Voltage Divider Verification

Discussion in 'General Electronics Chat' started by ELECTRONERD, Sep 17, 2009.

  1. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Hey Everyone,

    Just want to make sure I'm doing this right.

    Suppose I have a voltage divider like in the first attachement. You'll see that R1 = 680Ω and R2 = 20k giving me an output voltage of 8.7V [Vout = R2/(R1+R2)]. Now, to calculate the output current of the voltage divider, you do the following: Iout = Vin/(R1+R2). So my current output is approximately 435μA. Let's say I want to have 2.5μA for my output, I simply add a resistor like in the second attachment. What would the output current be now? Also, wouldn't there be a voltage drop across R3 so I would have to make the voltage a bit larger to compensate for that drop? Ignoring the voltage drop, this is what I did: 435μA - 2.5μA = 437.5μA. So, 8.7V / 437.5μA = 19885.71429Ω ≈ 20k.

    Please let me know if I'm doing this right! This is to get the correct base current and voltage to an NPN transistor.
     
  2. hobbyist

    Distinguished Member

    Aug 10, 2008
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    I'm working on it,
    but first if you put a resistor in parrallel with your 20K that will lower the total resistance, which then would be a lower voltage out across there.

    I'll see if I can come up with an equation to solve for this, then you can caompare notes and see if I got what you came up with.

    I'm assuming the R3 is going to ground to allow current to flow?
     
  3. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Sounds good! No, R3 goes to the base of the 2N3904 transistor.
     
  4. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Ok that would be a 0.7v. drop to ground.

    Assuming no emitter resistor?

    This should be able to solve using thevinen equivalency.

    I'll work this out and get back to you.
     
  5. hobbyist

    Distinguished Member

    Aug 10, 2008
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    I calculated to get 2.5uA you would need around 3.2 meg ohms

    I will rework and double check this.
     
  6. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Right, since I want 8V at the emitter, I want 8.7V at the base. Assuming there is no emitter resistor.

    According to the 2N3904 specs, on the second page entitled "On Characteristics" you can see that with a hFE of 40 the collector current should be 0.1mA. According to the equation, β=Ic/Ib I can transform that making it to be: Ib=Ic/β. So To make it in its linear region according to the specs, I need Ib=0.1mA/40 = 2.5μA. One thing I'm not sure about is what the emitter resistor should be if I did have one?
     
  7. hobbyist

    Distinguished Member

    Aug 10, 2008
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    the emitter resistance will change the entire input resistance so you would choose a value then rework the base resistors.

    But now you have to take into consideration B x RE
     
  8. hobbyist

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    Aug 10, 2008
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    I think one way to do it would be to choose your RE value then take the Beta times it for a approx. resistance value then rework the base resistors.

    If you could drop the series resistor, then you could do it entirely with a voltage divider itself, reworking the base voltage.
     
  9. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    One thing after another, eh? Well, I suppose I would like Re. It affects the gain, right? You're saying that if I do take β x Re into consideration I should probably figure out the emitter resistor before the base resistors? Could you show me what to do?

    I know what my collector current should be, according to the 2N3904 specs. So, Rc=9V/0.1mA = 90k.

    I also know the base current according to the specs. Since they say in the "On Characteristics" that Ic=0.1mA and Vce=1V that will give you an hFE of 40. So Ib=Ic/β and so my base current should be 2.5μA. Also, since Vce=1V I need the base voltage to be 8.7V so that the emitter voltage will be 8V. Then (I think this is the right equation), Vce=Vc-Ve so that would give me 1V for Vce. I'll need that voltage divider to give me the voltage and that extra R3 to give me the necessary current at the base. Then I would think that I could calculate Re. What should the current be for Re? How do I calculate it?
     
  10. hobbyist

    Distinguished Member

    Aug 10, 2008
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    I need to look more at your last post, but to answer IRE this current is by equation
    IE = (IC +IB)
     
  11. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Let me do a quick calculation using RC as 91K ohms as you stated above and use a volt gain of 10 just to get in the ballpark with this.

    90K for Rc won't work with 9V source. thats the the whole supply fropped across that resistor.

    let me look at your post and get myself straightened out here.

    To be linear the voltage across RC needs to be around 1/2VCC.

    I'm still working on this.
     
    Last edited: Sep 17, 2009
  12. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Let me back up take it one step at a time.

    You waqnt IC=0.1mA.

    VCC=9v.

    So RC= 4.5V / 0.1mA
    (VC= 1/2VCC) to be linear.

    RC = 45K make iot 47K ohms

    VCE = 1v.

    since VCE is chosen then VE is fixed
    since Ve is fixed then volt. gain can't be chosen.

    VE =3.5v.

    RE= VE / IC (ruff estimate)

    RE= 36K
    not much gain

    VB= 4.2v.

    RB1=20K to ground

    RB2=22K to positive

    That's about what I can come up with. for now..

    If you change the VCE you can rework the base values to get closer to what you want.

    see if this makes any sense and let me know what you think or any questions on how I came up with anything that don't look right.
     
    Last edited: Sep 18, 2009
  13. hobbyist

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    Aug 10, 2008
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    {{I know what my collector current should be, according to the 2N3904 specs. So, Rc=9V/0.1mA = 90k.}}unquote

    Do you see where this is incorrect? The entire voltage is dropped across RC and not the transistor.




    {{Right, since I want 8V at the emitter, I want 8.7V at the base. Assuming there is no emitter resistor.}}unquote


    Do you see where this leaves no collector resistor ? all the supply voltage is across the emitter resistor RE (8v.) and the transistor (1v.).

    I just seen this also
    if there is no emitter resistor than you have 0V. at the emitter and base voltage would be 0.7v.
     
    Last edited: Sep 18, 2009
  14. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    What should I do then? I need those currents, what am I to do?

    Now I know what I did wrong, Vc=Vcc/2 so Voltage at collector is 4.5V. Therefore, Rc=45k, correct?
     
  15. hobbyist

    Distinguished Member

    Aug 10, 2008
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    I never really designed a transistor circuit using the spec. data sheets like your doing, so I'll have to go at this one step at a time.

    Yes. 45K is the correct answer. standard (47K or 43K)

    So lets use 45K for the calculations only.

    So far

    VCC=9v.
    IC=0.1mA
    IB=2.5uA
    RC=45K
    VCE=1v.

    Now

    VC=4.5v.
    and
    VCE=1v.
    so then VE=3.5v.
    Therefor VB will calculate to be 4.2v.

    IB=2.5uA
    IE=(IC + IB) = 102.5uA
    so RE calculates out to (VE / IE) = 34,146 ohms.

    Now this is a total guess at the next part I'm doing.

    So with 2.5uA branching to the base the divider current should be at least 10 times greater, so lets make ID (divider current) 20.5uA, that would make the grounded resistor in the divider = to { 4.2v. / (20.5uA - 2.5uA) } = 233.333K ohms.
    And the top supply resistor = ( (9v - 4.2v) / 20.5uA)= 234.146K ohms

    I think...
     
    Last edited: Sep 18, 2009
  16. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    That’s a good conjecture, for sure. That was my entire goal, to be able to look at any transistors specs and design a circuit to saturate it and also design a circuit so it’s a Class-A amplifier. That way, I can take any transistor and design it as efficient as possible for my particular application. I’m sure we’ll both be able to do this soon!

    I forgot about how Vc=Vcc/2 so that definitely changed the base voltage. If someone could verify that your conjecture is correct, then all we would have to do is work out the voltage divider problem. One thing I just thought of, if we include R3, should we have the input signal at point A or point B? If we had it at point B it would definitely be more sensitive. You’ll find an attachment showing our complete circuit now. I have most of the values in there. I added C1 which we can calculate depending on our voice frequency.
     
  17. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    If anyone else would like to join in please do! I'm sure both hobbyist and I would love to read your comments and suggestions!
     
  18. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Here is a simulation of the bias (static) voltages using the exact calculated values.

    These are not too far off from the calculations.

    I'll run a dynamic test on the simulator and see what it does.

    static voltages.jpg
     
  19. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Here it is with a 10mV. (red) signal in 1khz.
    and the output (blue) amd the symetry of the output wave.

    Notice hardly no amplifiocation, due to the ratio of RC / RE.

    Dynamic test.jpg




    See linearity has to do with the input signal and the available collector current, and the voltage at which the collector is set at, on the load line.
     
    Last edited: Sep 18, 2009
  20. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    It should be more than that shouldn't it? We should have a gain of 40. Red doesn't appear to be 10mV, your scope says 10mV per division so it is reading 20mV.
     
    Last edited: Sep 18, 2009
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