OCR Gateway Physics P4P5P6 - 18/06/13

Mutleybm1996

A transistor is a tiny electrical switch - so think of it like a light switch but it's turned on or off depending on whether an electric current is supplied to the 'base' or not. When an electric current is supplied to the 'base', the transistor 'closes' (like a light switch closing and completing the circuit, turning on the light) and allows a larger current to flow through the 'collector' and the 'emitter'. In this case the transistor is supplied with a small current (due to the resistor in series with it) which allows the current to flow through the LED because it has 'closed'.

If that didn't help then maybe my description of it on my mindmap will:

http://gyazo.com/cd32ec1db00f02b7d4dc79e48afc06c9

Spec guys!:
For physics Do we have to know that it can be used as an amplifier?

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Reply 321

Knowing

1

Original post by benwalters1996

I believe it's the same? Just use the equation :smile:

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ImageUploadedByStudent Room1371397878.742899.jpg

So for example if they give you this diagram, does that mean voltage across R1 or R2?

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Reply 322

BP_Tranquility

5

Original post by andersson

A transistor is a tiny electrical switch - so think of it like a light switch but it's turned on or off depending on whether an electric current is supplied to the 'base' or not. When an electric current is supplied to the 'base', the transistor 'closes' (like a light switch closing and completing the circuit, turning on the light) and allows a larger current to flow through the 'collector' and the 'emitter'. In this case the transistor is supplied with a small current (due to the resistor in series with it) which allows the current to flow through the LED because it has 'closed'.

If that didn't help then maybe my description of it on my mindmap will:

http://gyazo.com/cd32ec1db00f02b7d4dc79e48afc06c9

Ah, it makes more sense now, thanks :smile:

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Reply 323

BP_Tranquility

5

Original post by benwalters1996

Spec guys!:
For physics Do we have to know that it can be used as an amplifier?

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No, there's nothing about amplifiers in the spec :smile:

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Reply 324

Mutleybm1996

19

Original post by Knowing

So for example if they give you this diagram, does that mean voltage across R1 or R2?

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In Electronics we're just taught that the Vout is 1/2 Vin if R1 and R2 are the same. Otherwise use Vout= Vin x (R2)/(R1+R2) and from there it's a piece of pie! ImageUploadedByStudent Room1371398111.301869.jpg

ImageUploadedByStudent Room1371398111.301869.jpg

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Reply 325

Mutleybm1996

19

Original post by BP_Tranquility

No, there's nothing about amplifiers in the spec :smile:

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P4/5 are easy....P6...the last few bits like motors etc are ANNOYING.

Don't quite get lenses in p5-any help would be appreciated!

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Reply 326

Knowing

1

Original post by benwalters1996

In Electronics we're just taught that the Vout is 1/2 Vin if R1 and R2 are the same. Otherwise use Vout= Vin x (R2)/(R1+R2) and from there it's a piece of pie! ImageUploadedByStudent Room1371398111.301869.jpg

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What I mean is, if they ask you "calculate V(out)", do we calculate V(out) at R1, or do we calculate V(out) at R2? If I know which resistor it's referring to I know how to work it out, but it doesn't tell you which resistor to work it out for?

Reply 327

Mutleybm1996

19

Original post by Knowing

What I mean is, if they ask you "calculate V(out)", do we calculate V(out) at R1, or do we calculate V(out) at R2? If I know which resistor it's referring to I know how to work it out, but it doesn't tell you which resistor to work it out for?

It's the same regardless? Isn't it? There's only one Vout.

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Reply 328

BP_Tranquility

5

Original post by Knowing

What I mean is, if they ask you "calculate V(out)", do we calculate V(out) at R1, or do we calculate V(out) at R2? If I know which resistor it's referring to I know how to work it out, but it doesn't tell you which resistor to work it out for?

If I understand correctly, the Vout you work out will be the same regardless of which resistor it's referring to because Vout measures potential difference which is the difference in voltage between the two resistors... :redface:

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Reply 329

Knowing

1

Original post by benwalters1996

It's the same regardless? Isn't it? There's only one Vout.

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There can be a V out for R1, and a V out for R2 surely?

Reply 330

Elcor

14

Original post by Knowing

There can be a V out for R1, and a V out for R2 surely?

The Vout is the voltage between the two resistors, if that's what you're asking

Reply 331

Mutleybm1996

19

Original post by Knowing

There can be a V out for R1, and a V out for R2 surely?

Nope. Vout is in the middle. Otherwise it would be input voltage or 0v

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Reply 332

Knowing

1

Oh ok, I kind of get it. So V(out) is basically the voltage that remains after passing through R1

Reply 333

Mutleybm1996

19

Original post by Knowing

Oh ok, I kind of get it. So V(out) is basically the voltage that remains after passing through R1

Exactly! And if it helps: ImageUploadedByStudent Room1371398950.718392.jpg

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Reply 334

BP_Tranquility

5

Original post by Knowing

There can be a V out for R1, and a V out for R2 surely?

If they don't mentioned potential divider in the question, or don't give you the diagram for it, they may be asking you to use the formula current=voltagexresistance ...? :redface:

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Reply 335

Knowing

1

Thanks! I get potential dividers now :biggrin:

Sorry, it's because I haven't been taught this lol XD
Thank you everyone :woo:

Reply 336

DeadUnicorn

9

I don't get this bit about LDRs and potential dividers: "A bright light causes low output voltage because the resistance of the LDR is lower than the fixed resistor." I thought that a bright light means decreased resistance so more voltage can flow? And also can someone please explain the whole thermistor/LDR as inputs for logic gates thing to me? I really can't seem to visualise this section D:

Reply 337

Knowing

1

Original post by DeadUnicorn

I don't get this bit about LDRs and potential dividers: "A bright light causes low output voltage because the resistance of the LDR is lower than the fixed resistor." I thought that a bright light means decreased resistance so more voltage can flow? And also can someone please explain the whole thermistor/LDR as inputs for logic gates thing to me? I really can't seem to visualise this section D:

I think this is it:

You replace R2 with the LDR. When it's in bright light, the resistance is lower, so R1's share of resistance increases. Therefore, more voltage is used up by R1, so there is lower voltage when it gets to R2/the LDR.

I haven't reached logic gates yet :s-smilie:

Reply 338

Elcor

14

Original post by DeadUnicorn

I don't get this bit about LDRs and potential dividers: "A bright light causes low output voltage because the resistance of the LDR is lower than the fixed resistor." I thought that a bright light means decreased resistance so more voltage can flow? And also can someone please explain the whole thermistor/LDR as inputs for logic gates thing to me? I really can't seem to visualise this section D:

You're right but if the LDR was the first resistor then the opposite effect happens (when it is light the Vout decreases) because the first resistor's resistance is large (relative to the second one), meaning there is a large voltage drop.

Edit: I just confused myself. Forget what I said.

(edited 10 years ago)

Reply 339

BP_Tranquility

5

Original post by DeadUnicorn

I don't get this bit about LDRs and potential dividers: "A bright light causes low output voltage because the resistance of the LDR is lower than the fixed resistor." I thought that a bright light means decreased resistance so more voltage can flow? And also can someone please explain the whole thermistor/LDR as inputs for logic gates thing to me? I really can't seem to visualise this section D:

There are two main concepts we have to keep in mind:
When R1 is greater than R2 in a potential divider circuit, then output is 0 volts. If R1 is less than R2, then output voltage is nearly the same as the input voltage.

Essentially, in a potential divider circuit, you replace R2 with a thermistor so when there is a high temperature, then the resistance of the thermistor falls. Therefore, because the resistance of the thermistor is low, the resistance of R1 is higher so there is no output voltage. Similarly, if there is a low temperature, the resistance of the thermistor is high and because of this, the resistance of R1 is lower which means there is an output voltage (roughly equal to the input voltage).

I'll use an example for the second question. This is very similar to the above example where you have R2 replaced with an LDR for instance. There is an AND gate where there is an input of a normal current but for the second input, a potential divider circuit in connected. This potential divider circuit has had its R2 replaced with an LDR so there is only a 'high' signal when the light intensity is low (because a low light intensity means that resistance is high so like before, the resistance of R1 is lower than that of the LDR, so there's a current at the second input of the AND gate). Hence, because there's a current at the first input as well as the second input, the AND gate also has an output and switches on whatever it's connected to. :smile:

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(edited 10 years ago)

OCR Gateway Physics P4P5P6 - 18/06/13

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