Connect with us

Beginner's question -- voltage drop

Discussion in 'Electronic Basics' started by Rheilly Phoull, Feb 26, 2004.

Scroll to continue with content
  1. R2 is part of the load, so if there is no load there will always be a
    minimum volt drop across R1 R2. As the load increases the volt drop on R1
    will increase. This circuit is called a "Voltage divider" and is OK for
    small loads.
    Regards ........ Rheilly Phoull
  2. electricked

    electricked Guest

    What's the difference between the following:

    | |
    | + |
    --- Vcc O LOAD
    - |
    | - |

    and this one:

    | |
    | + \ R1
    --- Vcc /--------
    - \ O LOAD
    | - / R2 |

    How is that different from the first circuit. Say Vcc is set at 9V. Make R
    in the first circuit drop 4V. That means 5V are left for the load. In the
    second one, R1 drops 4V, Right? So what's the R2 for?

    Please explain.


  3. James W

    James W Guest

    First, I assume you understand that voltage drop across a resistor is
    caused by the current flow through the resistor.(V=IR)

    Now, consider what would happen in your second circuit if the value of
    R2 was ZERO ohms... i.e. it is a piece of wire.

    The Voltage across ZERO ohms is ZERO volts.(again, V=IR where R=0, gives
    V=0, regardless of I). Since your load and R2 are in parallel, they have
    the same voltage across each, so the voltage across your load is zero.

    Now, consider that R2 has infinite resistance... Then we have a circuit
    that matches your circuit #1.

    For more reasonable values of R2 ( say 0<R2<inifinity ) we have your
    load in parallel with R2. For two resistors in parallel, the combined
    resistance is lower than the resistance of the least resistive of the
    two ( Rtotal for TWO resistors in parallel can be found by divding the
    product of the resistances by the sum of the resistances).

    As their comibined resistance drops, more current flows through the
    circuit. As more current flows, more voltage is dropped across R1.

    Since the Vsupply is fixed, as more voltage is dropped across R1, less
    voltage is 'available' to the load.

    I hope that helps a bit.

    - jim
  4. The first circuit discards the extra voltage with no extra current.
    The second one discards both voltage and current, so it is less
    efficient. However, the second circuit produces a more stable voltage
    output with changes in load current because the load current is only
    part of the total current passing through R1.

    Both circuits are poor ways to drop voltage if stability is important
    and load current varies. The first one may be practical if load
    current is very nearly fixed, and the second one may be practical if
    the load current is very small.
  5. This one:

    | |
    | + \ R1
    --- Vcc /--------
    - \ O LOAD
    | - / R2 |

    Is equal to this one:

    | |
    | + |
    --- Vcc*R2 O LOAD
    - ------ |
    | R1+R2 |
  6. John Fields

    John Fields Guest

    If we redraw your circuits to look like this:
    (View with a fixed-pitch font like Courier New)

    | |
    | [R1]
    |+ |
    [BATTERY] |
    | |
    | [LOAD]
    | |

    | |
    | [R1]
    |+ |
    [BATTERY] +------+
    | | |
    | [LOAD] [R2]
    | | |

    It's easy to see that if R2 is removed from the second circuit both
    circuits will be identical. That is, if R1 and LOAD are the same in
    both cases.

    Now, with R2 out of the circuit, if we throw some numbers in there we
    can get an idea of what's going on in the circuit. Remembering that
    Ohm's law states that:

    E = IR

    where E = voltage in volts,
    I = current in amperes, and
    R = resistance in ohms

    And knowing that the battery voltage = 9V and that we want R1 to drop
    4V, we'll either have to know the resistance of the load or how much
    current it draws before we can figure out what R1 needs to be to drop

    Just to make it easy, let's say that the load draws 1 amp and that since
    we want to drop 4V across R1, the 5V that'll be left will be dropped
    across the load. Since R1 and the load are in series, the same current
    will be flowing through both of them, and we can determine what R1 needs
    to be rearranging E = IR like this:

    R = E/I

    and solving for R:

    R = 4V/1A = 4 ohms.

    Likewise, for the load,

    R = 5V/1A = 5 ohms

    Since the load and R1 are in series, the total resistance in the circuit
    will be the sum of both resistances, so we can represent the circuit
    like this:

    | |
    | |
    |+ |
    [BATTERY] [9 OHMS]
    | |
    | |
    | |

    And if we want to check our calculations we can say:

    E = IR = 1A * 9 ohms = 9V

    So everything works out fine, and we can conclude that by knowing what
    the battery voltage, the voltage across the load, and the current
    through the load are, we can figure out what the resistance of the
    series resistor needs to be in order to drop a particular voltage at
    that current.

    Now, if we look at the second circuit:

    | |
    | [R1]
    |+ |
    [BATTERY] +------+
    | | |
    | [LOAD] [R2]
    | | |

    And redraw it so that it looks like this, for convenience,

    | |
    | [R1]
    |+ |
    [BATTERY] +---+---+
    | | |
    | [R2] [R3]
    | | |

    we can see that we have the load (R3) and R2 in parallel, and that that
    parallel combination is in series with R1.

    If we remove R2 and R3 from the rest of the circuit and relabel them for
    a moment:

    | |
    [R1] [R2]
    | |

    There will be a resistance between point A and point B which be due to
    the parallel combination of R1 and R2, and that resistance will be less
    than the resistance of either R1 or R2.

    That resistance can be found by using:

    Rt = -------------------
    1 1 1
    --- + --- ... + ---
    R1 R2 Rn

    Which, in the case of two resistors, simplifies to:

    Rt = -----

    Going back to our former labeling, then, we'll say:

    Rt = -----

    and write it:

    Rt = R2R3/R2+R3

    Going back to our solved problem and relabeling it for convenience, we

    | |
    | [R1] 4 ohms
    |+ |
    [BATTERY] +---+---+
    | | |
    | [R2] [R3] 5 ohms
    | | |

    everything is the same except we now have R2 in parallel with our 5 ohm
    load, so now what?

    OK, let's say that R2 is also 5 ohms. Then the total resistance of R2
    and R3 will be:

    Rt = R2R3/R2+R3 = 5*5/5+5 = 25/10 = 2.5 ohms.

    Since the parallel combination of R2 and R3 is the equivalent of a
    single resistor, we can redraw our schematic:

    +---------------+---> 9V
    | |
    | [R1] 4 ohms
    |+ |
    [BATTERY] +---> 3.5V
    | |
    | [Rt] 2.5 ohms
    | |

    Solving for the current in the circuit we can say, from Ohm's law:

    I = E/R = 9V/4R+2.5R = 1.38 ampere

    Now, since there's 1.38 amps flowing in the circuit, the voltage dropped
    across R1 will be:

    E = IR = 1.38A*4R ~ 5.5V

    Which means that, since we started with 9V and R1 is eating up 5.5V of
    it, there'll only be 3.5V left over for the load. The reason is that
    the parallel resistor caused more current to flow through R1, which
    caused it to drop more voltage than it would have if the parallel
    resistor wasn't there.

    An easier way to think about it might be to consider R2 and R3 to be
    light bulbs. With only one of them in the circuit it will be fully
    bright, but when you put another one in there they will both be dimmer
    than the single lamp. And R1 will get hotter, but that's for another
    time... :)
  7. electricked

    electricked Guest

    One question. I'm looking at the first diagram. Say R drops 4 volts. So 5 is
    left to the load. Say 4 volts are dropped by the load so 1 volt is left.
    What would happen if I really connected and executed this circuit? Would the
    circuit run? If so what would the effects be?

  8. electricked

    electricked Guest

    Ok, so what's the difference between the two circuits? How are they
    different and what are the circumstances they'd be applied?


  9. electricked

    electricked Guest

    Thanks John! Very informative indeed. I like how you explain things.

    So basically, they can be considered the same in terms of practical
    application, right? But the first one will put more current through the
    load, and the second one will split the current (if both the load and the
    resistor in parallel are same resistance, it should split the current in
    half, right?) so less current goes to the load. So why is it called a
    voltage divider then? Shouldn't it be called a current divider? Or both
    voltage and current divider? The reason I'm asking is that if I wanted to
    divide the voltage only I could've used the first circuit. But if I wanted
    to divide the voltage and the current (because the load draws less current)
    then I'd use the second circuit. Am I close or far from the truth?



  10. James W

    James W Guest

    You need to study Ohm's law Kirchhoff's Voltage law(KVL) a bit more. I'm
    not being difficult here.. you really do need to sit down with a text
    book and study, work out some problems, etc..

    Any.. to answer your question.

    The sum of the voltages in a circuit MUST equal 0 (KVL). If you have a
    9V source ( a battery for example) then the total voltage drop across
    the 'loads' will be equal to 9V.

    The extra volt you talk about does NOT exist. A resistor does NOT drop a
    fixed voltage. It simply follows Ohm's Law (V=IR), for a given current
    I, there will be a voltage drop V/R.

    So.. trying to respond to your query.. Let's say the circuit was running
    , and you had a 4 volt drop across R and a 5 volt drop across Load, and
    something happenned to the load such that the voltage drop 'wanted' to
    go to 4 volts.. Well, from Ohms law, V=IR, we see that only two things
    effect the voltage drop. Either I or R must drop to lower the V across
    the load.

    Let's say the the resistance of the load dropped. Well, now, the total
    resistance of the circuit will have dropped, so we apply Ohms law AGAIN,
    and find that the current MUST have increased.. so the voltage drop
    across each load will 'balance' out to MATCH the 9V source..

    Whew... does that make sense?

    - jim
  11. James W

    James W Guest

    I should have added.. this is EXACTLY how amplifier's work.. the
    'effective' resistance of the transistor ( for example ) changes, which
    leads to a change in the current through the circuit, which changes the
    voltage across the Collector resistor ( for example )

  12. electricked

    electricked Guest

    Hi James,

    It makes sense now. I was trying to look at two quantities
    (voltage,resistance) while forgetting about the third one (current). I'm
    "getting" it now. I know the laws but I'm trying to visualize what's going
    on in a circuit.

    So if the load's resistance increases (say due to high temperature) and we
    have 5V coming after the first resistor drop, then the current will
    decrease, right? It's starting to make sense now.


  13. Guest

    The first circuit is used as the speed controller for Scalelectrix
    cars by varying R. The load is *small*.

    The second circuit is used as a volume control with the centre tap as
    the wiper. The load is *big*.

    The first circuit is passing a lot of *energy* into the load, so it is
    important to try and make it efficient.

    The (volume-control) second circuit is passing virtually no energy at
    all (because the load is virtually non-existant) this allows regular
    control of the output from maximum to zero (volume).

    The first circuit cannot do this, go "down" to zero, (it can if "zero"
    == car not moving).

    The first circuit is "non-linear".
    The second circuit is "linear".

  14. Bill Vajk

    Bill Vajk Guest

    When you're first learning parallel/series networks (that's
    what these are) you need to ignore real world effects such
    as heat until later when you've mastered the basics.
  15. James W

    James W Guest

    Yes, and no..

    Remember, Ohm's law and Kirchhoff's Voltage law must be true at all times.

    You cannot just say "We have 5v coming after the first resistor's drop".
    As your loads resistance increase, so does the total circuits
    resistance, since you have a series circuit, and resistances in series
    ADD to give you total resistance.

    So, if you loads resistance increases, and hence the total resistance
    increases, then the circuit current decreases ( per Ohms law ). Now,
    since the total current decreases, and the current is the same in all
    parts of a series circuit, we'll have a lowering of the voltage across
    the series resister R. Again, because of ohms law. So, if the voltage
    acrros the resistor drops, and Kirchhoffs law requires that the total
    voltages around the circuit add to ZERO, the voltage across your load
    will increase.

    i.e. A change in resistance through any component effects ALL of the

    Get your hands on twp potentiometers and a battery. Play with varying
    the resistance of the two pots and take lots of voltage and resistance
    measurements ( be sure to disconnect the battery each time you measure
    the resistance of the two pots). Make a chart of your data, and think
    about it until it all makes sense.

    BTW, in a series circuit, the higher the resistance in any component,
    the higher the voltage across that component and the lower the current
    through that component. But don't forget, since we have a fixed voltage
    source, if the voltage across one component increases then there must be
    an offsetting decrease in the voltage across some collection of other
    components (KVL)
  16. John Fields

    John Fields Guest

    Somewhere in between... :)

    Looking at the voltage divider first, classically it's configured like
    this, which is the same as your first circuit but with the battery
    replaced with E1 and 0V, which represent the output voltage from the
    battery and the return to the battery.


    The reason it's called a voltage divider is because if E1 is sitting at,
    say, 9V, then E2 must be somewhere between 9V and 0V, so the two
    resistors are "dividing" the total voltage into two pieces; one from E2
    to E1, and one from E2 to 0V.

    It's the same as if you had a hose hooked up to a faucet and the
    pressure at the faucet was at 9 PSI, and where the water was spouting
    out of the other end of the hose the pressure was 0 PSI; If you took a
    pressure gauge and ran it up and down the hose, the pressure would vary
    from 0 PSI to 9 PSI depending on where along the hose it was. Having
    two resistors is like having two hoses screwed together with a pressure
    gauge where they're screwed together. If they're the same length and
    the same diameter and the pressure at the faucet is 9 PSI with the water
    running, then the pressure gauge between the hoses will read 4.5 PSI.

    Same thing with the resistors. If E1 is 9 volts (9 PSI) and R1 is the
    same resistance as R2, (two hoses with the same length and diameter)
    then E2 will be 4.5 V (4.5 PSI)!

    If you know E1 and you know the values of the two resistors or the
    current you can find E2 using Ohm's law, and you'll find that if either
    R1 or R2 vary, E2 will also vary. Sometimes this circuit is used to
    drop the voltage from a source with too high a voltage to a voltage a
    load can use, but it's usually not a good idea unless the load current
    is small and the load always draws the same current no matter what and
    is never disconnected from the circuit or connected to the circuit with
    the power on. The reason for that is if it's disconnected the voltage
    at the botton end of R1 will immediately go to whatever the source
    voltage is, and then when it's reconnected the load will have that
    voltage on it until it starts drawing enough current for the voltage to
    drop down to where it's supposed to be. Problem is, it may never get
    the chance to work properly if it gets connected to 9V and 5V is the
    maximum it's ever supposed to see. This circuit is generally used to
    get a voltage to use as a reference for something when the reference
    needs to be somewhere between E1 and 0V and the device it's supplying
    the reference voltage to (the load) has a resistance much, much, higher
    than R2. Here's an example:

    | |
    [R2] [R3]
    | |

    If we assume for a moment that R3 isn't connected, we can find the
    voltage at E2 using Ohm's law, like we did before, or to make it easier,
    we can say:

    E2 = E1R2/R1+R2

    Let's make E1 = 9V and R1 and R2 both = 1000 ohms. Then we can write:

    E2 = (9V*1000R)/1000+1000 = 9000/2000 = 4.5 V

    Which is just exactly what we said about the two equal hose segments!

    Now, let's say that to be really sure of what E2 is we'll want to
    measure it, and what we'll use to measure will be a digital voltmeter
    (R3) with a resistance of 10 megohms. Since we know that the combined
    resistances of R2 and R3 are going to be less than the resitance of R2
    alone, and that if R2 isn't equal to R1 E2 won't be equal to 4.5V, we'll
    need to find the value of 1000 ohms paralleled with 10 megohms and use
    that to calculate the new R2. So:

    Rt = R2R3/R2+R3 = (1000R*10000000R)/1000R+10000000R = 999.9 ohms

    Now, using 999.9 ohms for R2 and plugging it into the voltage divider
    equation equation we get:

    E2 = E1R2/R1+R2 = (9V*999.9R)/1000R+999.9R ~ 4.4998V

    so there's only a 200 microvolt error there, which is about 0.004%.
    Not too bad...

    Now, let's say that we measure the voltage with a crappy voltmeter that
    has an input resistance of 1000 ohms. For the equivalent resistance
    we'll get:

    Rt = R2R3/R2+R3 = (1000R*1000R)/1000R+1000R = 500 ohms

    and if we plug that into the voltage divider equation well get:

    E2 = E1R2/R1+R2 = (5V*500R)/1000R+500R = 1.666V

    a HUGE 56.7% error!

    So, you can see that the uses for the circuit can vary from a gross way
    to set the voltage across, or the current through, a load to a precise
    way to generate a particular reference voltage.

    But just _how_ to generate the reference, starting from scratch? Just
    for fun, let's say that you have a 10V supply, you've decided to use a
    10k ohm resistor for R1 and that you'd like to generate a 2.5V reference
    for a comparator input. Going to the circuit:


    And rearranging

    E2 = E1R2/R1+R2

    to solve for R2, (since that's the only piece of the puzzle we don't
    have) we get:

    R2 = E2R1/E1-E2

    And plugging everything in:

    R2 = (2.5V*10k)/10V-2.5V = 3333.33... ohms

    Now we have a problem because, if we're limited to standard 1% resistors
    the closest we can get to 3333.33 ohms will be 3320 ohms on the low side
    and 3400 ohms on the high side.

    Just to see what we've got, plugging that 3320 ohms into the first
    voltage divider equation will get us:

    E2 = E1R2/R1+R2 = (10V*3.320k)/10k+3.32k ~ 2.4925V

    which is only about 7-1/2 millivolts low. If you can live with that,
    you're done. If you can't, you'll need to go back to the table of
    resistances and see if you can find a "gear ratio" that works. Or use a

  17. John Fields

    John Fields Guest

    On Fri, 27 Feb 2004 16:28:37 -0600, John Fields


    .... If you know E1 and you know the values of the two resistors or the
    value of one and the current you can find E2 using Ohm's law, and
    you'll find that if either ...
  18. DarkMatter

    DarkMatter Guest

    Analyse it. Determine the power in each component. If you use a
    250mW resistor, will it get hot from being asked to dissipate too much
    heat? Is a higher wattage capable resistor needed?

    Learn how to refrain from top posting your replies.
  19. DarkMatter

    DarkMatter Guest

    Give up already. You have taken up the wrong vocation.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day