Science Buddies: "Ask an Expert"

I have a question about how to determine the correct size of a resistor.

I have assembled the light-sensor described in Measuring Vibrational Frequencies with Light and it works fine, and I am getting good voltage readings from my circuit. I am using the suggested light-to-voltage converter from Mouser electronics (part 856-TSL14S-LF - same link has the data sheet)

What I don't understand is how would I have known to use a "10 kΩ, 1/4-watt resistor"? How do you calculate this? If I had used some other companies light-to-voltage converter, how would I have known what resistor to use? I think the point of the resistor in this circuit is to protect the light-to-voltage sensor from getting fried, but the instructions don't really say much about what it's for.

I'm glad it works, but I want to know why I'm supposed to be using this particular resistor, and how I could figure it out myself.

Thank you to anyone who can explain this to me.
Austin

Hi Austin,

The purpose of the resistor is to limit the current to the Absolute Maximum Ratings on the photodiode part as indicated in its datasheet. Ohm's law states that I=V/R. This is why R = 10 kΩ was chosen.

The power version of Ohm's law is P=V*V/R. All the calculations are handled at:
http://www.angelfire.com/pa/baconbacon/page2.html

I'm a bit puzzled in that with V=5 and R=10, Ohm's law says P = 2.5 (not 0.25). So instead of a 10 kΩ, 1/4-watt resistor, one may need a 10 kΩ, 2.5-watt resistor. I may be missing something here!

(Note P for power is often listed as W for watts.)

Current limiting for the Photo diode is NOT the reason for choosing a 10 Kohm resistor in this case.

The circuitry here is an operational amplifier (very high gain). The resistor is used as a feedback resistor. If the voltage at the + and - terminals is not equal, then in the absense of a feedback resistor, the output will be limited by the supply voltage.

Unless you look up or measure the light to current properties of the photo diode, there isn't enough information to choose a feed back resistor to provide a specific light intensity to voltage reading.

This experiment does not require a specific calibrated gain so picking a 10 Kohm resistor is just a seat of the pants starting point. A 1 Kohm resistor would draw 5mA of current so it would not be as battery friendly. A 100 Kohm resistor would provide more gain and be more noise sensitive.

I'm a bit puzzled in that with V=5 and R=10, Ohm's law says P = 2.5 (not 0.25). So instead of a 10 kΩ, 1/4-watt resistor, one may need a 10 kΩ, 2.5-watt resistor. I may be missing something here!

V*V = 25. Divide this by 10,000 and you get .0025 W or 2.5 mW.

Craig,

I'm a software type. Guess, it shows. The formulas looked straightforward. I guess, too much so. I did sense something was wrong, and I'm glad you cleared the matter up. Thanks.

Thank you both for your answers.

Craig_Bridge wrote:Current limiting for the Photo diode is NOT the reason for choosing a 10 Kohm resistor in this case.

The circuitry here is an operational amplifier (very high gain). The resistor is used as a feedback resistor. If the voltage at the + and - terminals is not equal, then in the absense of a feedback resistor, the output will be limited by the supply voltage.

Can you translate this into 7th-grade speak? If I don't complete the circuit with something (wire or resistor) from Vout to ground, then the circuit is open and I get zero volts measured (I hope that's right or I'm really lost). So the function of the resistor back to ground is what exactly?

Unless you look up or measure the light to current properties of the photo diode, there isn't enough information to choose a feed back resistor to provide a specific light intensity to voltage reading.

This experiment does not require a specific calibrated gain so picking a 10 Kohm resistor is just a seat of the pants starting point. A 1 Kohm resistor would draw 5mA of current so it would not be as battery friendly. A 100 Kohm resistor would provide more gain and be more noise sensitive.

So what I think I understand here is that I can't be expected to have known what size resistor to pick, some expert just knew what would work ok.

I'm a bit puzzled in that with V=5 and R=10, Ohm's law says P = 2.5 (not 0.25). So instead of a 10 kΩ, 1/4-watt resistor, one may need a 10 kΩ, 2.5-watt resistor. I may be missing something here!

V*V = 25. Divide this by 10,000 and you get .0025 W or 2.5 mW.

Ok, I understand that math, but how does the 0.0025W relate to the 1/4 watt resistor (0.25W).

I'm trying to understand!!
Austin

Excuse my confusion... I was looking at the circuit in Figure 1 of https://www.sciencebuddies.org/mentorin ... p015.shtml

The circuitry here is an operational amplifier (very high gain). The resistor is used as a feedback resistor. If the voltage at the + and - terminals is not equal, then in the absense of a feedback resistor, the output will be limited by the supply voltage.

My answer is correct for determining that internal feedback resistor (a component between the output and input of the amplifier is said to be a feedback component), unfortunately, you were asking about figure 3 load resistor.

So the function of the resistor back to ground is what exactly?

In electrical engineer speak, this is a "load" resistor which determines the output impedance (resistance) of the circuit. It is either chosen based on the seat of the pants answer (same reasoning and calculations apply to the load resistor as the feedback resistor, so my previous answer about battery current and noise reasoning applies) or was chosen to match the input impedance of the A/D converter. Maximum efficiency occurs when the output impedance matches the input impedance of the next circuit but I suspect it was a seat of the pants decision.

If I translated this to, "Current flows through this load resistor back to ground to develop a voltage for the A/D converter.", would that make sense to you?

Ok, I understand that math, but how does the 0.0025W relate to the 1/4 watt resistor (0.25W).

A 1/4 W resistor is rated to handle up to 0.25 W of power so it can easily handle the much smaller 2.5 mW in this circuit.

Craig_Bridge wrote:you were asking about figure 3 load resistor (in https://www.sciencebuddies.org/mentorin ... p015.shtml)

Yes, that is the one I am interested in.

Maximum efficiency occurs when the output impedance matches the input impedance of the next circuit but I suspect it was a seat of the pants decision.

If I translated this to, "Current flows through this load resistor back to ground to develop a voltage for the A/D converter.", would that make sense to you?

Yes, that makes sense (especially when I take my voltage measurements across the resistor!!) But thru some experimenting I found that if I take my measurements from the light-to-voltage converter itself, across pin 1 (gnd) and pin 3 (Vout) then it doesn't seem to matter if I have the resistor or not, I get the same output voltage measurements as I do measuring across the resistor. Does that make sense? It seems to me that both are measuring the voltage difference between ground and Vout, except one isn't going thru a resistor...

Thank you for your patience!
Austin

I get the same output voltage measurements as I do measuring across the resistor. Does that make sense? It seems to me that both are measuring the voltage difference between ground and Vout, except one isn't going thru a resistor...

The A/D converter (or a meter) has an internal "input impedance" (resistance equivalent) between its two measurement connections. When you attach one to pin 3 (Vout) and the other to pin 1 (Gnd), you are effectively putting the A/D converter's internal resistance between Vout and Gnd (in parallel to the load resistor or in place of the load resistor).

Your trying different things proved that the value of the load resistance isn't very important to the operation of the circuit which makes perfect sense to me.

You should look up Kirkoff's current law to better appreciate some of the subtle points. In particular: You measure current going through a component; however, you measure voltage accross a component.

Craig_Bridge wrote: The A/D converter (or a meter) has an internal "input impedance" (resistance equivalent) between its two measurement connections.

This is exactly the idea my Dad and I came up with last night, it was the only thing we could think of that made sense to us! We guessed that the digital voltmeter we were using must have an internal resistor that essentially played the same role.

You measure current going through a component; however, you measure voltage accross a component.

This sounds like important information, I will look into it more.

Thank you again for your time!
Austin