the art of electronics, voltage dividers and intuition

Thread Starter

ninjaman

Joined May 18, 2013
324
hello
i have, learning the art of electronics and the art of electronics books. i am reading the parts about voltage dividers and i am confused about the resistance readings that your supposed to get with meters that measure the outputs and how they load the circuit.
it talks about voltage dividers, for instance a 20v source with two 15k resistors. when the source is seen as zero ohms or ground the resistors are in parallel and half of one resistor which is 7.5k. the output voltage is half the input if both resistors are equal. thevenin equivalent circuit should be the output voltage as source with the parallel resistance as thevenin resistance. i have simulated this in multisim and understand it. but then it goes onto adding a meter which loads the circuit. it has a number of dividers all with the same input and output but different thevenin resistances. then you have to take into consideration the tolerance of the resistor, here it is 1%. the meter is put on the output and a voltage is taken. where the voltage is different than expected it is the loading of the meter that causes it. with a 10v source and 8 volts at the output there is 2 volts across the internal resistance. this should allow for calculation of the internal resistance which should be in proportion to the voltages. 8 volts across the thevenin resistance and 2 volts across the internal resistance, 8 / 2 = 4. so 4 times the thevenin resistance should give you the internal resistance.
how do you get to the point where you can look at the circuit and tell without calculating what the result is? is there a better explanation of this?
i dont like the idea that i have chosen a hard subject to learn, maybe a chose a poor quality book or the example is not clear enough? does anyone else feel that the art of electronics (student manual third edition) is a little tricky?
 

crutschow

Joined Mar 14, 2008
25,250
Not quite sure what you are asking. :confused:
Is this about how to determine if meter loading is a significant factor in your voltage measurements?
 

panic mode

Joined Oct 10, 2011
1,845
duplicate your circuit.... then remove all sources: replace all voltage sources by short circuit, replace all current sources by open circuit. compute equivalent resistance ...
 

wayneh

Joined Sep 9, 2010
16,399
Not quite sure what you are asking. :confused:
Ditto. I don't get the question.

As a practical matter, a modern multimeter has such a high impedance (my cheap Harbor Freight meter is at least 1MΩ, I believe I've heard a Fluke is 10M) that it does not significantly distort voltage divider readings, compared to other error sources, until your divider resistors are >~100kΩ.
 

hp1729

Joined Nov 23, 2015
2,304
hello
i have, learning the art of electronics and the art of electronics books. i am reading the parts about voltage dividers and i am confused about the resistance readings that your supposed to get with meters that measure the outputs and how they load the circuit.
it talks about voltage dividers, for instance a 20v source with two 15k resistors. when the source is seen as zero ohms or ground the resistors are in parallel and half of one resistor which is 7.5k. the output voltage is half the input if both resistors are equal. thevenin equivalent circuit should be the output voltage as source with the parallel resistance as thevenin resistance. i have simulated this in multisim and understand it. but then it goes onto adding a meter which loads the circuit. it has a number of dividers all with the same input and output but different thevenin resistances. then you have to take into consideration the tolerance of the resistor, here it is 1%. the meter is put on the output and a voltage is taken. where the voltage is different than expected it is the loading of the meter that causes it. with a 10v source and 8 volts at the output there is 2 volts across the internal resistance. this should allow for calculation of the internal resistance which should be in proportion to the voltages. 8 volts across the thevenin resistance and 2 volts across the internal resistance, 8 / 2 = 4. so 4 times the thevenin resistance should give you the internal resistance.
how do you get to the point where you can look at the circuit and tell without calculating what the result is? is there a better explanation of this?
i dont like the idea that i have chosen a hard subject to learn, maybe a chose a poor quality book or the example is not clear enough? does anyone else feel that the art of electronics (student manual third edition) is a little tricky?
Re: Is there an easy way... (If I understand your question)

Look at the ratio between the two resistors. The voltage will divide accordingly. If one resistor has twice the resistance of the other it will have twice the voltage (2/3 of the applied voltage).
Be mindful of other parts in the circuit that could have an influence.
 

Jony130

Joined Feb 17, 2009
5,180
Look at the ratio between the two resistors.
1a.png

And note that if R1 = R1 Vout is 1/2Vcc but if R1 = R and R2 = 2R we have Vout = 2/3Vcc
 

shortbus

Joined Sep 30, 2009
7,840
Not wishing to hijack the thread but have a similar question on the subject. How do you determine how large of a value of a resistor to use in a voltage divider? Like for scaling the voltage to work with a comparator. when playing with a voltage divider calculator program there are many different values that give the same/needed result but the due to the resistor 'ratio', but what is the "best" value?
 

AnalogKid

Joined Aug 1, 2013
8,534
how do you get to the point where you can look at the circuit and tell without calculating what the result is?
Practice and experience. The only way to get an answer with 3-digit accuracy without a calculator is to be able to do 4-digit arithmetic in your head, but you can make reasonable estimates based on reasonable assumptions. For example, if you see a voltage divider with 22 K and 100 K resistors, a quick estimate is to round the 22 K down to 20 K or up to 25 K, so the result is somewhere between 5:1 and 6:1.
is there a better explanation of this?
Not really. Ohm's Law reduced centuries of speculation, theory, and experimentation to a simple multiplication. It doesn't get any more simple than that. Many large circuits, when analyzed, can be broken down into a series of relatively simple calculations that build on one another to the final result.

ak
 

dl324

Joined Mar 30, 2015
11,240
How do you determine how large of a value of a resistor to use in a voltage divider? Like for scaling the voltage to work with a comparator. when playing with a voltage divider calculator program there are many different values that give the same/needed result but the due to the resistor 'ratio', but what is the "best" value?
You want the voltage divider to be "stiff" enough to not be "loaded" by whatever it's driving, but not so stiff that it wastes significant power unnecessarily.

An LM393 comparator has a worst case input bias current of 250nA. A "stiff" divider is usually considered to be 10X the load current, so 2.5uA would be the minimum current you would design for. With a 5V supply, total divider resistance would be 2MΩ. When I need VCC/2, I often use 2 51K 1% resistors because the divider will be stiffer, additional dissipation is negligible, and I have a reel of them.
 

dl324

Joined Mar 30, 2015
11,240
but then it goes onto adding a meter which loads the circuit. it has a number of dividers all with the same input and output but different thevenin resistances. then you have to take into consideration the tolerance of the resistor, here it is 1%. the meter is put on the output and a voltage is taken. where the voltage is different than expected it is the loading of the meter that causes it. with a 10v source and 8 volts at the output there is 2 volts across the internal resistance.
You'll get better answers if you post a specific example.
 

hp1729

Joined Nov 23, 2015
2,304
Not wishing to hijack the thread but have a similar question on the subject. How do you determine how large of a value of a resistor to use in a voltage divider? Like for scaling the voltage to work with a comparator. when playing with a voltage divider calculator program there are many different values that give the same/needed result but the due to the resistor 'ratio', but what is the "best" value?
Choose a current value. Too high draws too much current. Too low might get upset by local resistances. Do the resistor values come up close to standard values? If not change the current value.
 

GopherT

Joined Nov 23, 2012
8,012
Not wishing to hijack the thread but have a similar question on the subject. How do you determine how large of a value of a resistor to use in a voltage divider? Like for scaling the voltage to work with a comparator. when playing with a voltage divider calculator program there are many different values that give the same/needed result but the due to the resistor 'ratio', but what is the "best" value?
it is a balance between...
- acceptable current loses because low value resistors allow more amps to flow,
- noise generated by thermal effects

- slower switching speeds when high-value resistors are used (to overcome input capacitance)

A total resistance of the voltge divider in the range of 5k to 50k seems common for Op Amps and Comparitors for audio circuits with Bipolar input stages and 10k to 100k or more for JFET inputs. These values assume a stable voltage reference (e.g. virtual ground). Lower values are used if the "reference" is a constantly changing value.
 

wayneh

Joined Sep 9, 2010
16,399
Pretty much the same as already noted.
• If you need to provide regulated power to something, use a different method. Dividers are best for conditioning information signals, not power.
• Estimate the current load by whatever it is that you want to see the target voltage. Hopefully in the µA range or less, like a comparator.
• Approximate the combined resistance of the two resistors to give a current thru the divider of at least 10x the load current.
• Tweak the ratio of the two values to get to your target.
• If the divider current is not well over 10X the load, consider calculating the exact resistor values by including the load current distortion.
• If it really matters, sort thru the resistors you have to find one lower or higher, to further dial in - by testing - what you need.​
 

crutschow

Joined Mar 14, 2008
25,250
Not wishing to hijack the thread but have a similar question on the subject. How do you determine how large of a value of a resistor to use in a voltage divider? Like for scaling the voltage to work with a comparator. when playing with a voltage divider calculator program there are many different values that give the same/needed result but the due to the resistor 'ratio', but what is the "best" value?
The equivalent resistance of the divider should be such that any current drawn by the load will cause a voltage drop in that equivalent resistance less than any voltage error you can tolerate.

For example, for the comparator example of 250nA bias current, and if you wanted a maximum error of 10mV, then the equivalent divider resistance should be no more than 10mV / 250nA = 40kΩ.
Thus if you wanted ½ source voltage, you could use two 80kΩ (or smaller) resistors in series.

An interesting note is that, for an arbitrary voltage divider ratio, some 1% resistor selections give you closer to the desired ratio than others.
To determine the optimum values I wrote a short Visual Basic program years ago to try all the pairs of resistors near a given desired ratio and pick those with the best match.
Here's a website that performs a similar calculation.
 

shortbus

Joined Sep 30, 2009
7,840
Thought of another question pertaining to this. Is the correct resistance to use the total resistance, R1 + R2, or do you just worry about R1?
 

crutschow

Joined Mar 14, 2008
25,250
Thought of another question pertaining to this. Is the correct resistance to use the total resistance, R1 + R2, or do you just worry about R1?
The "correct" resistance, when referring to the effect of any current drawn from the divider, is the equivalent output resistance, which is the value for the two resistors in parallel or (R1*R2) / (R1+R2).
 
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