Dropping 4.5V to 1.2V

Discussion in 'The Projects Forum' started by KiwiKid, Jun 23, 2011.

  1. KiwiKid

    Thread Starter New Member

    May 12, 2011
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    I need to figure out what value fixed resistor to use to drop 4.5V to 1.2V. The current is 20mA. (its is coming from a 08M picaxe)

    I know this seems like a stupid question because its very basic but i'm confused because our teacher told us we need a 470Ω resistor.

    The circuit is a subsystem for a motor where a transistor is used as a switch to activate the motor, the base of the transistor is connected to a picaxe output pin. NO OTHER COMPONENTS ARE USED other then the motor, the transistor, the IC and the fixed resistor placed between the picaxe and the transistor.
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    If the current is 20ma then Ohm's Law rules. You need to subtract 1.2VDC from 4.5VDC, or drop 3.3V. Feeding these values into Ohm's Law it works out to

    3.3V / 0.02A = 165Ω. Using standard resistor values this works out to either 150Ω, 160Ω, or 180Ω. You can always parallel resistors to get closer to the desired value.

    It sounds like some information is missing if you need 470Ω. Schematics are always nice. I would also double check with that teacher to see if you left something out.
     
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  3. KiwiKid

    Thread Starter New Member

    May 12, 2011
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    That's what I got. The transisotr i'm using is a BC337 which has a base-emmiter on voltage of 1.2, although earlier when he was drawing the circuit he for somereason was dropping it to 0.7V, but even that dosn't work out with a 470Ω resistor.

    heres a pic http://s1195.photobucket.com/albums/aa391/pieterthegreat/?action=view&current=circuit21.png
    Its a bit messy but i'm using a old mac which is horrifically laggy and i hate.


    Thanks for quick response.
     
  4. electro_yas

    New Member

    Dec 20, 2010
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    There is a problem in the calculations. According to the datasheet of that transistor the maximum current it can pass through is 1A and the DC gain is about 100. Then if you are passing 20mA to the base it can exceed the maximum collector current rating of the transistor. So better to use a higher resistor value as your teacher said. That is my idea.
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    No, transistors in general have a BE drop of 0.7VDC (personally I use 0.6V). This is pretty definate for all BJT types. A Darlington Pair, which this isn't, would drop 1.2V. I looked up the data sheet.

    There is a rule of thumb (but it is a hard rule) that the CE current is 10X the base current. So if the CE current is 20ma, the BE current is 2ma.

    (5V-0.7V) / 0.002 = 2150Ω ≈ 2.2KΩ

    If he meant to have 20ma drive to the transistor it would handle up to 200ma, or 0.2A.

    (5V-0.7) / 0.02 = 215Ω ≈ 220Ω

    To me it sounds like he is using something other than the 1/10 base current rule.

    *********************

    Just took another look at your schematic, it shows the power supply as 4.5V, so...

    (4.5VDC - 0.7VDC) / 0.02A = 1190Ω ≈ 1.2KΩ
     
    Last edited: Jun 23, 2011
  6. electro_yas

    New Member

    Dec 20, 2010
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    This can be true for some power transistors where high current is used but for my general purpose transistors its normally around 100, so better to check the gain of the transistor experimentally using a variable resistor and two ammeters .
     
  7. Wendy

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    Mar 24, 2008
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    Nope, the 10X rule is universal. We are talking digital mode, saturation in other words. The beta you are talking is for analog use. Since you are turning a motor on/off this is a digital application.

    The two most efficient modes a transistor has in terms of wattage is on or off. When on it has very low voltage across the CE, and when off it has no current. Since power = voltage X current, and you are taking one of the variables to zero, the transistor stays cool. This is why PWM (pulse width modulation) is so effective.

    It occurs to me to turn your problem around. If the resistor was 4700Ω the current through the BE would be

    I = (4.5VDC - 0.7VDC) / 4700Ω = 8.1ma

    This means your transistor can reliably handle up to 81ma. It could probably handle more, but we are talking reliable operation.

    I would be interested in finding out your teachers logic when you get a chance to talk to him again.
     
  8. electro_yas

    New Member

    Dec 20, 2010
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    Yes , since this is just for switching purpose transistor can operate in the saturation region and your reply about gain of 10X is correct to get the minimum minimum power dissipation in the transistor but the IC driving the transistor may get hot.
    What is the motor's maximum current rating?
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    We go with the specs the OP gives us. He defined the 20ma. It does seem a bit high, but it is what was given.
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    I didn't read all the replies.
    You and your teacher do not understand simple electrical circuits.
    1) The output of a loaded MPIC is not 5V. Its voltage depends on its load. it might be only 3V.
    2) Where is the schematic??
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    The schematic is post #3, but I'll repost it in an easier to view format...

    [​IMG]

    Like I said earlier, the voltage shown for the BE drop is way too high. This voltage will go up with current, but 20ma isn't that high (though for this application it is a bit much).

    Here is the transistor datasheet...

    http://www.fairchildsemi.com/ds/BC/BC337.pdf

    Given the information provided I still think the current is closer to 8ma.
     
  12. Audioguru

    New Member

    Dec 20, 2007
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    Why don't you look at a datasheet for a BC337? It shows that the base-emitter voltage is typically 0.9V when it is saturated with a load of 200mA and is a max of 1.2V when its collector current is 500mA.

    We don't know the current used by your motor which is the highest when it starts and when it is stalled.

    The output of the PIC is about 4.8V with a load of 20mA and the voltage actross the base resistor is 4.8V - 1.1V= 3.7V. For a current of 20mA its value is 3.7V/20mA= 185 ohms. With a base current of 20mA the BC337 saturates well with a load of 200mA.

    If it had a 470 ohm base resistor then the base current is only 9mA and the transistor might not saturate if its load current is higher than only 90mA.

    Maybe your teacher does not know that the base current must be 1/10th the collector current for the transistor to saturate well regardless of its beta.
    Its beta is spec'd when it has a collector to emitter voltage of 1.0V so it is not saturated.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    The PICAXE uC that the OP is using has an I/O pin limitation of around 20mA source or sink, which is where the 20mA came in. It can source less than 20mA current.

    So, if our OP's Vcc is 4.5v, and Vbe is ~0.7v and 20mA is the maximum current, then the base resistor can be as low as (4.5v-0.7v) / 20mA = 190 Ohms, allowing up to 200mA collector current with the transistor in saturation.

    If 470 Ohms is used as a base resistor instead, then you will have (4.5v-0.7v) / 470 Ohms ~= 8.1mA, which will allow up to 81mA collector current.

    However, 1.2v has been mentioned - you didn't mention whether your transistor is a Darlington or a standard transistor. However, if the transistor is a Darlington, then there will be two base-emitter drops, which is where the 1.2v comes in (it'll actually be closer to 1.3v). Darlingtons require far less base current than a standard transistor, as the gains of the two transistors in the package multiply each other.

    A 2N2222/PN2222 is an example of a standard transistor.
    A TIP120 is an example of a Darlington.
     
  14. Wendy

    Moderator

    Mar 24, 2008
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    How about that Wook, we agree on our math. I'm going to be interested if the OP comes back with something else. Even if the BE drops with the extra current the change won't be that dramatic.
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    Hi Bill,
    We're sort of in agreement, but it seems that you are looking at a collector current of 20mA, but I'm looking at the actual limitation in the PICAXE I/O pin.

    It would help a lot to know what the motor current requirement is.
     
  16. Audioguru

    New Member

    Dec 20, 2007
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    The OP is wrongly using a Vbe of 1.2V in his calculations because it is the max Vbe of the BC337 transistor when it is saturated with a collector current of 500mA and a base current of 50mA.
     
  17. WellGrounded

    Member

    Jun 19, 2011
    32
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    KiwiKid,

    I have a feeling that the 470 ohm value for the circuit is correct by all the info supplied in the thread.

    Since this a breadboard layout then the easiest thing to do is to:
    (1) - Put a 1.0 ohm precision resistor between the motor and the collector. (2) - Place a 500 ohm pot in series with the 470 ohm resistor.

    Slowly turn the pot backwards from 500 ohm. Measure the voltage on the motor and notice when there seems to be no more change in the motor voltage and stop turning the pot.

    Measure the voltage on the 1.0 ohm resistor using the millivolt scale. The reading you get will be your current. If the motor won't run properly because it has a high current draw then a 0.1 ohm resistor has to be used and your reading should be multiplied by 10 to get the correct current rating.

    Measure the total resistance reading of the pot and 470 ohm resistor to get the maximum resistance reading to make the motor work properly. It could be 25% to 50% higher than the suggested 470 value.

    For measuring use an analog meter with a high Volts/Ohms rating.

    If we get the motor current and work backwards we can verify the base current and base resistor values and everything will fall into place with our formulas that have been shown without having to worry too much on exact specifics.

    Sometimes "hands on" is the best way to solve a problem and your breadboard layout makes it simple to make alterations to the circuit to do measurement testing and then confirmation calculations.

    Danny

    To all members - How do start a new line in the layout of your reply if you you want to show items in a list?
     
  18. SgtWookie

    Expert

    Jul 17, 2007
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    Not exactly certain what this request is, but if you use the "Go Advanced" button below the text reply box, you will get a host of options to use.

    • This is an
    • example of
    • the Unordered List option

    However, please avoid selecting a custom font or custom colors; it really makes life difficult for those of us who reply w/quotes. I wind up having to manually edit and remove the custom font/colors out just so I can read the reply while editing.
     
  19. WellGrounded

    Member

    Jun 19, 2011
    32
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    "Sarge",

    Thanks for the help. It was just what I was looking for although I didn't describe properly the list structure in my question.

    Danny
     
  20. castley

    Member

    Jul 17, 2011
    31
    0
    You need to lose 3.3 volts at 20 Ma Using Ohms Law R=V/I, Divide 3.3 by .02= 165 ohms . A 1/4 watt rating should be adequate.
     
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