I have been working on a project that is supplied by 8.4v battery. I then used the 7805 voltage regulator to get 5v and supply the electronics.

Later, I figured out that I need to have 3.3v along with with 5v to supply a different module in the project that operates on 3.3v. I looked for the LD1117V33 and a smiliar voltage regulators but I couldn't find any in my country.

I then decided to make a voltage divider as follows:


simulate this circuit – Schematic created using CircuitLab

Now, I've tested the output voltage and it was a perfect 3.3v.

Next, when I loaded the module.. it didn't power on!

I measured the voltage and it was around 1.5v. I came to realize that the voltage drop was due to the parallel resistance that was inserted with the 2Kohms which decreased the overall resistance and caused a voltage drop.

What solutions are there? Take in note that I do have the (Op amp 741), but I couldn't know for the sack of me to know the exact connection. I saw few schematics, but I was not certain and didn't want to take the chance and blow up the expensive module I have.

Could you help in giving the right way to make a buffer in parallel with the 2Kohms?

Edit: Is is possible to reverse engineer the R2 such that the overall parallel resistance = 2Kohms. Whihc will make the output voltage = 3.3v.

I know that after connecting my module the voltage drops to 1.5v! So I can know the exact resistance of the module, or I can simply find it using an ohm meter.

Edit: Based on a quick Matlab run with the following code:

close all
clear all

RL = 544; %This was calculated. Because the Vout = 1.5v after adding load.
%Then Rparallel = 428ohms Therefore Rl was found to be 544.

R1=[1:10:10000]'; %Vareying values for R1

for i=1:1:1000
    Rp(i,1) =  (33/17)*R1(i,1);
    R2(i,1) = -(Rp(i,1)*RL)/(Rp(i,1)-RL);


for i=1:1:1000
if(solution(i,1) <0 || solution(i,2) <0)
    solution(i,:) = 0;

Now when finding the solutions, The best as far as I see is at R1 = 271ohms and R2 = 15951 ohms.

Will this configuration work, or will it have any further effects or complications as voltage is linearly probational with current?

  • \$\begingroup\$ This business of trying to make a voltage divider with resistors and use it as a voltage source, is a classic lesson, and I am upvoting your question since you worked through it and found the limitations yourself. \$\endgroup\$ – gbarry May 7 '14 at 17:15
  • \$\begingroup\$ I have a question, is it possible to reverse engineer this? I mean to calculated the required voltage as a sunsitue to the 2KOHMS in a way that the overall voltage at the output will be 3.3v? Will this idea work? \$\endgroup\$ – Adel Bibi May 7 '14 at 17:26
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    \$\begingroup\$ Yes, in theory, but what you will find is that if the current drawn by your load changes, the voltage will change too. The "current drawn" changing is the equivalent of the load resistance changing. \$\endgroup\$ – gbarry May 7 '14 at 17:55
  • \$\begingroup\$ Please have a look at the last edit. If the current drawn was constant, would this do the trick for me? \$\endgroup\$ – Adel Bibi May 7 '14 at 19:58
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    \$\begingroup\$ Keep in mind other ways of getting 3.3v with common parts. 2 silicon diodes will produce a 1.4V drop at current draws over 5~10 mA, for 3.6V. If you need to drop it more, an extra germanium diode will drop it 0.2V more for 3.4V, well within the tolerance for any 3.3v device I can think of. The other option is a 3.3V zener diode. Or a any common adjustable regulator can do 3.3V without issues. USB car chargers are typically 12v in adjustable switching regulator that can be changed with two resistors. \$\endgroup\$ – Passerby May 7 '14 at 20:15

You said you wanted to make a buffer using 741 op amp. The 741 won't give you the 40mA you desire. A transistor on the output of the 741 will solve problem.

enter image description here

  • \$\begingroup\$ Do I have to supply the 741 with a -8.4v as well as the +8.4v? \$\endgroup\$ – Adel Bibi May 7 '14 at 17:15
  • \$\begingroup\$ No need for negative voltage. The 741 will operate well here in single supply of +8.4 volts \$\endgroup\$ – Marla May 7 '14 at 17:17
  • \$\begingroup\$ Thank you so much! I'll be upvoting your post. I will then try it out tomorrow. If everything worked fine and there were no good answers, I'll accept yours. \$\endgroup\$ – Adel Bibi May 7 '14 at 17:18
  1. To be able to use a voltage divider you need to power a device that draws a constant amount of current.
  2. The drawn current should be very low or else you'll lose power by heating the resistors (actually no matter how low your current is you'll still lose power).
  3. If your module is expensive, use a voltage regulator (it's cheap and safe) or a converter that has a reliable design with no risks of overvoltage.

Since you know V, have an idea of what the unknow load R3's current should be, you can figure it out.

Remember, Voltage is the same in parallel circuits, and current is the same in series circuits. that said, the current going through R1 had to be the same going through R2 plus R3. The Voltage going through R2 is the same going through R3.

So, using Ohm's law, if for R1 we have r 1000, and v of 3.5, then I must be 0.0035 or 3.5 mA. Obviously it's not working for 40mA load.

So knowing that I between r2 and r3 is 3.5 mA, and v is 1.5, Rt is about 428 ohms. And Rt for parallel resistors is 1 / ( (1/r2) + (1/r3) ). In this case, its about 0.9 k {maybe, I didn't do the math}. 1.5 volts and 900 ohms means 2 mA through the regulator and 1.5ma through r2.

That said, the total current through r1 will never be higher than V/R or 5/1000 or 5 mA... Adjust r1 and r2, solve for r3 where v is 3.3 and I is 0.045

  • \$\begingroup\$ I know my math is wrong in some places but I'll have to correct it later when I'm on a computer \$\endgroup\$ – Passerby May 7 '14 at 19:10
  • \$\begingroup\$ I've been trying to do the same exact math for a time now. All I'm getting are really small values of resistances! Most of them are not available. Almost of a short circuit. I will try to plot them in matlab to find a good solution, and post it here. Will be waiting your reply and edits. \$\endgroup\$ – Adel Bibi May 7 '14 at 19:12
  • \$\begingroup\$ @adelbibi by my math, for r1 dropping 1.7v and passing 45mA, it needs to be 37 ohms, with R2+R3 being a combined 73 ohms. If R2 is sized for 5mA and your load is a constant 40mA, that means 3.3 / 0.005 = 660 ohms. R3, your load, is basically a 82 Ohm resistor. If you can't find exact values, you need to combine resistors. That's another con about voltage dividers as regulators. But I think 36 and 660 are common enough, or easy to find in a pack of standard 10% resistors. \$\endgroup\$ – Passerby May 7 '14 at 20:02
  • \$\begingroup\$ Sorry Mr @Passerby I've just checked I mistakenly said 40mA. It is not specfiied exactly the current drawn by the module. So the part of 40mA is not reliable. But the voltages are exactly as I mentioned earlier. 3.3v before connecting the load and 1.5v after connecting the load. That's all the 100% correct information I have regarding the module. \$\endgroup\$ – Adel Bibi May 7 '14 at 20:06
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    \$\begingroup\$ @adelbibi a resistor divider only works well when exact current draws are known. Otherwise it will not work. If the module isn't too picky about its input voltage, two silicon diodes in series will drop 5v to 3.6v. \$\endgroup\$ – Passerby May 7 '14 at 20:22

Adel Bibi - It's good you figured out what you did wrong. What you now need is a source of 3.3 volts. Since you are already using a linear regulator to get 5 volts, and were willing to accept the power losses associated with it, the easiest approach is to use an LM317 to regulate 12 volts to 3.3 volts.

If you don't want to lose the power, use a buck switching converter to produce the 3.3 volts. These are available on eBay for very cheap money, already manufactured and ready to go.

  • \$\begingroup\$ The problem is that I have a very limited time to finish it. The components are not available in where I live! I will have to utilize the available resources such as the OpAmp 741 as a buffer. Won't this work? \$\endgroup\$ – Adel Bibi May 7 '14 at 16:11
  • \$\begingroup\$ Marla's answer will probably work, but it may not. The problem is that the 741 is only specified to work at supply voltages of 10 volts total. At 8.4 volts it may well work - but not necessarily. \$\endgroup\$ – WhatRoughBeast May 7 '14 at 18:33
  • \$\begingroup\$ That is a problem/ It's kinda weird how a small problem is pulling 3.3v out of 5v can be that complicated. \$\endgroup\$ – Adel Bibi May 7 '14 at 19:37
  • \$\begingroup\$ @ WhatRoughBeast : 741 specifications for maximum voltage is + and - 18 volts. ti.com/lit/ds/symlink/lm741.pdf \$\endgroup\$ – Marla May 8 '14 at 16:00

You could use another linear regulator either in parallel or series with the 7805 (depending on how much current you need to draw). The link below is a 3.3V linear regulator that is pretty cheap, and can supply 500 mA.


I had success using the above regulator using a 12 V source.

  • \$\begingroup\$ The problem again is I have no access to similar regulators. If I wanted any of them I will have to order them from outside the country, and I have a very limited time. The only way is to connect a buffer. Shouldn't it work? if yes, could you provide a schematic for the connection? \$\endgroup\$ – Adel Bibi May 7 '14 at 16:13

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