Measuring LED forward voltage

Thread Starter

zazas321

Joined Nov 29, 2015
936
Yeah . Im trying to understand things slowly... Apparently i tryed doing something MikeML suggested me, but my multimeter sucks and i cant measure the accurate current.... I have built a simple circuit with a 470 Ohm resistor, 9.4Volt battery and a led. My multimer works only choose 10 option as u can see in the photo and it shows 0,02 but sometimes changes to 0,01. I cant understand why my multimeter wont show anything when i put it in 200m mode..
 

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dl324

Joined Mar 30, 2015
16,841
Adding good quality multimeter photo because its not clear from those 2 photos below what modes i was using
You can't use an ammeter to measure current in your test circuit because the meter will "disrupt" things. Mike assumed you had a power supply with an accurate current meter, but you're using batteries and wall warts.

It would be better for you to a current source.
 

Thread Starter

zazas321

Joined Nov 29, 2015
936
You can't use an ammeter to measure current in your test circuit because the meter will "disrupt" things. Mike assumed you had a power supply with an accurate current meter, but you're using batteries and wall warts.

It would be better for you to a current source.
So if i want to measure anything in my circuit i power up with my batteries i will get not accurate vallues ? you mean that?
 

dl324

Joined Mar 30, 2015
16,841
He doesn't understand how many LEDs you want to measure. I do. Build the current source I posted; it'll make your task easier.
Yes. R2, the 33 ohm resistor, sets the current. If you assume the BE will be around 0.7V at 20mA, the LED current will be I = V/R = 0.7V/33 ohms = 21.2mA. That's close enough to 20mA for your purposes.

If your power supply voltage is high enough, you can put 2 or more LEDs in series; but that would make measuring the forward voltage more tedious.
 

dl324

Joined Mar 30, 2015
16,841
So if i want to measure anything in my circuit i power up with my batteries i will get not accurate vallues ? you mean that?
No. You can use batteries if you want. Their series resistance won't affect the current source as long as the voltage is high enough and the batteries have sufficient life left in them. I prefer to only use batteries when they're required; especially if they're not rechargeable.
 

Thread Starter

zazas321

Joined Nov 29, 2015
936
How did you calculate that with this circuit we will have about 0.7V at the BE ? :eek:
I know how to calculate the voltage to the BE if there is voltage divider, i mean 2 resistors. But i dont know how is BE calculated on this one :/
 

dl324

Joined Mar 30, 2015
16,841
How did you calculate that with this circuit we will have about 0.7V at the BE ?
When the transistor is conducting 20mA of current, it's on (completely). I use 0.7V for that condition; some use 0.6V, but it's actually closer to 0.7V.

The value is in the BC547 datasheet. For your transistors, it's 0.77V max at 10mA. Typical value wasn't given, but the voltage will be higher at higher current.
upload_2015-12-10_8-34-45.png

For the divider circuit, I used the 'on' value for the BE junction to determine that it would be limited to 0.7V; and not what the divider ratio would have given.
 

Thread Starter

zazas321

Joined Nov 29, 2015
936
When the transistor is conducting 20mA of current, it's on (completely). I use 0.7V for that condition; some use 0.6V, but it's actually closer to 0.7V.

The value is in the BC547 datasheet. For your transistors, it's 0.77V max at 10mA. Typical value wasn't given, but the voltage will be higher at higher current.
View attachment 96273

For the divider circuit, I used the 'on' value for the BE junction to determine that it would be limited to 0.7V; and not what the divider ratio would have given.
Does a transistor saturation means its in its off state? so if its 0.77Voltage max at 10mA so if we have 20mA it will be way higher?
 

dl324

Joined Mar 30, 2015
16,841
Does a transistor saturation means its in its off state? so if its 0.77Voltage max at 10mA so if we have 20mA it will be way higher?
Now you're headed down a path that will give you more information that you need to check forward LED voltage...

When a transistor is used as a switch, that's it's saturation mode; full on. In that mode, both junctions are forward biased. If you had been ready to absorb that information, you would have noticed that the IV curves I posted in your LDR thread mentioned the 4 modes of operation; and I mentioned saturation mode.

At a current of 20mA, the BE voltage will be higher than at 10mA, but not necessarily as high as you're thinking. The 0.77V you're referencing is the maximum that it could be at 10mA. It could be less, and for most devices it will be. All you can safely say is that at 20mA the voltage will be higher than at 10mA. It's within the operating specs of the transistor, and that's all you need to be concerned about.

The manufacturers give minimum, typical, and maximum specs to aid designers in designing circuits that will work for the entire range of values.
 

Thread Starter

zazas321

Joined Nov 29, 2015
936
I will try to build a mini circuit like 3 different colour LED lighting when its dark, and calculate everything so maybe it will be easier for me to understand.. I just need to know the values of my LDR when its dark and when there is light in room. I will measure after they fix my multimeter.. The current gain of the BC547 transistor in datasheet is said to be 110-800 sooo how do i know which one do i use to get maximum accuracy? To calculate everything in the circuit i will also have to find out what is the transistor base resistance. Can i measure it with multimeter?
 

dl324

Joined Mar 30, 2015
16,841
The current gain of the BC547 transistor in datasheet is said to be 110-800 sooo how do i know which one do i use to get maximum accuracy?
I already gave you an appropriate beta when using the transistor as a switch; 20.

But, in answer to your question:
upload_2015-12-10_11-34-53.png
You are operating the transistor at 20mA. I've already told you that beta decreases with higher current (but didn't have curve tracer data to post). But you can see it from the 2mA and 100mA data above. If you are using the B or C varieties, your beta will be higher and the circuit will work "better". In the case of your circuit, a "good" design would be one that would work for a transistor that had the minimum beta.
To calculate everything in the circuit i will also have to find out what is the transistor base resistance. Can i measure it with multimeter?
The datasheet for your LDR will give you the light and dark resistance. If you don't have a datasheet, you can measure the resistance; when you get your meter repaired.
 

Thread Starter

zazas321

Joined Nov 29, 2015
936
Yeah okaay my badd i have jjust found out that im using bc 547C so if i take average current gain it will be like 40~?
 

Thread Starter

zazas321

Joined Nov 29, 2015
936
I dont want to go to deep with the those measurements before i start measuring by myself, but going thorugh the forums and reading about LDR working circuits i have found one thing i just couldnt understand :eek:

http://forum.allaboutcircuits.com/blog/step-by-step-design-of-a-simple-ldr-circuit.577/
if you can find those lines where the guy writes:
We will assume that β=100, so we need a base current of 50μA. The Vbe voltage will be somewhere close to 0.7V, so we will work with that value.

At 0.7V, the current in the LDR, namely Id, will be 8.75μA. The voltage across Rb will be 4.3V and we need a current in it of 58.75μA, resulting in a nominal value of 73kΩ. We could use a standard E24 value of 75kΩ or go with one of the surrounding E12 values of either 68kΩ or 82kΩ. Since the LED will probably be visibly "on" well before even 5mA of current, it is probably reasonable to go with the 82kΩ value, so let's use that.

He is assuming that the voltage accros BE junction will be 0,7V. Okay, i understood that, but why did he calculate the current in the LDR using the same voltage? Why he assumes that voltage acros BE junction will be the same as across the LDR?
 

dl324

Joined Mar 30, 2015
16,841
He is assuming that the voltage accros BE junction will be 0,7V. Okay, i understood that, but why did he calculate the current in the LDR using the same voltage? Why he assumes that voltage acros BE junction will be the same as across the LDR?
I agree with everything stated in the article. The LDR and BE junction are in parallel, so they will have the same voltage across them. At the currents mentioned, the BE junction will control that voltage.

You can study KVL here. I haven't read that page, but it should be accurate. Current direction might be backwards, but that's a known issue.
 

Thread Starter

zazas321

Joined Nov 29, 2015
936
Yeah i was thinking about that at first too, but then i noticed that there is a another resistor in series with LDR , so i thought the BE junction is parralel to both of theese resistors, but not only LDR itself? Why in this case we imagine that there isnt another resistor in series with LDR?
 

Thread Starter

zazas321

Joined Nov 29, 2015
936
I have just realised that the thing i said was a big mess... I hope you understand something from the photo. I have calculated the voltages across the componens. Are my calculations correct? I just didint know what will be CE voltage.? I guess it should be also 0,7V isnt it?. Cause if we have 4,3V at V3 and Vled and assuming that those 2 components are in series with CE , the voltage must add up and be 5V
 

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