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Little help needed (circuts and voltage drops)

  1. Sep 5, 2012 #1
    I've already read plenty on the subject, for some reason i just don't get it. Now the saying is, or how i understood it...

    :From the book :

    "As charge moves through the external circuit, it encounters a loss of 1.5 volts of electric potential. This loss in electric potential is referred to as a voltage drop. It occurs as the electrical energy of the charge is transformed to other forms of energy (thermal, light, mechanical, etc.) within the resistors or loads. If an electric circuit powered by a 1.5-volt cell is equipped with more than one resistor, then the cumulative loss of electric potential is 1.5 volts. There is a voltage drop for each resistor, but the sum of these voltage drops is 1.5 volts - the same as the voltage rating of the power supply"

    I'm trying to figure out how to regulate porwer, Say you have a 9vDC - 200mA, and you are trying to run that threw a circut of resistor(s) to acheive a power supply of 9vDC @ 20mA. I see the simple solution of adding the resistence, i would need a 1.8watt-1.8(is that right?)ohms resistor. What i can't figure out is the voltage drop, in this case would be 4.5volts after the current is past the resistor, i think.

    So, am i even right in thast calculation, or is it 0v.
    Say i'm right with the 4.5v, that means i have a power source of 4.5vDC @ 20mA, when i need 9vDC, what would i do to solve this problem.

    i imagine the volts are important, if you have something calling for 9v@20mA, you wouldn't want to give it half the voltage, the true amount of power is ohms correct? Ohms is the product of current in amps and voltage. so half the voltage at the same current is half the ohms/power. again, i ask if i'm right (just started learning about ohms law last week)
  2. jcsd
  3. Sep 5, 2012 #2


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    Staff: Mentor

    If you want to supply a regulated 9V then you might use an unregulated source of 13.5V. This leaves a good margin that you can drop across the regulating element to leave exactly 9.0V for your voltage-sensitive circuit.
  4. Sep 12, 2012 #3
    i've been trying to work my problems on a circuit simulator....as long as its princibles are correct...i think i understand it.

    say i had a 9vDC power supply, to get 9v @ 25mA. by this program i'm using, i would want to put my 360ohm resistor at the end of my circut, before returning to ground. The current will flow @ 25mA on either side of the resistor, but after the resistor is 0 voltage. Before the resistor is my 9v @ 25mA

    I would like to know if this is correct, it seems odd. But say i wanted to protect an LED that would like my 9v-25mA. i would set it up like this

    9v+ --------- 9v-25mA fuse --------LED(100ohm-resistance)-----resistor(260ohms)------ground

    adding total resistance = 360ohms out of 9v = 9v-25mA entering the LED with a voltage drop after the LED
    is that correct?
    Last edited: Sep 12, 2012
  5. Sep 12, 2012 #4


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    I see, you are using the word "regulate" to have a meaning different from its usual use in electronics.

    To set the current through your LED to 25 mA the arrangement you propose will work. Somehow you must have determined that the LED resembles a 100 ohm resistor at 25 mA. Although that is not a good model for a LED, it is sufficient here.
  6. Sep 12, 2012 #5
    Hello Chris:

    I may be reading between the lines - but are you trying to use a 9Vdc/200mA supply to light a LED with 20mA(first post) 25mA(last post) max current limit?

    The supply will provide 9V under all normal (not overloaded) conditions, BUT the 200mA is its current limit, it will only put out 200mA if you connect a circuit to it that will flow 200mA when 9V are applied.

    Thinking of an LED are resistance is not ideal, OK for a regular Light bulb - yes, but an LED works differently and it should have a Vf (forward) and Current (mA) and/or Power (mW) rating.

    Given my assumptions - the LED probably has a Vf of 0,8 to 1.5v - but if you use 0.8 to start with it will light up, just not be as bright. So you have a 9V sully and 0.8 V drop across the LED, you have to then find a resistor that will have a 8.2V Drop when flowing 20/25mA.

    R = V/I .... 8.2V/0.020A = 410Ohm resistor; not a common size try a 470 Ohm - to start or recalulate based on the actual LED parameters ( there are also online / free calculators for this).

    In your "circuit" where you use the LED as resistor you used 100Ohm - if you calculate the V drop on the LED ( @ 100 Ohm) it comes to 2.5V - this seems a little high, at least for a single LED. The LED is not linear like a Resistor - in fact it is opposite, the higher the current flowing the LOWER the "apparent resistance" across it - that is why I do not like the "thinking" of it as a resistor.
    Last edited: Sep 12, 2012
  7. Sep 12, 2012 #6
    Thanx both of you for your help, i just got my snap-circuits in to get a better understanding. I'm gonna play with them to learn more. Also the LED specs seem out of wack because its made up specs, i have no LED that requires 9v-25mA, i just used it as an example, also gave it a resistance value because i know it will affect the current which means i needed to reduce my resistor based on the amount the LED will resist. I would ask about the forwad voltage, but i'm probably gonna learn about it with my Snap-Circuits.

    Thanx again, if i have anymore future ?'s, which i'm sure i will, i'll be back.

    Just found out the higgs particle may have finally been observed....pretty kewl.
  8. Sep 12, 2012 #7
    Hello Again CA -

    Looks like you are just starting out - I got into electronics with one of the ol' radio shack "spring connector" kits - basically the equivalent to the snap kits today. I was 10 Yrs old.

    I have been happy to see that Radio Shack is getting back into the electronics kits and hobbyist market, so they will have a lot of stuff - they even have arduino control boards. With these you can actually build some neat things - and with each project learn a little about the circuits. Otherwise I would say start with the simple things - Batteries, resistors and LEDs are good - and get a small Digital meter - separate from the kit. In your first post you seemed to ask about all of the circuit at once - but the beauty of electronics is that you really can study and learn about each component individually - trying to look at 4 devices all at once only make it much more complicated.

    V=I*R is your friend.

    A great lecture is the into to EE at MIT - and discussed abstraction - do not let the big word bother you - in short you do not have to know ALL of the details of how components work to be able to apply, use and understand them very effectively. ( http://ocw.mit.edu/courses/electric...tronics-spring-2007/video-lectures/lecture-1/ )
  9. Sep 13, 2012 #8
    The radio shack kit is the one i really wanted, the snap-circuits is not what i expected, it only shows the layout of different circuits and short discriptions of the parts.

    sadly the closest radioshack sux, they don't have much int the kit department. i thought it was odd they didn't carry the kit they produce to sell. i'm going to just buy a bread board, maybe a couple of printed circuit boards an alot of resistors & stuff.

    It all started with certain gadget i want to make, though i like learning, so i figured i'd learn a little more. I'm only left with one problem with my project.

    I have a low current, voltage sensitive circuit. I need the starting voltage to run threw my circuit(9v) with a 9k-resistor(9v@1mA) after my circuit, and a 9v-2mA fuse before my circuit.

    I also need to be sure the circuit is complete, as in a LED. the problem i run into on my circuit simulator is the current is always the same between resistors. like say i had 4 resistors, total resistance = 9k, now the only current i can get at any location is 1mA., witch isn't enough current to run any LED i can find. even if i put the LED before my sensitive circuit, i have the same equal current problem, equal current threw the whole circut.

    i need 9v@1mA to run threw a circuit, with a light indicating the circuit is complete. as in, if power dosent make it threw my circuit, no light turns on.

    Last edited: Sep 13, 2012
  10. Sep 13, 2012 #9


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    Staff: Mentor

    Fuses don't routinely have a voltage associated with them (9V@1mA), they have a current rating at which they are designed to melt. Are you saying you have purchased a fuse rated at 2mA?

    For a series circuit, all elements have identical currents, that's how it works, and you can't change that. :smile:

    If your need was to have a LED indicate 1mA of current in a circuit, then you could devise a transistor amplifier to drive that LED with 20 mA of current to ensure good brightness (assuming there is a power supply to continuously power it). If battery operated, maybe use a flashing LED so the average current is very low, nothing like 20 mA.
  11. Sep 13, 2012 #10
    voltage dosen't matter in a fuse? i dont have that fuse by the way, i assumed voltage mattered. like when i replace the main fuse in a salt box for swimming pool, it has a 15amp-250v fuse. why does it iclude the voltage? why dosent voltage matter? its pressure, higher voltage = more pressure which = more power within a given point.

    where have i gone wrong? my understanding is:

    Voltage is how big the ball of energy is, curent is the rate at which the balls flow, resistance is the energy required to restrict the flow. example:

    it takes 9k of resistance to reduce the flow of 9v(balls of energy) to a current(speed) of 1mA.?

    is it...the amount of energy doesn't matter as far as heating things up, only the rate at which it flows? or am i just all screwed up on my way of thinking?
    Last edited: Sep 13, 2012
  12. Sep 13, 2012 #11


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    Not really. Not for the low voltages you are using. 24 V, 100 V, 250 V, whatever, they all work.

    Where voltage rating is relevant is after the fuse has melted. The purpose of a fuse is to melt if a sustained current exceeds some defined value. After it has melted, it is intended that no further current flows. But if the voltage across the ends of the open-circuit fuse is sufficient to jump the air gap then the fuse won't be breaking the current. Higher voltage fuses are longer to present a longer path to block that higher voltage after the narrow part of the fuse element melts, and they are sturdier to withstand the higher energy spark when melted in a higher voltage circuit. Yet, sometimes the glass still shatters.
    So that you don't try and shove in a 12 V car fuse, maybe.
  13. Sep 14, 2012 #12
    Hello Again C A

    A 1 mA fuse ? That is pretty small - and probably easy to blow. If you start with batteries - no need for fuses. ( You just may burn a finger if you - but that lesson you will remeber better:smile:)

    Anyway - your analogy is close, but always remember it is an analogy & not a perfect model of electricity. However I would think of Voltage as the FORCE to push the flow of electricity.

    So Force x Distance (movement ) = work is similar to Voltage x Current = power ( work)

    Voltage pushes current through a resistor - this generates heat ( power). ( actually that is exactly how a fuse works, the material in the fuse is made to heat up to the point of melting at the rating (can be more complicated that that - depending on the type of fuse.

    The voltage rating on a fuse is the maximum voltage that the fuse can still interrupt the current. If you have a 9v supplying a circuit - the fuse need to be rated 9V or higher. ( For example the 0.001A Fuse I find on Newark is rated to interrupt 50A at 125 VAC or VDC)
  14. Sep 15, 2012 #13
    if the actual aim is to see if the circuit is complete insert a galvanometer/ammeter in series.
    Or if you really want to light an led without any drop in your circuit, insert a RELAY. In the secondary of relay, incorporate the led circuit.
  15. Sep 18, 2012 #14
    yea, thats what i came up with electricallov.

    I finally got my problems solved.

    9v+---10mA/fuse---PowerSwitch---(+) terminal\9v/(-)Terminal---Ammeter(0-10mA)---ResistorPOT(1.8watt/0-10k)---Resistor(1.8watt/1.8k)---ground.

    though i may have to adjust my resistance, depending on how much resistance a 9v battery has. i can figure that out now. thats the setup for a 9v source with 0 resistance.

    got to use the fuse also, very sensitve circuit im going to run current threw, also learned it to has a resistance value, currently unkown to me, thats why i added a varible resistor and ammeter. after more research, i'll know weather i should go witrh a 10mA or 5mA fuse. I'm thinking of playing with some tDCS. Looking for a good source of stimulation areas, along with functions related to area. Also looking for information on savants brain activity to explore some ideas of mine.
    Last edited: Sep 18, 2012
  16. Sep 18, 2012 #15


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    You may have to contend with elements not as ideal as you may wish. From the data at the Newark site, for example, the 10mA fuse has a resistance of 80Ω. If current rises to 15mA, the fuse (nominally 10mA) may take 30 mins to melt. Also, at a price of $25 per unit, you would hope failures to be very infrequent. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon9.gif [Broken]
    Last edited by a moderator: May 6, 2017
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