Photocell - I/V characteristics

[Mutaja]

Photocell -- I/V characteristics

Homework Statement

A photocell as shown in the attachment has an I-U characteristic also as shown. The highest effect (W) that can be reached in the photocell is 5mW.

What's the least allowed value of R that won't overload the photocell in any way?

Calculate the voltage across the cell at 0.1, 0.2 and 0.3 lumen.

Homework Equations

Ohms law?

The Attempt at a Solution

The answer to the first question is 1.25k ohm.

The answer to the 2nd question is 4V, 3V and 1.6V (0.1Lumen, 0.2 Lumen and 0.3 Lumen.

For the first problem, I'm assuming that the answer is 1.25k ohm because R = U/I -> 5V/4mA = 1.25kohm.

But why those numbers?

The 2nd question I'm not sure what to do.

Any input of guidelines are greatly appreciated.

 

[rude man]

Attachment?

 

[Mutaja]

Attachment?

Oops, sorry. It's there now.

 

[gneill]

For a graphical approach you need to identify the region on the IV diagram that corresponds to photocell power being less than the maximum allowed (the given value being 5 mW). Then determine a load line that stays within that region (just "kisses" the edge of the region, minimizing the load resistor value).

You can translate that into an algebraic solution too, with a bit of effort.

 

[rude man]

OK, I finally noticed that the max allowable power dissipation is 0.5mW. At least that is my interpretation. The wording "the highest effect" is a bit puzzling still however.

The operating point is the intersection of the luminance curve with the load line, or i(V,L) = (E-V)/R where L = luminance. So if we pick for example luminance of 0.5 lumen we can construct the load line & find the intersection point. The product of that voltage and current is of course the diode's power dissipation, and that equals 0.5 mW.

You can now adjust the slope of the load line (= -1/R) and see how low you can make R without exceeding 0.5 mW dissipation.

 

[Mutaja]

Thank you both for replying!

I'm not sure I follow, but I've drawn a graph of the power (ignore the word effect - I meant power) - see the attachment.

What I'm thinking here is that on this line, the power will be 5mW. To get the lowest possible resistance (R=P/I^2) the power has to be as small as possible, and the current as big as possible. I find this to be when P is 5mW and I is 4mA. But then the resistance is 312.5 ohm. However, I can see that 312.5 ohm is exactly 1/4th of what the answer should be, meaning that if I use my current formula, the answer will be P = 5mW and I = 2mA -> R = 5mw/(2mA^2) = 1250 ohm.

Am I onto something here?

 

[gneill]

Thank you both for replying!

I'm not sure I follow, but I've drawn a graph of the power (ignore the word effect - I meant power) - see the attachment.
<snip>

:
:
:
​

Am I onto something here?

We'll need to see the attachment... it seems to have gone missing.

 

[Mutaja]

We'll need to see the attachment... it seems to have gone missing.

Must've forgot to press 'upload'. Should be ok now.

 

[gneill]

Okay, the curve you've drawn corresponds to the photocell handling 5 mW of power. It forms the "envelope" of the safe operating region of the photocell. Next you want to draw a separate load line (which will be a straight line, blue in the figure below) that will lie within that region. To minimize the corresponding resistor value you want that line to be as steep as possible.

You could also solve for this line algebraically, finding the slope such that the line just touches the 5 mW curve at one point.

 

[Mutaja]

Kind of obvious when you explain it like that. Thank you so much, once again. Thanks to both of you, I really appreciate it!

 

[Mutaja]

I might add that I'm unfortunately still stuck with my 2nd problem. How does the light intensity affect the voltage across the cell? Is it because higher voltage = stronger light? Still confused about where lumen comes into the equation, though.

 

[gneill]

I might add that I'm unfortunately still stuck with my 2nd problem. How does the light intensity affect the voltage across the cell? Is it because higher voltage = stronger light? Still confused about where lumen comes into the equation, though.

The photocell is constrained to operate along the load line. Draw the load line on the characteristic curves (note how they're labelled). Is that enough of a hint?

 

[Mutaja]

The photocell is constrained to operate along the load line. Draw the load line on the characteristic curves (note how they're labelled). Is that enough of a hint?

Alright, so it's where the load line (how is that load line calculated by the way? I'm fully aware that I should be able to see that, if not know that, but right now there's a mess in my head) intercepts the lines for lumen? If so, 0.1 lumen = 4V, 0.2 lumen = 3v and 0.3 lumen = right over 1.5V?

If it's not too much to ask, how is this calculated (I assume it's possible as the graphs are written based on some numbers)?

Thanks for you help once again.

 

[gneill]

Alright, so it's where the load line (how is that load line calculated by the way? I'm fully aware that I should be able to see that, if not know that, but right now there's a mess in my head) intercepts the lines for lumen? If so, 0.1 lumen = 4V, 0.2 lumen = 3v and 0.3 lumen = right over 1.5V?

Looks right!

If it's not too much to ask, how is this calculated (I assume it's possible as the graphs are written based on some numbers)?

You should be able to write an equation for the curve that defines the power boundary, and an equation for a line "anchored" at its x-intercept at 5V (the load line must pass through that point since it's set by the source voltage). Think point-slope form for the line equation, where the slope is an unknown to be determined. How might you solve for the slope where the line just intersects the curve?