the right because, if I tried of join it with the Mains, simulation yields an error.

So, here is the sine wave;


At 12div and 10V per division = 120V.

Here is the signal on the chanel blue:


And finally here is the square singnal;


If we slow down;



At 10Div and 0.1mV we got a square signal of 10mV.

My question is, can be this signal able to be seen in the yellow channel?
It is well connected the circuit?

Thanks for the help and comments, Im a industrial engineer, but I play a lot with Arduino, not an Electronic, so bear with me.

Thanks.
-Alex.
 

AlbertHall

Joined Jun 4, 2014
12,396
The voltage generator in the simulator will maintain its stated voltage regardless of what happens so it will simply absorb the square wave with no effect on its voltage. This sytem relies on the fact that the mains has some non-zero impedance - i have no idea what value to expect. Add a resistor between the voltage source and the rest of the circuit, say 1 ohm.
 

Thread Starter

AlexLPD

Joined Aug 22, 2010
60
The voltage generator in the simulator will maintain its stated voltage regardless of what happens so it will simply absorb the square wave with no effect on its voltage. This sytem relies on the fact that the mains has some non-zero impedance - i have no idea what value to expect. Add a resistor between the voltage source and the rest of the circuit, say 1 ohm.
Hi Albert, thanks for the kind response, I made a lot of experiments, finally this is the setup I got;



The resistance is either 1 or 10 Ohms, first I will show the 1Ohm case;

The only problem I can see with this is the wave amplitude is ;
13 * 5 = 65VAC.

Yet the 10Khz wave is very clear;

11divisions *0.5 = 5.5V

And the filter output is;


4div at 1V per division = 4V.

So, this means the circuit is working ok, and the lost of VAC on the input is due the simultion?

Thanks.
-Alex.
 

Thread Starter

AlexLPD

Joined Aug 22, 2010
60
In the other hand If I put a 10Ohm resistor, the signal is much more small;
The principal sinewave is 110VAC, witch is smaller than in the Orginal. I belive the software is calculating a resistor divider.



This give me a smaller and weaker signal;


In this case, using the same settings as before the signal Peak to peak is around 1.8V,
and once it pass the filters is; 0.8Vpp.


So I got some questions, If I use the 1 oh resistor, the main voltage decrease but the signal strength is quite high, but If I use the 10ohm resistor the mains voltage seems more accurate, yet the signal is quite small.

I believe, in the first approach, yet I need some help to clarify, before I start to design the detector circuit.

Thanks in advance.
-Alex.
 

AlbertHall

Joined Jun 4, 2014
12,396
First you need some way to measure the actual line impedance at the modulation frequency then you can simulate the actual conditions.
You could do this by connecting a grounded signal generator via a resistor and two Y-rated capacitors (Y- rated capacitors are designed to be safely connected permanently across the mains). Then you can measure the generator output voltage and compare it to the voltage across the resistor to calculate the mains impedance.
 
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