Hello, I know the basics of capacitive and inductive reactance, such as in an AC circuit pertaining to a capacitor, As the frequency increase the capacitor has less capacitive reactance and at a certain point will act as a short. and as frequency decreases the capacitive reactance increases showing more resistance to a circuit and at a certain point will act as an open circuit.

I like to learn by visualizing the circuit in use, let's say (speaking very broadly) that i have a circuit with a capacitor In an AC circuit and we were to look at the waveform on an oscilloscope and let's say the voltage lags the current by 90 degrees at a certain frequency. If i was to increase the frequency the capacitive reactance would decrease and the voltage would be lagging the current LESS than 90 degrees??

So in other words would the two waveforms come closer together?

Thanks

tjones9163:

Hello, I know the basics of capacitive and inductive reactance, such as in an AC circuit pertaining to a capacitor, As the frequency increase the capacitor has less capacitive reactance and at a certain point will act as a short. and as frequency decreases the capacitive reactance increases showing more resistance to a circuit and at a certain point will act as an open circuit.

I like to learn by visualizing the circuit in use, let's say (speaking very broadly) that i have a circuit with a capacitor In an AC circuit and we were to look at the waveform on an oscilloscope and let's say the voltage lags the current by 90 degrees at a certain frequency. If i was to increase the frequency the capacitive reactance would decrease and the voltage would be lagging the current LESS than 90 degrees??

So in other words would the two waveforms come closer together?

Thanks

That's the way I understand it, but I want to hear from someone who actually knows. Since I don't come by as often anymore I'm replying so this will be in my updated topics section.

You seem to be describing a pure capacitor. As such the phase angle doesn't change but the value of the current changes.

Thus at any point on the waveform, I = C dv/dt

Once you add a resistor in series with the capacitor then the current lead will start to change with frequency.

I'd have to go back and verify the details but the lead will be a phase angle determined by the vector sum of the 0° resistance and the 90° current.

Just use a divider made from a resistor and capacitor, drive with a signal generator and put a 'scope on the top of the divider and on the middle. At the breakpoint frequency the lead or lag will be 45 degrees. Depending on which way up the R and C are it will be high-pass or low pass.

Breakpoint frequency = 1 / (2πRC)

As above , ohms law applies to a resistor v=ir , or i = v/r. and that is independent of frequency. A capacitor ( ideal) doesn’t have a “resistance” except to AC and that is called its reactance ( which is still v/i) The law for a capacitor is i = C ( dv/dt) . Where dv/dt is the rate of change of the voltage . For a sine wave , dv/dt is the cosine - hence the phase difference of 90 ( or lag) of the current (i). That lag is independent of frequency , but the reactance is changing with frequency .

Google will give you some good explanations too

If you have a resistor and capacitor in series , then will see a changing phase /voltage with frequency as the reactance of the capacitor changes but the resistance of the resistor remains fixed .

You can simulate this sort of thing quite easily in LTSPICE just sitting at your PC.

For AC stuff, there is a bit of a learning curve but I would recommend it.

This: SINE(0 5 100 0 0 0 100)

just means 5 volt amplitude, 100Hz and we stop it after 100mS

This: .tran 20mS

We are doing a transient simulation for 20mS

The graph should be clear, based on the circuit below. I(R1) is just the current.

perfect thanks and i have downloaded LTSPICE and i will start practicing with it, seeing as i dont own an oscilloscope.

perfect thanks and i have downloaded LTSPICE and i will start practicing with it, seeing as i dont own an oscilloscope.

I’m attaching the LTspice (.asc) file that I used to create that example, although it is very simple to set up from that picture I posted.

Sometimes, if you are simulating a particular components, say a rail to rail opamp, you have to find the equivalent Analog Devices part and simulate that.

Also, I admit to being no guru, but I believe that it is worth investing some time to be able to do basic things.

RC_AC_behaviour.zip (394 Bytes)

If you google "Bode Plot" you will find a lot of information. You will see that for a simple RC circuit phase angle vs frequency has a very simple relationship. Once you understand this relationship you can draw amplitude and phase plots in less than a minute. Once you understand the relationship then you can use spice to reinforce your knowledge.

6v6gt:

perfect thanks and i have downloaded LTSPICE and i will start practicing with it, seeing as i dont own an oscilloscope.

I'm attaching the LTspice (.asc) file that I used to create that example, although it is very simple to set up from that picture I posted.

Sometimes, if you are simulating a particular components, say a rail to rail opamp, you have to find the equivalent Analog Devices part and simulate that.

Also, I admit to being no guru, but I believe that it is worth investing some time to be able to do basic things.

Hello again, i have spent the last couple of hours messing around with the circuit that you sent me and here is what i came up with pertaining the original post i made and tell me if i make a wrong statement here or approached it the wrong way.

I opened the project and decided to change one variable(hertz) about the circuit at a time. The first thing that i varied was the frequency[0-50v-(1HZ/10HZ/100HZ)-0-0-0-100] so in other words i tested the circuit each time with these variables and the first time=1HZ, second time=10HZ, third time=100HZ. When i run the simulator i measure the distance from the (voltageprobe2) and the current(IR1) at the peak of each wave. I get back a time in ms.

What i noticed is when i increase the frequency the time between the two peaks of the waves decreases which makes sense to me because capacitors off less reactance at higher frequencies.

Another observaton that i made was that as the frequency increased so does the current going through the circuit, which makes sense for the same reason above.

1)Was the way i approached it correct by changing one variable and also measuring between peak to

peak so i get a consistent basis?

2)Is my observatons correct?

Thanks again, I see how important this program is and i want to get really good at it.

OK. About LTspice it self, it sounds like you've got the hang of it. Setting a grid on the graph and maybe changing the interval to make it finer also helps, then you can take measurements on the 0V crossing instead of on peaks. Sparkfun has a series of tutorial videos which I found quite useful: Getting Started with LTspice - learn.sparkfun.com .

As to what you are simulating, you are interested not in the absolute times, but the phase shift which you have to derive from those times and the frequency (e.g. a 90 degree phase shift at 100Hz is 2.5mS) so you can see the answer the original question:

If i was to increase the frequency the capacitive reactance would decrease and the voltage would be lagging the current LESS than 90 degrees??

You could then also go on to test the assertions regarding phase shift in posts #3 and #4. I'd also be curious to see what results you get.

Thanks again, what is the equation that you found that 90 degree phase shift?

when i typed into google- phase shift equation this popped up

"The phase shift equation is ps = 360 * td / p, where ps is the phase shift in degrees, td is the time difference between waves and p is the wave period. Continuing the example, 360 * -0.001 / 0.01 gives a phase shift of -36 degrees."

But i believe how you solved it is different?

So your calculation says the 1mS (at 100Hz) is a 36 degree phase shift, which is correct.

In my example, I said that a 90 degree phase shift is 2.5mS at 100Hz.

Using your calculation, it gives the same answer:

ps = 360 * 2.5 / 10 = 90 (where 2.5 (mS) is the time shift, 10 (mS) is the period of a 100Hz signal and 90 (degrees) is the phase shift.

But, yes, I did it like this.

A 360 degree phase shift is as if you shifted the signal by its period. A 100Hz signal has a period of 10mS. So shifting it 360 degrees means a 10mS time shift. So shifting it 90 degrees gives a 90/360 part of the period, that is 2.5mS.

6v6gt:

So your calculation says the 1mS (at 100Hz) is a 36 degree phase shift, which is correct.In my example, I said that a 90 degree phase shift is 2.5mS at 100Hz.

Using your calculation, it gives the same answer:

ps = 360 * 2.5 / 10 = 90 (where 2.5 (mS) is the time shift, 10 (mS) is the period of a 100Hz signal and 90 (degrees) is the phase shift.But, yes, I did it like this.

A 360 degree phase shift is as if you shifted the signal by its period. A 100Hz signal has a period of 10mS. So shifting it 360 degrees means a 10mS time shift. So shifting it 90 degrees gives a 90/360 part of the period, that is 2.5mS.

Wow that was a perfect explanation, I am also researching SOH CAH TOA which is the relationship between sine cosine and tangent.

Thanks again for the explanantions.

Maybe this also helps to visualise it, especially the creation of a sine wave, and at least explain where the 360 degrees comes from.