New to electronics and struggling

Thread Starter

charliedurrant

Joined Jan 6, 2012
18
Hi All,

I'm old by students' standards having been a programmer for 21 years I suddenly woke up and wanted to try electronics. I've started to do up an old car and have fitted a mappable ignition system and wanted to learn how things work.

I got a book from the library the other day 'Electronics made simple' by Ian Sinclair and amd finding a good but hard read. Almost every page has to be read twice bu that's part of the fun.

I wanted to check the circuits in the book and thus being a programmer I downloaded Multisim as components are sparse at the moment (that will come in time). I thought my understanding was correct about inductors until I tested a very simple circuit.

1 resister
1 inductor
12v power supply (AC)
scope across the resister
scope across the inductor.


I get a wave for the reading across the inductor but the scope shows vertical lines for the reading across the resister? Oh dear.....am I being stupid?


I was expecting the 2 voltage readings to be 90 degress out of phase as the book seems to say that the inductor will pull back the potential difference due to inductance.





Thanks in advance

Charlie
 
Last edited:

Adjuster

Joined Dec 26, 2010
2,148
Almost all of the voltage will be dropped across the resistor. The oscilloscope reading is therefore off the scale most of the time, apart from some apparently vertical lines where the trace rapidly crosses the screen.

You may be able to get a better display by increasing the scale voltage per division, but there may not be sufficient range available.

Note that for a real oscilloscope, depending on its specification inputs of this level may well be dangerous. (Some models might well handle 120V rms all right, but generally connecting mains to a scope is dangerous territory).

AC mains in my country is 230V, which is a lot worse.

Note also, the common lines of many real oscilloscope inputs are connected together, and also to mains ground, so that the kind of setup you have here would create a short-circuit. Simulated oscilloscopes often behave as if they had effective differential inputs, but this is not usual in practice.
 
Last edited:

Thread Starter

charliedurrant

Joined Jan 6, 2012
18
Almost all of the voltage will be dropped across the resistor. The oscilloscope reading is therefore off the scale most of the time, apart from some apparently vertical lines where the trace rapidly crosses the screen.

You may be able to get a better display by increasing the scale voltage per division, but there may not be sufficient range available.

Note that for a real oscilloscope, depending on its specification inputs of this level may well be dangerous. (Some models might well handle 120V rms all right, but generally connecting mains to a scope is dangerous territory).

AC mains in my country is 230V, which is a lot worse.

Note also, the common lines of many real oscilloscope inputs are connected together, and also to mains ground, so that the kind of setup you have here would create a short-circuit. Simulated oscilloscopes often behave as if they had effective differential inputs, but this is not usual in practice.
Much more to learn! I've adjusted:

1) the resister (much lower)
2) the power supply voltage (down to 12v)
3) the inductor (small change)

and I now have two waves 90 degress out of sync!




Much appreciated...

Just so you don't think I'm a total dim wit, none of the book examples quote any values as it is just explaining the theory.

Charlie

p.s. note, I read the manual on the virtual scope and it did mention that in the real world it would be grounded.
 

joeyd999

Joined Jun 6, 2011
4,331
...and I now have two waves 90 degress out of sync!
That would be 'out of phase' for the initiated. 'Out of sync' means something entirely different.

By the way, your original images clearly showed two signals 90 degrees out of phase. With experience, you'll be able to recognize such things on sight.
 

Thread Starter

charliedurrant

Joined Jan 6, 2012
18
That would be 'out of phase' for the initiated. 'Out of sync' means something entirely different.

By the way, your original images clearly showed two signals 90 degrees out of phase. With experience, you'll be able to recognize such things on sight.
Yes, teminology is important, approx 90 degress out of phase.

I am struggling to see why the original image's plots are out of phase. They must be as all I did was change the resister, inductor and input voltage but the near vertical lines seem to cross at the +- peaks and not 90 degrees before?

Charlie
 

joeyd999

Joined Jun 6, 2011
4,331
Yes, teminology is important, approx 90 degress out of phase.

I am struggling to see why the original image's plots are out of phase. They must be as all I did was change the resister, inductor and input voltage but the near vertical lines seem to cross at the +- peaks and not 90 degrees before?

Charlie
When two signals are 90 degrees out of phase, the dv/dt of one is the highest when the dv/dt of the other is 0. Look at where the peaks are when the vertical line crosses the 0V point.

There is a little bit of a (optical) resolution issue here, too, both with the software you are using, and the picture.

Even so, close enough for gov't work. :)
 

crutschow

Joined Mar 14, 2008
24,380
I am struggling to see why the original image's plots are out of phase. They must be as all I did was change the resister, inductor and input voltage but the near vertical lines seem to cross at the +- peaks and not 90 degrees before?
If the signals were in-phase, the vertical lines would cross the other sine-wave at 0V. When the veritical lines cross the sine-wave at it's peak, then the phase is 90°.

Just reduce the sensitivity of the oscilloscope input in you first simulation and you will see it plainly.
 

Thread Starter

charliedurrant

Joined Jan 6, 2012
18
If the signals were in-phase, the vertical lines would cross the other sine-wave at 0V. When the veritical lines cross the sine-wave at it's peak, then the phase is 90°.

Just reduce the sensitivity of the oscilloscope input in you first simulation and you will see it plainly.

Yes! Sorry I was being daft, it is zero when the other wave is at its peak....

Charlie
 
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