# Voltage divider 101 help, any suggestions?

Alright, I will try and be brief with this, I have created a voltage divider network displayed below in the picture. I am using a one 100k resistor and one 68K resistor because I used this voltage divider calculator to determine the values of the resistors. I have input voltage of 5VDC with a current of 500mA coming in on the red wire on the right side of the bread board, and the black being the GND. I need a fairly accurate output voltage of 2VDC going across the D+ wire located on the left (bottom) portion of the picture. I also need 2VDC going across the D- wire located to the right of the white wire. The D+ coming in from the top also supplies 5VDC according my multimeter, so I decided to setup the same voltage divider network using the 100K to + 5VDC and 68K to GND (D-) I am getting really close to 2VDC on the D+ (white) wire on the vout portion of the voltage divider network, but I am getting 3VDC on the D- with the same setup. I do not understand how I can be getting 3VDC? I do not need 3VDC but rather 2VDC. I have tried several different resistor scenarios but have been unable to get 2VDC on both the D+ (white) and 2VDC on the D- (blue). If anyone has any thoughts/suggestions on this subject I would greatly appreciate it. Also these four wires are being feed into a four wire plug, and I am taking these reading from the other side of the plug. The second picture describes what I am talking about.

What is your green wire marked as D- connected to. One end of the 68k resistor appears to be connected to it but where does the other end of the wire actually go to. This end needs to go to your ground reference point so that the 100k/68k connection point (blue wire) can develop the correct 2volts.

You also need to establish how you are measuring the 3 volts. I presume your +ve lead is connected to the 100k/68k junction (blue) but where does your -ve lead connect (the other end of the 68k ?)

When doing breadboarding like this you should draw out your circuit and mark exactly what wire connects to where. That way, when you build your breadboard you have a means of verifying that you have connected the components correctly.

By the way your 5 volt supply is not providing 500ma to the circuit. When a power supply has a current rating specified it simply means it is capable of supplying that amount of current

And finally, if the circuit that you are connecting your 2 volts to draws any current then your 2 volts will drop since the external load is effectively in parallel with your 68k resistors. You can minimise this effect by reducing the source resistance. Simply change your resistor chains by a factor of 10 or even 100 so that you have 10k plus 6.8k or even 1k plus 680ohms. With the latter components the power dissipation across the resistors will be minimal (0.025watts at worst) the loading effect of your external circuit will be negligible.

jack

I don't calculate if your divider has correct values, but there is a huge problem; the current is too high. This kind of divider work with very low loads only. You need to understand that your load needs to added as a parallel resistor for the divider, so your circuit changes drastically when load is connected.

When this kind of divider is the only way to get suitable voltages, there needs to be power supply that can handle the total load, and some kind of mechanism to keep partial voltage stable, like zener diode. Not recommended, makes heat, and your resistors are taking lots of the power and will burn.

You say the current is 500mA? It is not possible with that voltage. Only the first resistor 100k with 5V would take I=U/R=5V/100000=0.00005A=0.05mA

What is the device that is connected at the end of the line?

Cheers,
Kari

5VDC with a current of 500mA coming

I think you have made a slight mistake. Your supply is probably 5Vdc and can PROVIDE (source) 500mA. However, the amount of current that flows is dependent on the load.

In the case of your voltage divider, your total resistance is 168Kohm. So 5V / 168Kohm = 29.76uA. Using ohm's law you can verify that 29.76uA flowing through your 68Kohm resistor is equal to 2.024V.

Why is your setup are you using 2 of each resistor?

You connection would be simple: +5V --- R1 ---*--- R2 --- GND.

R1 = 100KOhm
R2 = 68KOhm

The voltage present at the * (to GND) would be 2V.

jackrae:
What is your green wire marked as D- connected to.

The D- wire is connected to a plug that I have salvaged from a sound card, identical to the one displayed in the second picture. (Not sure exactly what the style of connector is called), but then the male portion of the connector is soldered to four wires which ties into a USB standard Type A plug.

jackrae:
One end of the 68k resistor appears to be connected to it but where does the other end of the wire actually go to.

The four wires (red, yellow, black, & green) plugged in to the far right of the left side of the bread board are connected to a USB standard A plug.

jackrae:
You also need to establish how you are measuring the 3 volts. I presume your +ve lead is connected to the 100k/68k junction (blue) but where does your -ve lead connect (the other end of the 68k ?)

The four wires (red, yellow, black, & green) on the right portion of the picture are connected to USB standard A plug, which draws power from a MacBook Pro. The MacBook Pro provides +5VDC on the red wire and yellow wire, and the black corresponds to the red, and green corresponds to the yellow. If I understand correctly both the black and green wires act as grounds for the circuit.

jackrae:
When doing breadboarding like this you should draw out your circuit and mark exactly what wire connects to where. That way, when you build your breadboard you have a means of verifying that you have connected the components correctly.

I originally tried following this schematic but for whatever reason I could not get 2VDC on the white (D+) and 2VDC on the (D-).

jackrae:
By the way your 5 volt supply is not providing 500ma to the circuit. When a power supply has a current rating specified it simply means it is capable of supplying that amount of current

Yeah I understand, if the load (iPod Touch 2G) on the circuit does not require 500mA (the max amperage a USB port from a computer can provide) then the USB port from the computer will not provide the max current, but that the USB port has the capability to provide up to 500mA of current to a load on the circuit. I guess I should have been a little more clear with this, not that 3AM posting is any excuse.

Why is your setup are you using 2 of each resistor?[/quote]

I am using two of each because I want ~ 2VDC traveling to the D+ (white) wire and I want ~ 2VDC traveling to the D- (blue) wire.

Is it possible to divide the voltage from the red (+5VDC) and the black (GND) to two separate wires and provide ~ 2VDC to each wire?

I originally tried following this schematic but for whatever reason I could not get 2VDC on the white (D+) and 2VDC on the (D-).

If wired as shown in that schematic, you will get 2.0V. If you didn't, what did you get?

Have you connected the free end of the green wire to ground (or where). The schematic you used is correct (using your resistor values) so there is no reason why you aren't getting the correct voltage unless a) it's not breadboarded correctly b) not wired correctly or c) component values are not as stated. Since you can measure the voltage (by a meter I hope) then you can use the same meter to verify the component values. Commercial resistors are almost 100% correct (within their tolerance band) but it is possible that the 100k or the 68k are faulty.
jack

Well, I do not have a 75K ohm resistor, nor do I have a 50K ohm resistor (75 + 50 = 125) but I do have a 82K ohm resistor that I am substituting for the 75K and a 47K resistor that I am substituting for the 50K (82 + 47 = 129) gives me a difference of 4K ohms. Below is a picture of the current voltage divider network I have setup, and I am using two 82K ohm resistors and two 47K ohm resistors to divide the voltage between the +5VDC (red) wire and the GND (black wire) to the two wires (white) and (blue). As I just measured the voltage, I am getting 3.4VDC on the D- (blue) and I am getting 1.8VDC on the D+

When I take a VDC reading from the wires I hook up my red probe (+) from my multimeter to the red wire on the circuit, and I hook up my black probe (COM) to the GND (black) wire. The multimeter will read 5VDC. When I want to measure the voltage of the D+ (white) wire I hook up my red probe to the white wire and my black probe to the (black) GND wire. When I want to take a voltage reading of the D- (blue) wire I hook up my black probe to the blue wire and my red probe to the red wire.

With my current setup, as displayed in the picture, the multimeter is displaying 1.8VDC on the D+ wire and 3.4VDC on the D- This is obviously not what I need, and I need 2.0VDC on both the D- and D+

What happens to your voltages when you disconnect the connector to the podbreakout device

For me it is impossible to see, what you are trying to do... where do you need that 2V, and how much load there is going to be?

Can you draw a picture of your current circuit + what you about to connect to it?

Kari

jackrae:
What happens to your voltages when you disconnect the connector to the podbreakout device

not sure, I haven't tried that, but I will definitely give that try. What is the reason/logic for disconnecting the PodBreakout v1.5?

GaryP:
For me it is impossible to see, what you are trying to do... where do you need that 2V, and how much load there is going to be?

Can you draw a picture of your current circuit + what you about to connect to it?

Kari

SERIOUSLY?

For me it is impossible to see, what you are trying to do... where do you need that 2V, and how much load there is going to be?

The original poster is trying to build a charger for their iPod.

Can you draw a picture of your current circuit + what you about to connect to it?

They have already done that and explained multiple time what they are doing. If you would like more information check out this page: MintyBoost - The mysteries of Apple device charging

(82 + 47 = 129)

5V / 129K = 38.76uA

82K * 38.76uA = 3.18V
47K * 38.76uA = 1.82V

You are seeing 1.8V on your D+ connection because that is what voltage your divider has created. The 3.4V on D- is because your voltage divider is backwards. Swap the position of the two resistors you are using, and you'll see 1.8V on D- as well.

If you want to get closer to 2V, you'll need to fix your divider.

Another way to Calculate:
Vout = Vin * (R2 / (R1+ R2))

Vout = 5 * (47 / (82 + 47)) = 1.822V

Is it my turn to ask; "SERIOUSLY"? iPod was mentioned in reply #4 without extra information. I don't know anything about iPod, so it was quite hard to understand what that spider web was about to be...

Mintyboost...? Sounds like fun...

Cheers,
Kari

James I really appreciate the information on the subject matter, as you can tell my knowledge is quite limited. I did manage to switch the resistors on the D- and GND. I now have the 82K resistor going from the GND to the D- and the 47K resistor going from the +5VDC to the D-.

I took a new voltage reading and now I am getting ~ 1.830VDC - D+ and ~ 2.080VDC on D-

James one more thing, if I understand voltage divider networks correctly, according to the wiki article the R1 or Z1 is the resistor next to the +V source and R2 or Z2 is the resistor connected/next to the GND, no?

Oh, how the fun never stops. I managed to get ~ 2VDC on both D+ and D- I used a Rt resistor of 68K and a Rb resistor of 47K. This setup game a Vout of 2.04 with a Vin of ~ +5VDC. I ended up with the same Vout of 2.04 with the resistors flipped, e.g. I used a 47K resistor for Rt and a 68K resistor for Rb. I began to assemble an assembly, and tested all the points, but for some reason I am getting +5VDC when I connect my leads to D- and +5VDC (pin 23) on the PodBreakout v1.5. I am little confused at this point, and tired to say the least so I am going to take a break and hopefully try and figure this problem out in the morning.

Wow, that article was too technical for me to read.

But hey, why don't you make it easy way;

5V is 100% of the voltage. If you need 2V out of that, it is 40%, right?

Find resistors with ratio 40% / 60% and you should get as close to 2V and 3V than those resistors tolerances are, and 5V is what it is expected to be.

40k /60k is as good as 49,9k / 75k

This is probably stupid question, but do you know Ohms-law?

Cheers
Kari

GaryP:
This is probably stupid question, but do you know Ohms-law?

No such thing as stupid questions but stupid people (that's a joke, so don't take it personally). As for knowing Ohm's law, yes. Do I know how it can be applied to resistors in a voltage divider network/circuit, not the best, but I do have a basic understanding of how Ohm's law works. I used just the other day to understand what was going on with the garbage disposal.

And as for the question you asked in an earlier post here is a diagram of the wire harness assembly I have assembled to try and get +2VDC on both D+ and D- circuit.

Finally, I do appreciate the constructive comments about the resistor ratio to develop the proper voltage.