# Photo - This Voltage Divider is not doing what I expected - why?

Discussion in 'General Electronics Chat' started by Lumenosity, Dec 19, 2017.

1. ### Lumenosity Thread Starter Member

Mar 1, 2017
398
40
Hello and thanks for stopping to look at this thread.

In the photo, I have created a voltage divider that mathematically should give me 3.3v at the red wire between the two resistors.
But instead, I'm getting 1.79v. I tested the resistors with a VOM and they are accurate.

Can you see what I have done wrong? I'm missing something obviously.

Thanks

PS
Are these resistors printed correctly?
How is one supposed to know which end to begin reading the bands?
I have to test each one just to verify it's value.

2. ### Dodgydave AAC Fanatic!

Jun 22, 2012
7,640
1,270
Only two reasons for it,

1) wrong value resistors.

Lumenosity likes this.
3. ### Lumenosity Thread Starter Member

Mar 1, 2017
398
40
Thanks Dave.
I've carefully checked the resistors as mentioned and they are spot on.

The IR of the meter appears to be 00.4 ohms (zero,zero,point four)
What I mean is, when I set the meter at it's lowest value and touch the leads together.

4. ### WBahn Moderator

Mar 31, 2012
23,356
7,079
Where's the other probe on your voltmeter going when you measure the 1.79 V. Seems telling that the expected voltage across the 180 Ω resistor would be 1.76 V.

5. ### Alec_t AAC Fanatic!

Sep 17, 2013
8,938
2,105
As for reading resistor values, try to match the band colours with those of the standard E12 or E24 series. If left to right makes more sense than right to left, then L to R is the most likely sequence. But if in doubt, use an Ohmmeter.

6. ### WBahn Moderator

Mar 31, 2012
23,356
7,079
How is 00.4 Ω different from 0.4 Ω?

A voltmeter should have extremely HIGH resistance, like 10 MΩ.

Are you sure you are on a DC V range and not on a current range?

7. ### Lumenosity Thread Starter Member

Mar 1, 2017
398
40
Hello,
See the red wire coming from between to two resistors? I check the voltage at the end of the red wire with the other VOM lead connected to the 5v+

Hello,
Hmmm......I set the meter to the 200ohm range then touch the leads together. Result= 00.4 ohms
(I used 00.4 just to clarify. Overkill perhaps. 0.4 would have been sufficient)
I tried another VOM and the results were precisely the same.

Yes, I'm positive I'm using the DC volts range.

Am I connecting the +5v and ground leads to the correct places?

Last edited by a moderator: Dec 19, 2017
8. ### panic mode Senior Member

Oct 10, 2011
1,640
454
are you sure you have correct polarity for supply? what is red output wire connected to? how about disconnecting it and trying measuring voltage just across voltage divider?

Lumenosity likes this.

Mar 1, 2017
398
40
10. ### Lumenosity Thread Starter Member

Mar 1, 2017
398
40
Hello Panic Mode,

I have +5v from a DC power supply going to the lead marked +5v
The Red output wire between the resistors is not connected to anything.
I just use it to test the output voltage.

Where should I place the leads to do that?

11. ### AnalogKid AAC Fanatic!

Aug 1, 2013
7,126
2,015
In that case, 1.79 V is exactly correct. 1.79 + 3.3 = 5.09 V. You are measuring across the wrong resistor.

With your meter set to the 20 volt range, connect one lead to the junction of the two resistors, and the other lead to the ground end of the 330 ohm resistor.

Also, a meter cannot measure its ow input resistance. What you measured in post #3 is the offset error of that range of the ohms function.

ak

Lumenosity likes this.
12. ### Lumenosity Thread Starter Member

Mar 1, 2017
398
40
When I measure across both resistors, I get 5v.
But when I measure across just the 180ohm resistor I get 1.79v

But if I measure across the 330ohm resistor, there is no voltage drop at all.
However, measured by itself, the 330ohm resistor measures 330ohms resistance?

It's as if they are doing nothing. I'm really cornfused.

13. ### Lumenosity Thread Starter Member

Mar 1, 2017
398
40

OK.
When I do this I get 3.3v

So I have 5v coming out of the power supply and 3.3v going back TO the power supply ground (negative) side?

Now I have to try to figure out what I've learned

14. ### ebeowulf17 Distinguished Member

Aug 12, 2014
2,608
495
Ideas:
1. User error
3. Meter error
There's no way you have 5V across the two resistors together, 1.79 across one of them, and nothing across the other.

15. ### ebeowulf17 Distinguished Member

Aug 12, 2014
2,608
495
Sorry, cross post there - I was replying to one post earlier.

16. ### Lumenosity Thread Starter Member

Mar 1, 2017
398
40
User error is a given (see avatar)
Post 11 seems to have pointed me in the right direction....but I still have to figure out where I'm going

Mrdouble, panic mode and ebeowulf17 like this.
17. ### Lumenosity Thread Starter Member

Mar 1, 2017
398
40
Thanks to everyone who replied.

It is REALLY appreciated

18. ### ebeowulf17 Distinguished Member

Aug 12, 2014
2,608
495
Nothing wrong with a little user error; that's usually part of the learning process (certainly has been for me!)

Lumenosity likes this.
19. ### WBahn Moderator

Mar 31, 2012
23,356
7,079
And therein lies you problem. If you want to measure the voltage across the 330 Ω resistor, then you need your voltmeter leads to be across the 330 Ω resistor. You are measuring the voltage across the 180 Ω resistor. In fact, with the black lead connected to 5 V and the red lead at the junction, you should be reading about -1.76 V. That's assuming that you have the leads plugged into your meter correctly (black to COM and red to VDC or similar labeling).

That's the resistance of the probe leads, not the resistance of the voltmeter.

No. Connect the leads across the component you want to measure the voltage across.

Lumenosity likes this.
20. ### AnalogKid AAC Fanatic!

Aug 1, 2013
7,126
2,015
Nope. Voltage doesn't "go to" something, current does.

In conventional terms, current leaves the power supply + terminal, goes through both resistors, and returns to the power supply - terminal. A critical truth of DC circuits sigh as this is that all of the current goes through all of the components all of the time. That current causes a voltage drop across each component in the loop, and the size of the voltage drop is directly proportional to the size of the resistance. Ohm's Law: E = I x R

With the sum of the two resistors you can use OL to calculate the current through the loop. With the current you can use OL to calculate the voltage drop across each resistor.

Last edited: Dec 19, 2017
Lumenosity likes this.