Confusion about voltage drop

[Krooger]

Hello, probably a dumb question, I'm super new.

I made the circuit I drew and don't understand the voltage drop

Expected:
Led voltage drop to be 2V because datasheet said that's best
Pair of resistors voltage drop to be 7V because the battery is 9V

My results:
Led voltage drop 2.2
Pair of resistors was 6.3
Then found 0.5v from after the LED to just before the battery, all just wire.
I don't understand why there is a voltage drop there?

Also, the battery terminals actually read to be 10v so where did that extra volt go?

Am I way off? Could someone clear my confusion or maybe point me in the right direction on what to study?

[IanB]

A 9 V battery is not 9 V. If it is a fresh alkaline battery it will start out something like 9.6 V, and if a heavy duty battery more like 10 V. The voltage will go down under load, and will go down as it drains.

There should not be 0.5 V drop in the wire. Check for poor connections or bad quality wire.

[ledtester]

A battery has internal resistance. It won't affect the battery's voltage that you measure if there is no load but will when the battery supplies current.

Also, the battery terminals actually read to be 10v so where did that extra volt go?

The 10V is the "no load" reading of the battery. But once the battery supplies current there is a voltage drop across the internal resistance.

More info:

http://www.learningaboutelectronics.com/Articles/Battery-internal-resistance

[Krooger]

When I say the 0.5 drop is through wire, I really mean through a breadboard and some of the jumper cables. Could bad breadboard quality and cables cause this reading?

[nightfire]

Breadboards are not known for best and reliable connections, so the joints of a breadboard will contribute to some additional resistance.

[ledtester]

Possibly. Assuming your other measurements are valid it would seem that connection between the LED and battery has a resistance of about 31 ohms.

Try making a more robust connection between the LED and battery -- i.e. twisting wires and leads together or even soldering them.

[Krooger]

Wow, after adding many more unnecessary cables through the breadboard the voltage drop through it went all the way to 1V.
This seems significant, do people account for this normally?

[IanB]

It could be a poor quality breadboard, or poor connections where you plug the wires into the holes.

Many low cost breadboards are unreliable. Quality varies a lot.

[ledtester]

Wow, after adding many more unnecessary cables through the breadboard the voltage drop through it went all the way to 1V.
This seems significant, do people account for this normally?

They don't account for it - they avoid it by either:

1. using higher quality breadboards -- e.g.:

https://www.eevblog.com/forum/chat/3m-breadboards-for-cheap/

2. not using breadboards

[EEVblog]

You need some really high contact resistance to get 0.5V or 1V drop, that seems rather unlikley.
I am not convinced that your battery is actually precisely 9V.
All the voltages around the circuit must add up.
Pick one ground reference point and measure other points from that.

[WattsThat]

Please take a photo of your setup and show us the points where you’re measuring.

There is an explanation for your question. The problem is, we don’t know the where, the what and the how of what you’re asking.

[RJSV]

Yeah, looks like that Dave guy has it right...(oh wait, he's the head of eevblog !)
   You need to maintain some consistent context, meaning that saying '10 volts' needs to be more like:
   '10 volts with nothing attached'.  Then, response is, you needed to have it in circuit, so that little bit of load will bring your battery output down, slightly, maybe down to 9.7 v, to guess what it might settle at.
Now, am I trippen, or isn't the little end '+' on battery ?
And bigger line is your minus (-).  LED runs with current
as the symbol points, battery + goes into LED, and LED points to battery minus (-).
   Also looks, maybe, like you've added some real measured voltages, adding up to 8.5 V, but then substituted 'nominal' 9.0 V to get that .5 left over...You shouldn't mix your contexts; use an ACTUAL batt voltage, then subtract the measured 8.5.
Example, might be battery at 8.7 with the 2 components dropping 8.5 V.  That way your result would be 0.2 V being dropped through your wires and breadboard.  I suppose you could try soldering a little test set, to see if that 0.2 V gets 'better', i.e. lower, but I would tend to just shrug, at 0.2 V, just a judgement call (Whatta ya gonna do?), especially for some novice explorations.  The half-volt would start to get my notice.
   The stuff you can look up, relates to a circular examination: Going all the way, round, should add up to Zero; or looking separately, your 'load' components reading, should match, (equal) your immediate battery reading...that is, the battery voltage reading when in circuit with that exact load.
For example, you wouldn't want to use your running battery reading you remember from yesterday, or on some other LED, it has to be all measured at one sitting.
   For the circuit current draw, simply divide I= V/R where V is measured across resistor.
For more reading, I think try 'Kirkoff's Law', that analizes around a circle.  Makes sense, but not sure what a Google search might show.
Oh, and avoid the trap, of overusing the LED forward voltage spec. That is a voltage specification, on paper, and not a good starting point as the forward voltage is more like a 'reaction', to current flow.  It's confusing, but you might notice, using nominal LED voltage, as you make plans, you can set your LED current, approximately, to match the spec, like, say, 20 mA (that's a lot, for a higher power LED like in flashlight).
THEN, you can go in, after building, and measure everything; that will give you the exact LED voltage, at the exact current that you've ended up with.
Suppose you tried for 20 mA, and end up reading 19.5 mA, that's pretty good. (Btw, I almost always prefer to read current by way of using the resistor drops, along with ACTUAL ohm reading, of your resistor pair.  Two resistors 330 ohms, in parallel maybe would read as '146 ohms' (rather than purely '150').

[RJSV]

Oops, sorry, I meant to say the two resistors would be '165' ohms, as 330/2 .
--Rick B.

[golden_labels]

How are you measuring the voltages? I mean: how do you attach probes to the circuit? Perhaps you are pushing/pulling legs of the components with probes, causing a poor connection. Current across the LED is already much above what is needed for it to shine near its maximum, so you would not notice the increase in resistance by a dozen ohm.

[RJSV]

So, here is a more specific example, (for more clarity):
  You look at the spec, for your LED part number, and it says '2 volts nominal' at 20 mA'.
You take a 'best guess', at battery voltage.  A nmh 9V battery example might be 8.4 volts nominal. So the two nominal voltages, from spec, gives a difference of 6.4 volts. This is approximate, for a first round.
Then, you divide, the two 'nominal' items; that would be
  R = V/I.    That's 6.4 V drop needed, divided by 20 mA,
giving 320 ohms.  If that equation came out to 310 ohms you have to use closest value in stock, which is 320.  Then, having established the components you can set out to recalculate using the real values; real battery voltage goes to, as example, 8.1 volts, as those nmh (nickle metal hydride) batteries run at a low nominal.  Then, you measure actual resistor, say that's at 318 ohms.  So, the second time around with calculating,  you have to drop 6.1 Volts; that's a refined version, from starting at needing 6.4 volts drop.  But not that different, just more realistic.
Now, to set your current to 20 mA, use the actual drop needed, 6.1 V, divided by 20 mA, indicating that a more accurate R needing to be 315 ohms, more exactly.
   See the minor dilemma, there ?  When you're stuck with non-standard value, you have to just use the 320...as that's closest resistor in stock.
   Then, I guess as a 'third round' you calculate your stuff keeping the nominal R=320, actual 318 ohms, with measured battery output voltage, if you want really close numbers.  That last process is getting a bit obsessive, as now you are close to done, but you have to accept the resistor inaccuracy, either that or add a parallel 'trim' to bring R= 320 down to 318 ohms, but you STILL won't get, exactly, the nominal 2 volts LED drop.  Oh and at this point you need to bring in that 0.2 volts 'wiring' drop.  Still won't be perfect, but your eyesight can't distinguish so accurate as to notice any brightness differences.
Notice, the spec (probably) says 'nominal 2 Volts', rather than '2.0 volts', a hint about what 'nominal' is meaning.  You could spend time getting everything accurate, but best approach is to adjust the resulting current, and let the nominal LED drop deviate slightly, from 2.0 volts.  Later, anyway, the room temperature itself is going to change, causing a (very) slight shift in LED drop.  But that's normal, as best consistency can be had by targeting the current, rather than voltage.

[RJSV]

   I think, looking at your schematic again, I would try circuit using 330 ohms in parallel with something like 10 k, that trims the final value. But, see, the process doesn't have to be so super accurate, it's just educational, to know what you'd have to do, and how much that extra little accuracy improvement is going to affect the outcome.  Usually, you can just use the nominal values, along with realistic value for battery under similar load.  Of course, temperature affecting the nominal LED voltage is surely going to change the actual current, but, again, that's a fine point not usually bothered with.
   Your wiring really should be expected to be below about 0.5 v drop, though.

[Siwastaja]

Wow, after adding many more unnecessary cables through the breadboard the voltage drop through it went all the way to 1V.
This seems significant, do people account for this normally?

I don't recommend breadboards for this reason, among others (such as high parasitic capacitance and coupling of AC signals, and general wear and tear causing intermittent contacts). High quality breadboards of course exist, but I just prefer soldering and desoldering things. You get quick at it the more you do it.

Also note LEDs are rated for a certain current. Even if you had exactly the same current over the LED which the datasheet assumed (which you don't have), the voltage across the LED still is a poorly controlled variable, and varies significantly depending on the manufacturing batch, and temperature. So measuring 2.2V when you expected 2.0V (I'm assuming it said "typical" in the datasheet) is completely normal. What is important is you don't exceed the recommended current of the LED.

[CaptDon]

I see by your numbers about 40ma. of current through the led. That will kill your 9v battery quickly and seems like about twice the current most leds use.

[james_s]

Yikes, yes 40mA is way more than a typical 5mm LED is rated for, 20mA max is typical, 5-10mA is plenty for most applications, modern LEDs are really efficient. If the current is too high it will not only cause the battery voltage to sag but the LED temperature will rise and the forward drop will vary.

[Krooger]

Wow, thank you everyone for all the information, very much appreciated. I have learned a lot just by reading the comments.

I'll have to further research the things everyone mentioned and will look at soldering to reduce errors.

I know most LEDs are meant to run at 20mA but the data sheet for my red LED says to run it at 50mA which I thought was weird but that's why the current is higher in this circuit.

Very fun hobby but a bit difficult to start up without formal training.

[james_s]

It's not that difficult, there are lots of books and tutorials out there. EE is a vast topic and even veteran professional engineers don't know everything but the basic stuff like this you can pick up with a little patience. Not that much different from many other hobbies, sewing, crafting, cooking, playing a musical instrument, woodworking, gardening, etc gotta learn the basics and then you go from there.

There are certainly high powered LEDs out there that run at higher currents, it's helpful to post details like that in such cases, most are probably going to assume a bog standard 5mm indicator LED.

[EEVblog]

I know most LEDs are meant to run at 20mA but the data sheet for my red LED says to run it at 50mA which I thought was weird but that's why the current is higher in this circuit.

That's the maximum continuous forward current spec.
Light output is basically proportial to current, and most leds will still operate down to 1mA or so.
There is usually never a need to run a LED at it's maximum forward current, unless it's a torch or bulb or something.

[Nusa]

Remember multimeter leads and probes/clips, especially the crappy ones that come by default on cheap meters, are likely to have resistance. So your voltage measurements may be a bit lower than reality.

[IanB]

Remember multimeter leads and probes/clips, especially the crappy ones that come by default on cheap meters, are likely to have resistance. So your voltage measurements may be a bit lower than reality.

Surely you mean current measurements? Voltage measurements are not affected by probe resistance.

[Ian.M]

Remember multimeter leads and probes/clips, especially the crappy ones that come by default on cheap meters, are likely to have resistance. So your voltage measurements may be a bit lower than reality.

Errrrrr.......

Would you like to explain how in a 9V supply circuit, the less than 1 uA current drawn by a typical DMM with 10 Meg input impedance on a voltage range  can result in enough voltage drop to be noticeable on that meter? 

Lets be generous and assume a 5 digit meter, and you are measuring the battery voltage, so its autoranged to single digit volts (i.e. n.nnnn V).   The lowest digit increment is 100 uV, so with the meter presenting a 0.9 uA load to the nom. 9V battery, it would need over 110 ohms of lead resistance to drop the reading by one least significant digit.

I can believe possible lead resistances of a couple of ohms each, but higher than that would be a defect even on a Harbor Freight 'free' DMM.

Edit: IanB beat me to it, but I'll let this stand anyway.