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Voltage divider, High Ohm resistors vs. Low Ohm

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burners

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It seems after reading 2 electronics books and lots of internet researching I still haven't had a definite answer to this question.

What are the differences when building a voltage divider and using high ohm resistors vs. low ohm.

Example 1 Low Ohm Resistors
Input: 14v
R1:180
R2:100
Output:5v

Example 2 High Ohm Resistors
Input: 14v
R1:18k
R2:10k
Output:5v

Which is preferred and why?
 
It seems after reading 2 electronics books and lots of internet researching I still haven't had a definite answer to this question.

What are the differences when building a voltage divider and using high ohm resistors vs. low ohm.

Example 1 Low Ohm Resistors
Input: 14v
R1:180
R2:100
Output:5v

Example 2 High Ohm Resistors
Input: 14v
R1:18k
R2:10k
Output:5v

Which is preferred and why?

Example 1 circuit current as it stands is 50 mA

Example 2 circuit current as it stands is 50 uA

So this brings us to the intended application of the divider? You build a divider based on the intended application for the divider. Yes, stand alone as drawn they both output 5 volts but what happens to each when additional circuits are added to that 5 volt output?

Ron
 
Example 1 circuit current as it stands is 50 mA

Example 2 circuit current as it stands is 50 uA

So this brings us to the intended application of the divider? You build a divider based on the intended application for the divider. Yes, stand alone as drawn they both output 5 volts but what happens to each when additional circuits are added to that 5 volt output?

Ron

I think my goal is to get as much current out as possible, so it sounds like your saying I should go with the low ohm resistance to get the most current correct?

Thanks
 
Low ohm will distort less if current is being sourced by the divider due to the current being drawn from the divider is a smaller percentage of the total current in the divider. For the proper voltage ratio, currents in the top and bottom resistor must be equal and drawing current out from the center-tap will reduce current available in the lower resistor. The net effect is that as more current is drawn from the divider, it's output voltage will sag more and more in proportion.

To get a voltage that doesn't droop by more than 5%, the current in the rest of the divider must be 20x higher than the current being drawn from it. That's a lot of wasted power and a lot of heat so you need larger power resistors and the voltage will sag more the more current is drawn. And the worst part is that the power is wasted and heat is generated even when small amounts of current are being drawn since most of it is just flowing through the resistors to ground. For example, for 100mA, you need 2A in the rest of the divider. If you want 1% droop, then that's 10A. Not a great way to go about things.

If you really want no distortion, you should just a high resistance divider and buffer it with an op-amp buffer. Then you have a steady voltage output with no distortion from current draw. The voltage will be much more stable with current draw and heat will be reduced with reduced current draw.
 
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As an addendum to dknguyen's post:

The equivalent resistance of any voltage divider is Req = (R1 x R2) / (R1 + R2). This equivalent resistance appears in series with any output load and will reduce the divider output voltage by a factor of Req x Iout.
 
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You need to stop been so vague, and tell us exactly what you're trying to do.

Why is it that everyone assumes that when a question is asked it is for a production / final circuit design and not just for knowledge or testing?

I have a small project where I am sort of reverse engineering the old Apexi SAFC fuel controller for fuel injected cars. Its more of just curiosity in the sense of @ 3k RPM with a setting of +3 how much does the Apexi adjust the incoming signal.
So the Apexi can read several sources including RPM, Map, TPS, Hot Wire, Flap which are mostly 0-5v but can be up to 12volt.
The project is very temporary since this only needs to be done once, that is why I am interested in something simple and cheap as long as its accurate.
The way the project is setup is I use 1 analog output which scales from 0-5 volt and that loops back into the micro for a reading, this reading is sent via serial to a PC to later be parsed.
The other output from the same pin goes through the Apexi SAFC and back into the micro which is then sent via serial to the PC.
I also have a hard drive with 4 magnets and a magnetic pickup which simulates RPM, I use 4 magnets so I can get a simulated 8k RPM when the drive is spinning at 2k. I adjust the speed of the drive with external 12volt source controlled by the micro which it reads the rpm and then sends it to the PC via serial.
Both of these readings are compared in Excel with a nice Line graph showing the difference at 0-5v in 255 increments over 0-8000 RPM.
I know most of the signals the Apexi reads are 5 volt but I know some are 12 volt so I was just getting a game plan in place for reading a 12 volt signal.
Once I am done with the project I will share the results on my website but I currently have no desire to re-make the Apexi SAFC since I already own one :)
 
Why is it that everyone assumes that when a question is asked it is for a production / final circuit design and not just for knowledge or testing?

I made no such assumption, nor did I imply it.

Your question was so vague as to be meaningless, it's like asking how long a piece of string needs to be - it depends entirely on the circumstances, exactly as potential dividers do.

I have a small project where I am sort of reverse engineering the old Apexi SAFC fuel controller for fuel injected cars. Its more of just curiosity in the sense of @ 3k RPM with a setting of +3 how much does the Apexi adjust the incoming signal.
So the Apexi can read several sources including RPM, Map, TPS, Hot Wire, Flap which are mostly 0-5v but can be up to 12volt.
The project is very temporary since this only needs to be done once, that is why I am interested in something simple and cheap as long as its accurate.
The way the project is setup is I use 1 analog output which scales from 0-5 volt and that loops back into the micro for a reading, this reading is sent via serial to a PC to later be parsed.
The other output from the same pin goes through the Apexi SAFC and back into the micro which is then sent via serial to the PC.
I also have a hard drive with 4 magnets and a magnetic pickup which simulates RPM, I use 4 magnets so I can get a simulated 8k RPM when the drive is spinning at 2k. I adjust the speed of the drive with external 12volt source controlled by the micro which it reads the rpm and then sends it to the PC via serial.
Both of these readings are compared in Excel with a nice Line graph showing the difference at 0-5v in 255 increments over 0-8000 RPM.
I know most of the signals the Apexi reads are 5 volt but I know some are 12 volt so I was just getting a game plan in place for reading a 12 volt signal.
Once I am done with the project I will share the results on my website but I currently have no desire to re-make the Apexi SAFC since I already own one :)

As you've now made it a little less vague, you need to consider the source and load impedances (or resistances) - you may well find a simple series resistor (of the correct value) is all that you need (and works better).

For a potential divider the 'rule of thumb' is to pass at least five times the current through it that the load needs.
 
Why is it that everyone assumes that when a question is asked it is for a production / final circuit design and not just for knowledge or testing?

I don't think that is quite what it is. People just feel the more they know about what you are looking to do, the better they can help you. This is why in my initial post I mentioned that a divider circuit is built around the intended application.

Actually what you have is a data acquisition device that likely does an A/D conversion. The sensor feeds this device and as you mentioned the voltages can vary. I screw around with several variations of this device for example. If you scroll down that page and note the specifications for the Analog Input you will see it has an input impedance of 200 KΩ. The input voltage range is 0 to 10 Volts.

So for example if I have a sensor output of 0 to 14 volts I need to accurately scale it down. Now in your example you are scaling 0 to 14 volts = 0 to 5 volts. However, keep in mind that your A/D input device impedance will be in parallel with the lower leg of your divider (your 0 to 5 volts output). That changes everything as to my (or your) divider ratio. You don't want inaccurate data right?

This brings us back to the divider being designed for the intended application. You want your divider to be accurate.

<EDIT> As usual Nigel was quicker on the keyboard than I. :) </EDIT>

Ron
 
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Yeah I hear what you guys are saying but the reason I wasn't being specific was because I didn't want a specific answer. I didn't want to know which voltage divider was preferred for my exact application. I kinda wanted to be more broad and hear / read what type was better for different applications including my own.

I think you guys pretty much summed it up well though and I appreciate that, Thanks again
 
Wel there you have it:

full-time energy/heat dissipation vs voltage droop/sag
 
It seems after reading 2 electronics books and lots of internet researching I still haven't had a definite answer to this question.

What are the differences when building a voltage divider and using high ohm resistors vs. low ohm.

Example 1 Low Ohm Resistors
Input: 14v
R1:180
R2:100
Output:5v

Example 2 High Ohm Resistors
Input: 14v
R1:18k
R2:10k
Output:5v

Which is preferred and why?


The answer is "it depends". Consider the circuit of your voltage divider. This is the diagram of the basic voltage divider.

Please view in a fixed-width font such as Courier.

+12V------+
|
|
+++
| |R1
| |
| |
+++
|
+++
| |R2
| |
| |
+++
|
-+-
-
You can calculate the voltage at the R1-R2 junction with Ohms law.


Now, consider what happens when you connect something to the divider output, which is shown as Rload.

+12V-------+
|
|
+++
| |R1
| |
| |
+++
+-----+
+++ +++
| |R2 | |Rload
| | | |
| | | |
+++ +++
| |
-+- -+-
- -

Now, you must recalculate the value of the bottom part of the divider, which is now R2 in parallel with Rload. After recalculating, you can again use Ohm's law to calculate the divider output, which is the voltage that Rload will feel.

So, you can see why there is no absolute answer to your question... it all depends on what you want to connect to it.

Hth,
Dave M
 
As has already been stated here before, it all depends on what you're trying to do.

Now...a voltage divider comprised of low resistance resistors will draw more current from the supply than one comprised of high resistance resistors. This is the actual current flowing through the divider itself. This current must be subtracted from the total available from the supply.

The upside of voltage dividers comprised of low resistance resistors is that they're harder to load them. The downside is they draw more current from the supply. Vice versa for a voltage divider comprised of high resistance resistors.

Every time you add a branch to the supply, total current draw from the supply increases because each branch drops the total circuit resistance. If you drop it so far as to tax the available current supply, you're overloading the supply.

Now...when you place a resistor between the positive side of a circuit and the positive rail you form a voltage divider. The circuit itself becomes the "bottom resistor" in the divider. The amount of voltage dropped across said resistor is equal to the product of the current drawn by said circuit and the resistor's value in Ohms (Ohm's Law...E = I x R).

However, when you add a circuit across the bottom resistor in a voltage divider, you decrease its effective value. The effective value it drops to is determined by the total circuit resistance and the value of the bottom resistor itself. Adding the circuit across the bottom divider resistor is electrically equivalent to adding a 2nd resistor in parallel with the bottom resistor. This increases current draw through the top resistor, which in turn increases the amount of voltage dropped by the top resistor, which in turn decreases the amount of voltage available at the divider's output.

Therefore, to keep the voltage output the same when a circuit gets added, you must drop the value of the top resistor in the divider to allow more current through without dropping more voltage across the top divider resistor.
 
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