Light bulb & Diode: linear or non-linear elements.

AI Thread Summary
The discussion centers on the non-linear behavior of light bulbs and diodes in response to varying potential differences. The resistance of a light bulb is not constant because it increases with temperature as the filament heats up, affecting electron flow. In contrast, a diode exhibits high resistance in one polarity and low resistance in the opposite, allowing current to flow only in one direction. The participants also debate the accuracy of experimental readings and the interpretation of current-voltage graphs, emphasizing the importance of analyzing data carefully. Overall, the conversation highlights the complexities of electrical components and the significance of experimental methodology in understanding their behavior.
kamhogo
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Homework Statement


These questions are related to an experiment where we had to record the current for varying values of potential difference (10 readings), then repeat the process with the applied potential difference was reversed

a) Why is the resistance of a light bulb not constant?
b) Does the resistance of diode change when polarity is reversed? Why?

Homework Equations


· ∆V=IR (linear resistors)

· R = 1/ (dI/dV) (non-linear elements)

· I = eIe

3. The attempt at
a solution

a) The light bulb is non-linear because its resistance varies with increasing/sustained current.
The resistance is not constant because as you increase the current (by increasing the voltage difference), the filament of the bulb heats up. This has for effect to increase the resistance. Higher temperature makes the ions in the filament vibrate more, making it more difficult for electrons to pass between them (—› reduced electrons flow —› reduced drift speed —› reduced electron current —› reduced conventional current). As the resistance decreases, the current flowing through the filament increases. However, the light bulb does behave like a linear element when the current flowing through it is not sustained (the filament does not have time to heat up enough to affect conductivity). This explains why the experimental current-vs-voltage graph of the light bulb appears to be a linear one with slope 1/R. Indeed, during the experiment, the switch allowing the current to pass through the light bulb was maintained in a closed position only long enough to obtain the voltage and current readings. Thus the current was not sustained

b) The resistance of the diode seems to change when the polarity is reversed. For negative values of applied potential difference, the current was zero. (Is this normal or an error from the experiment)
If someone could tell me if my thinking makes any sense and why the diode resistance changes when the polarity changes + why it increases so abruptly , that would be well-appreciated :)
 

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Hi kam:

I am unable to answer all your questions, but I think I can help with some.

I believe that for the bulb, resistance increases with temperature, which increases with power, which increases with current. I am not sure why resistance increases with temperature, but my guess is that it is because the atoms in the filament get further apart.

For the diode: they are designed so that current will flow only in one direction/polarity. That is, resistance is intended to be very high with one polarity, and very low with the other.

Hope t his helps.

Regards,
Buzz
 
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Your thoughts about why a lightbulb might display nonlinearity in its resistance with respect to current are along the right lines, although your conjecture regarding there being insufficient warmup time before reading the current value is questionable. You don't have any information on the thermal mass of the filament or what sort of time constant might be involved. Is it possible that your equipment is just not sensitive enough to see the effect?

You should do a web search on "diode VI curve" to see what its normal behavior is.
 
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Buzz Bloom said:
Hi kam:

I am unable to answer all your questions, but I think I can help with some.

I believe that for the bulb, resistance increases with temperature, which increases with power, which increases with current. I am not sure why resistance increases with temperature, but my guess is that it is because the atoms in the filament get further apart.

For the diode: they are designed so that current will flow only in one direction/polarity. That is, resistance is intended to be very high with one polarity, and very low with the other.

Hope t his helps.

Regards,
Buzz
Thanks!
 
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gneill said:
Your thoughts about why a lightbulb might display nonlinearity in its resistance with respect to current are along the right lines, although your conjecture regarding there being insufficient warmup time before reading the current value is questionable. You don't have any information on the thermal mass of the filament or what sort of time constant might be involved. Is it possible that your equipment is just not sensitive enough to see the effect?

You should do a web search on "diode VI curve" to see what its normal behavior is.
Thanks for the feedback!
 
kamhogo said:
the light bulb does behave like a linear element when the current flowing through it is not sustained (the filament does not have time to heat up enough to affect conductivity). This explains why the experimental current-vs-voltage graph of the light bulb appears to be a linear one with slope 1/R.
I'd expect that the filament would change temperature faster than you could take a meter reading, so can't understand why you missed out on the expected graph.. But I'm not able to comment specifically because I can view only JPG attachments.

Can you attach your results as a JPG screenshot?
 
NascentOxygen said:
I'd expect that the filament would change temperature faster than you could take a meter reading, so can't understand why you missed out on the expected graph.. But I'm not able to comment specifically because I can view only JPG attachments.

Can you attach your results as a JPG screenshot?
tmp_12192-20160407_1843071282769602.jpg

Here's a snapshot of the graph...
 
The plotted points should always be made visible on a graph.

It looks like you may have allowed yourself to draw a large deviation in the curve based on a single oddity? On experimental plots, one out-of-line point is rarely justification for departing from a smooth curve---any oddly placed data value is more probably a mistake in experimental procedure, so should be investigated or smoothed over in the final analysis.

In the region up until it glows cherry red is where a bulb is significantly non-linear, and your plot demonstrates this.
 
NascentOxygen said:
The plotted points should always be visible on a graph.

It looks like you may have drawn a large deviation in the curve based on a single oddity? On experimental plots, one out-of-line point is rarely justification for departing from a smooth curve---one oddly placed data value is more probably a mistake in experimental procedure, so should be investigated or smoothed over in the final analysis.

Thanks for the feedback, I'll rearrange the graphs. One more question: is R= ( dI/dV)^-1 the resistance of the bulb?
 
  • #10
kamhogo said:
Thanks for the feedback, I'll rearrange the graphs. One more question: is R= ( dI/dV)^-1 the resistance of the bulb?
That's the incremental resistance, the tangent to your curve. I don't think you are asked for it here. It can be gleaned by inspecting the V-I plot.

The resistance of the bulb is V/I.
 
  • #11
NascentOxygen said:
That's the incremental resistance, the tangent to your curve. I don't think you are asked for it here. It can be gleaned by inspecting the V-I plot.

The resistance of the bulb is V/I.
I am asked for the dynamic resistance (which I guess is the same as the incremental resistance). I was asked to draw two tangents at two different points on the graph and then find R at those points based on the slope of the tangent. I got 34.364 and 6.5359 (slopes of tangents). Are those the resistances or the inverse of the resistances?
 
  • #12
Where the slope has units of volts ÷ amps that's ohms.

As I pointed out, your tangent of 6.5 is based on a nonsense datum.
 
  • #13
NascentOxygen said:
Where the slope has units of volts ÷ amps that's ohms.

As I pointed out, your tangent of 6.5 is based on a nonsense datum.
Thank you for the feedback. It's really appreciated.
 
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