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Basic Electronics » Voltage Divider (Resistors)

 December 16, 2012 by GenHornet18 Hello, I'm currently working on interfacing a new graphics LCD with the ATmega168(LCD Datasheet. I don't have a lot of extra components sitting around (outside of the NerdKit), although I really wish I did now. The first challenge I am currently working on is stepping down the MCU voltage from +5V to the LCD typical voltage of +2.8V. Since I don't have any other voltage regulators (besides the 7805) I searched for a way to lower the voltage with the components I have at hand. This has lead me to the resistor voltage divider. The math mostly makes sense to me and all seemed to be well until I applied a load (the LCD) to voltage-out from the divider. Using a multimeter and checking the voltage all is well (V-out is approximately +2.8V) however when the load is connected the voltage drops (approximately +2.1V). Therefore I'm not actually sure if I fully understand the voltage divider (I had assumed irrelevant of load the voltage would still be +2.8V at V-out). Could someone enlighten me to how exactly this is supposed to work, or more so what I'm missing? The current circuit as it sits is essentially: `````` 1000 1000 333 +5V-----/\/\/-----/\/\/-----/\/\/----0V | |----LCD(Load)-----0V `````` P.S. The long term plan is to run the entire circuit around 3V (utilizing the MCU's internal oscillator), however I think this would work nice just as well and at bare minimum would be a good learning exercise for me. Thanks, Brad The load will always have an effect on the voltage levels on a voltage divider. This is because the load itself is also adding a component of resistance to the circuit. In the circuit that you have shown the resistance of the load (LCD) is in parallel to the 1333 ohm resistance made up of the two resistors in the divider. Placing resistors in parallel will result in a lower combined resistance across that portion of the circuit. The load itself effectively lowers the resistance on the right side of the divider, which results in a voltage level across the load closer to 0V, because a larger portion of the total resistance of the dividers is connected to the left of the load. The voltage divider is all about the ratio of the resistance on either side of the point being measured. The relative values of the resistors in the divider compared to the effective resistance of the load is important. The smaller the effective resistance of the load when compared to the values of the resistors in the voltage the more significant this effect will be. Thinking of an extreme example, if a load that is near 0 ohms was connected to a voltage divider it would bring that side of the divider to near 0 ohms. In this case the resulting voltage across the load would approach 0 volts. By using relatively small value resistor in the voltage divided the effect can be minimized, however this would also result in higher current draw thru the voltage divider, producing heat and requiring larger components in the circuit. If you were to make your voltage divider up using two 10 ohm resistors and a 3.33 ohm resistor the current draw across the divider would be 10 times what it is now. The voltages at any point on the circuit would be that same because the ratios would be the same. If you placed the same load in the same location on that divider the change in the voltage measured would be much smaller. With regards to the voltage to your LCD I would suspect that placing resistors in series with the LCD would act as current limiting resistors. Because the LCD is a semi conductor component this may not be the case I am certain other people will have a better handle on that idea. Darryl I noticed errors in my last post. What I meant to say was. If you were to make your voltage divider up using two 10 ohm resistors and a 3.33 ohm resistor the current draw thru the divider would be 100 times what it is now. Hey, Thanks Darryl for the insight. If I'm understanding correctly due to the addition of a load I've turned this circuit form being in series to being a combinatorial circuit. ie. `````` 1333 1000 |----/\/\/----| +5V------/\/\/---| |-----0V |----LOAD-----| `````` Therefore to apply Ohm's law correctly I would have to calculate the circuit as a parallel circuit existing within the series circuit. In order to do this I think I need more pieces to the puzzle, specifically I need to know something about the load. The datasheet for the LCD doesn't mention the resistance of the LCD and gives ranges for the current and voltage (I'm starting to see why voltage regulators are used more commonly for this purpose). How exactly would I go about calculating the voltage for this circuit with load (and therefore choose an appropriate resistance to obtain that voltage)? Thanks for any insight you can provide, Brad Brad - If you measured 2.1v with the load and 2.8 without the load at the same point, you should be able to calculate the resistance of the load from that information. I think it comes out to 1585 ohms. However this might not be constant and may depend on various factors. A variable voltage regulator would be the easiest approach, or possibly an op-amp. 3 diodes in series would drop .7v each and cut the 5v down to 2.9v (if that helps). Brad I agree with pcbolt, I also calculate the load resistance now to be 1584.7 ohms and it will not likely be constant. We should clrarify what voltages we are dealing with. You will have number of logic or signal connections between the LCD and the Micro as well as at least one power connection to the LCD. I believe you could deal with them differently. Since the signal lines should be very low current (very high resistances) a simple voltage divider should work, It may even be possible to drive the logic lines at 5V depending on the data sheet for the LCD. I have reconsidered my comment about a resistor in series with the LCD power, I am not certain that would work. pcbolt gave some better ideas, if you happen to have a few diodes even LEDs should work. Since the current draw from the LCD is quite low with about 1580 ohms and 2.8 volts. Hey, It appears this is more complex then originally thought. My original plan was to step down the voltage from +5V (running the MCU) to +2.8V (typical voltage for the LCD). Then using the +2.8V I would power the LCD and use transistors on the clock/data lines(serial interface) routed from the same +2.8V source. From what I'm understanding though my ideal solution of using a voltage divider (which worked nice in theory until a load was applied) doesn't seem to be so ideal. Just to ensure I understand why exactly: The LCD is variable in the current in requires (between simply running the logic controller, having the display on, etc.). This variable current draw would cause the voltage to be variable (and could theoretically fall out of range). Is that correct? Given my understanding above, it seems that this voltage divider won't work as expected (I do have some LED's from the NerdKit so I think I'll try a few of them in series). This makes me ponder the question; Where exactly is this type of voltage divider utilized? Would this simply be used when the load is constant (ie. LED, resistor)? Brad I believe the most likely use of a voltage divider is when the load is constant as you said and also if there is very little load as is the case in the data or signal logic connections. You could take a look at this link. voltage level link The thread may get a little off topic in spots, as they often do, but overall I think it covers a number of possible options There is a good discussion of voltage level differences and how to approach them. Noter has posted a picture of a voltage divided type logic level converter he put together near the end of the thread. Darryl Hi, Putting an LED in series worked fine and brought the voltage down to approximately +3.1V (low enough to be within the LCD's range). Later I'll see to using a more permenant solution (regulator, didoes, etc.). Thanks for all the help, Brad