If you don't have a data sheet for your LED that has a chart of voltage across the LED versus current through it you can always make one. This exercise is also good forgetting to know the performance of the LED.
excel is handy for doing the calculations for you and making a chart.

 

Expect the first part (the low current portion) of the resulting chart to deviate from the actual characteristic significantly because most meters have about 10 MΩ of resistance when measuring voltage, thus the measurement-induced errors will be pretty major until you get down into the 100 kΩ range or so.

 

Expect the first part (the low current portion) of the resulting chart to deviate from the actual characteristic significantly because most meters have about 10 MΩ of resistance when measuring voltage, thus the measurement-induced errors will be pretty major until you get down into the 100 kΩ range or so.

Not too relevant is it? A 10 M meter resistance across the LED's resistance seems meaningless. You are technically correct, but it isn't enough to matter.

 

Not too relevant is it? A 10 M meter resistance across the LED's resistance seems meaningless. You are technically correct, but it isn't enough to matter.

I didn't read your instructions closely enough. For some reason I thought you were measuring the voltage across the resistor. My bad. Though even at these currents it might be enough to matter some.

Let's use a supply voltage of 5 V. With a 10 MΩ resistor the maximum diode current would be 0.5 uA. If the LED has a quality factor of 1 and a forward drop of 2 V at 5 mA, then it will lose about 60 mV for each decade decrease in current. Four decades of current drop would lower the voltage to about 1.250 mV, which would result in a current in a 10 MΩ meter of about 0.125 uA, which is a quarter of the current that should flow in the LED. That's quite possibly enough to skew the resulting curve by a noticeable amount, but not nearly as much as I was envisioning in my previous understanding of your method.

 

I didn't read your instructions closely enough. For some reason I thought you were measuring the voltage across the resistor. My bad. Though even at these currents it might be enough to matter some.

Let's use a supply voltage of 5 V. With a 10 MΩ resistor the maximum diode current would be 0.5 uA. If the LED has a quality factor of 1 and a forward drop of 2 V at 5 mA, then it will lose about 60 mV for each decade decrease in current. Four decades of current drop would lower the voltage to about 1.250 mV, which would result in a current in a 10 MΩ meter of about 0.125 uA, which is a quarter of the current that should flow in the LED. That's quite possibly enough to skew the resulting curve by a noticeable amount, but not nearly as much as I was envisioning in my previous understanding of your method.

There are some good lessons in doing the exercise. You realize you don't need to run the LED at the stated current ONL You realize the voltage changes and is not constant. It is a good exercise in using Microsoft Excel.
Any type of diode can be used in the exercise with similar benefit. Very suggested.
This used to be a classroom exercise, but the school is closed now.

 

This exercise is also good forgetting to know the performance of the LED.

10Meg is too big of a resistor to start for an led. 10K more likely.

 

10Meg is too big of a resistor to start for an led. 10K more likely.

He's trying to map out the I/V curve and you want to get down pretty far in order to really see the knee. Since you only need to take a few measurements per decade, it's not that much more effort.

 

Not too relevant is it? A 10 M meter resistance across the LED's resistance seems meaningless. You are technically correct, but it isn't enough to matter.

Using R = 10 MΩ, a 5 volt supply, a meter with 10 MΩ input resistance and a green LED whose voltage drop measures 2.00 volts, Step 8 of your exercise would have us calculate the LED current as 300 nanoamps. Wrong: 200 of those nanoamps are going through the meter, leaving only 100 nanoamps as the actual LED current.
Perhaps to you, being off by a factor of 3 isn't "relevant." To me, it's a gross error caused by sloppy technique.

 

Using R = 10 MΩ, a 5 volt supply, a meter with 10 MΩ input resistance and a green LED whose voltage drop measures 2.00 volts, Step 8 of your exercise would have us calculate the LED current as 300 nanoamps. Wrong: 200 of those nanoamps are going through the meter, leaving only 100 nanoamps as the actual LED current.
Perhaps to you, being off by a factor of 3 isn't "relevant." To me, it's a gross error caused by sloppy technique.

??? With the meter across the resistor, yes,
How much does it change the voltage across the LED?
The results are not off by a factor of 3.
You are technically correct, but It does not change the lesson of the exercise and is thus irrelevant. An error, yes. Gross error, no.
The objective of the exercise is that the voltage across the LED changes with current. there is no devil in the details in this case.
Are you one of those guys that says we must run the LED at the rated current the data sheet suggests?

I will rewrite the exercise to start at 100K to avoid this question.

 

If you don't have a data sheet for your LED that has a chart of voltage across the LED versus current through it you can always make one. This exercise is also good forgetting to know the performance of the LED.
excel is handy for doing the calculations for you and making a chart.

I have a very low cost Data Acquisition module from DATAQ which is wonderful for doing this kind of thing. The nice thing about using a DAQ is that you can take a lot more data points...at least conveniently. Nice work.

 

How much does it change the voltage across the LED?

it is not uncommon for a diode to have very high resistance at low current levels, well into the Mohm range.

So yes, at the kind of levels you are talking about, the meter becomes an important part of the circuit.

But fundamentally, for what you wanted to do, there is no reason for you to start measuring at 10Mohm. Pick a more reasonable number and your approach will be fine.

 

I have a very low cost Data Acquisition module from DATAQ which is wonderful for doing this kind of thing. The nice thing about using a DAQ is that you can take a lot more data points...at least conveniently. Nice work.

Very preferable but not within the budget of the casual student.

 

??? With the meter across the resistor, yes,
How much does it change the voltage across the LED?

No, I'm talking about the meter being placed across the LED, exactly as in Step 5 of your exercise.

The results are not off by a factor of 3.

Go back to what I wrote above, re-read it and do the math: of the 300 nA flowing down through R, fully 2/3 of it is flowing through the meter's 10 MΩ input resistance instead of through the LED.

 

Very preferable but not within the budget of the casual student.

If part of the educational process, it's well within the budget.

 

Very preferable but not within the budget of the casual student.

https://www.dataq.com/products/di-145/
Can't get much cheaper than this.

 

https://www.dataq.com/products/di-145/
Can't get much cheaper than this.

So nice! Thanks for the info.
I went to buy one and the store was closed for the holidays. :-(
It is on my shopping list if I still have the money on the2nd.

 

it is not uncommon for a diode to have very high resistance at low current levels, well into the Mohm range.

So yes, at the kind of levels you are talking about, the meter becomes an important part of the circuit.

But fundamentally, for what you wanted to do, there is no reason for you to start measuring at 10Mohm. Pick a more reasonable number and your approach will be fine.

Your evaluation is correct. For circuit design that is good. For year one basics a secondary objective is to get the student familiar with resistor values and reading resistors. Starting from 10,000,000 is beneficial.

 

https://www.dataq.com/products/di-145/
Can't get much cheaper than this.

Okay, I got my DI-145, running in oscilloscope mode for the moment. It runs continuously. What I want for the LED exercise is just to take one quick reading at a time.
Read LED voltage.
change the ballast resistor.
Take the next reading.

I have a 74221, mvb, that gives me a pulse (about 0.2 sec), positive or negative. what I was trying to do was to take a reading while the pulse was high.
What am I missing?

My breadboard has + and - 15 V. I added regulators to bring that down to + and - 9 V.