Rod's Physics 3/4 Questions Thread

[Rod]

OK your unit conversion is slightly iffy. 1 A = 1000 mA
Multiplying both sides by 0.027, we get 0.027 A = 27 mA
That's a bit closer to the answer you have. In fact, if you didn't round 5.3/200=0.0265, you would end up with the same answer.

Just remember that 1 A = 1000 mA and that you'll always have more mA than A.

Ahhh I see, thank you!

[Rod]

Don't understand 'using voltage amplifiers' and 'clipping' AT ALL. I'm reading the theory, and I'm just lost. Going to try and watch some videos and animations now, does anyone have any advice/explanations?

'Using amplifiers' means when the transfer characteristic of an amplifier is turned into a Vin vs time graph or Vout vs time graph. And also some other stuff which I am having a hard time getting my head around.

Thanks guys

[hongkyho]

'Using amplifiers' means when the transfer characteristic of an amplifier is turned into a Vin vs time graph or Vout vs time graph. And also some other stuff which I am having a hard time getting my head around.

I'll give you a very simple explanation of amplifiers to give you a starting point so you can then ask questions.

In simple (and slightly inaccurate) terms, amplifiers increase the voltage of an electronic device. So, for example, when the sound coming from a speaker is too small, an amplifier can be used to increase this volume. How amplification works is not on the VCE course, so don't read into too much of the transistor stuff.

Therefore, you have three graphs: a Vin vs time graph (before the device is amplified), a Vout vs time graph (after the device is amplified) and a Vout vs Vin graph.

A Vout vs Vin graph gives you the corresponding values of Vout and Vin. Basically, what an input voltage amplifies to.

When you look at a Vout vs Vin graph, there are parts where the line has a constant gradient and other parts where it is horizontal. The part where the gradient is constant is called the linear region. This is the region where the amplifier actually amplifies the signal. Beyond this, you will reach a horizontal line, meaning that the output signal no loner increases/decreases proportionally to the input signal, hence it is no longer being amplified, hence it has been 'clipped'.

[Rod]

I'll give you a very simple explanation of amplifiers to give you a starting point so you can then ask questions.

In simple (and slightly inaccurate) terms, amplifiers increase the voltage of an electronic device. So, for example, when the sound coming from a speaker is too small, an amplifier can be used to increase this volume. How amplification works is not on the VCE course, so don't read into too much of the transistor stuff.

Therefore, you have three graphs: a Vin vs time graph (before the device is amplified), a Vout vs time graph (after the device is amplified) and a Vout vs Vin graph.

A Vout vs Vin graph gives you the corresponding values of Vout and Vin. Basically, what an input voltage amplifies to.

When you look at a Vout vs Vin graph, there are parts where the line has a constant gradient and other parts where it is horizontal. The part where the gradient is constant is called the linear region. This is the region where the amplifier actually amplifies the signal. Beyond this, you will reach a horizontal line, meaning that the output signal no loner increases/decreases proportionally to the input signal, hence it is no longer being amplified, hence it has been 'clipped'.

Okay awesome, understood. But what is the point of a Vin vs time graph and a Vout vs time graph? Why do we have to learn them? I understand why we learn a vout vs vin graph, because we can find out whether it is an inverted or non-inverted amplifier, we can figure out what the signal is being amplified (like how you said) and also we can find out the gain. And I still don't understand what clipping is.

Thank you

[Rod]

Here  is your really useful guide that I have been using;

 Amplifier circuits (which are based on transistors) convert a small (possibly time-varying) voltage into a larger, proportional (possibly time-varying) voltage. - understood
 There is a linear range, in which the amplifier circuit can amplify the voltage proportionately. Outside of this range, clipping or distortion occurs, where the output voltage is not proportional to the input voltage.
 Biasing a transistor circuit means setting it up so that when there is no AC input, the output voltage is the value in the middle of its possible output range. This allows for the greatest possible variation in Vin, without distortion occurring. This mid-point is called the quiescent point or the bias voltage.
 Voltage gain is the factor by which the input voltage is amplified to obtain the output voltage:
Av = ΔVout/ ΔVin = gradient of the ΔVout vs. ΔVin graph.
 If the voltage gain is negative, then the amplifier is described as inverting.
 If the voltage gain is positive, then the amplifier is described as non-inverting.

The bald stuff is exactly the theory that I'm having trouble with.

[hongkyho]

But what is the point of a Vin vs time graph and a Vout vs time graph? Why do we have to learn them?

You don't have to learn them, but they complete each other. These graphs come in three (Vin vs time, Vout vs Vin and Vout vs time) for a certain situation. What VCAA and VCE physics like to do is give you two to work out the third. They have little education value beyond this but they are there to complete the picture. I mean, an amplifier is useless when you have nothing to amplify.

[hongkyho]

And I still don't understand what clipping is.

Okay, let's have a look at this graph:


Remember that the x-axis always represents the independent variable, i.e. the variable that we are changing.

In this case, if we change the input (or original signal) to any value between -40 and 40mV, the amplified voltage will decrease (since this is an inverting amplifier) by similar increments, i.e. 10V in the output voltage for every 40mV in the input voltage.

Once we get to 40mV, and increase the input voltage beyond that, the output voltage no longer decreases by this increment. Hence, it is no longer being amplified. This means that up to this point (40mV) the amplified voltage had changed accordingly. Beyond this point, the amplified voltage will no longer change as it is stuck at -10V even though the input may be 100mV.

[hongkyho]

Biasing a transistor circuit means setting it up so that when there is no AC input, the output voltage is the value in the middle of its possible output range. This allows for the greatest possible variation in Vin, without distortion occurring. This mid-point is called the quiescent point or the bias voltage.

I'm not sure if this is exactly in the study design, but I put it in because it was in one of my school SACs.

This point refers to the amplification of an AC signal which, as you know, is typically a sinusoidal graph. This mean that the graph varies in the positive and negative direction by the same amount from the average voltage (typically 0V)

When amplifying an AC signal, you want to centre this average voltage in the centre of the linear region. Why? So that you hit non-linear range at both ends of the graph at the same time. Using the same example above, you would want the average voltage to be exactly 0V. This is because there is the same amount of voltage that the input voltage can vary by before you hit the non-linear range. In this case, you can both increase and decrease the input voltage by 40mV before you hit the non-linear range.

If your Vout vs Vin graph was something like this:
(I don't think this is an amplification graph but just using it as an example)
you would want the average voltage of your input signal to be 2.5V, so that it can vary in both direction by the same amount (1.5V) before clipping. In order to achieve this with your typical sinusoidal AC voltage (where the average is 0V), you would have to add 2.5V to the signal so the whole Vin vs time graph is moved by by 2.5V.

[Rod]

I'm not sure if this is exactly in the study design, but I put it in because it was in one of my school SACs.

This point refers to the amplification of an AC signal which, as you know, is typically a sinusoidal graph. This mean that the graph varies in the positive and negative direction by the same amount from the average voltage (typically 0V)

When amplifying an AC signal, you want to centre this average voltage in the centre of the linear region. Why? So that you hit non-linear range at both ends of the graph at the same time. Using the same example above, you would want the average voltage to be exactly 0V. This is because there is the same amount of voltage that the input voltage can vary by before you hit the non-linear range. In this case, you can both increase and decrease the input voltage by 40mV before you hit the non-linear range.

If your Vout vs Vin graph was something like this:
(Image removed from quote.) (I don't think this is an amplification graph but just using it as an example)
you would want the average voltage of your input signal to be 2.5V, so that it can vary in both direction by the same amount (1.5V) before clipping. In order to achieve this with your typical sinusoidal AC voltage (where the average is 0V), you would have to add 2.5V to the signal so the whole Vin vs time graph is moved by by 2.5V.

Thank you Hong! All that really helped!

PS - I absolutely hate you dp, I'm a full blooded collingwood supporter, I've been raised to hate anything that is Carlton so I get this innate feeling of anger whenever I see your profile ! Hope Daisy snaps his ankle!

Rod

[Rod]

Hey guys, just a question here....

So when I do circuit analysis questions this is what I do; I break up all the resistors in the circuit into one, this one resistor gives me the effective resistance. I can then find the total current of the circuit using the effective resistant and voltage supplied by the battery. So, here is my confusion, when I trace BACK to find the voltage drops of each resistor, I get different values. Despite these values being correct, I am confused why. Since year 10, we have learnt that the voltage drop of resistors in a parallel circuit will always be the same according to Kirchoff. So why are we starting to ditch that rule, and why are we starting to figure out that each resistor has a different voltage drop. Am I missing something? Am I thinking the wrong way? Am I using the rule incorrectly?

Once again, this is more of a confusion question, I'm getting the questions right just confused with the answers I'm getting.

Thanks again

[Nato]

Hey guys, just a question here....

So when I do circuit analysis questions this is what I do; I break up all the resistors in the circuit into one, this one resistor gives me the effective resistance. I can then find the total current of the circuit using the effective resistant and voltage supplied by the battery. So, here is my confusion, when I trace BACK to find the voltage drops of each resistor, I get different values. Despite these values being correct, I am confused why. Since year 10, we have learnt that the voltage drop of resistors in a parallel circuit will always be the same according to Kirchoff. So why are we starting to ditch that rule, and why are we starting to figure out that each resistor has a different voltage drop. Am I missing something? Am I thinking the wrong way? Am I using the rule incorrectly?

Once again, this is more of a confusion question, I'm getting the questions right just confused with the answers I'm getting.

Thanks again

Yeah, according to Kirchhoff the voltage drop of resistors in parallel will be the same.
Can I ask, when you are subbing these values in to find the voltage drop of the individual resistors (where you say you are getting different values) are you using the total current in the circuit, or the current through that individual resistor?

[SocialRhubarb]

Voltage drop across parallel resistors will be the same.

Voltage drop across resistors in series can vary.

[Rod]

Yeah, according to Kirchhoff the voltage drop of resistors in parallel will be the same.
Can I ask, when you are subbing these values in to find the voltage drop of the individual resistors (where you say you are getting different values) are you using the total current in the circuit, or the current through that individual resistor?

I use the total current .

[Nato]

I use the total current .

Alright, I suggest you use the current going through the individual resistor. Because by using the total current, you can't get the actual voltage drop of that resistor, because the TOTAL current isn't actually going through the individual resistor.
does that make sense? haha

[Rod]

Alright, I suggest you use the current going through the individual resistor. Because by using the total current, you can't get the actual voltage drop of that resistor, because the TOTAL current isn't actually going through the individual resistor.
does that make sense? haha

Sorry for not being clear! I don't use the total current for the individual resistor, but the resistors broken down into one! So for example, when I do a circuit question I break down the circuit until I have one resistor. With that I can find the total current. Then I back track until the last stage in which the resistors are in series, so I use the total current to figure out the voltage drop of each 'broken down' resistor. Then, after finding the voltage drops, I back track again to the normal circuit I was given by the question, the voltage drop values do not change because resistors that were previously in parallel would have the same voltage because of Kirchoff's laws. And the answers I have been getting are correct so I think my method is working. I'm just thinking a bit more and confusing myself, and want to know more about why this is happening.

Rod