Easy circuit analysis questions

Discussion in 'General Electronics Chat' started by tracecom, Jan 10, 2013.

  1. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
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    I built the little circuit shown in the schematic below. As you can see, I took a lot of measurements, some of which I understand and some I don't.

    The circuit is powered with a 6V lantern battery. The base bias for Q1 comes from another circuit which is also powered from the lantern battery, but only supplies 3.33V as an output signal. When I attach it to R4, the voltage drops to 1.75 and then there is a .93V drop across R4, so the final bias voltage for Q1 is .81V. The base current is .935mA (less than 1 mA,) and the collector current is 55mA, so β for Q1 in this circuit is 58.8.

    I understand that R1-R3 limit the collector current to 55mA, but what I would like to know is whether Q1 is in saturation, and if so, is it considered "soft" or "hard" saturation.

    My other hope is that someone will show me the math for Ohm's law as it applies to biasing Q1. (Yeah, I am a little embarrassed to ask.)

    Thanks in advance.
     
    Last edited: Jan 11, 2013
  2. ScottWang

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    Aug 23, 2012
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    According to the datasheet on page 2, the β for Q1 (Vce sat) should be 20, how many β is good that related with what frequency are you using.

    If you just using with DC, that is not a big problem, and how deep will you get into the saturation, in my personal application, I love to cross the threshold and adding some more, if the threshold is point to that when the Ib just make the Ic get into the status of saturation, and if you really need to do that, then increasing Ib some more that I will do.


    http://www.datasheetcatalog.org/datasheet/MicroElectronics/mXuwzwr.pdf
     
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  3. vrainom

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    Sep 8, 2011
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    First off, the voltage drop across r4 doesn't add up with your current measure, .93v / 460 = ~2ma.

    The current gain (β or hfe) for your transistor is a minimum of ~200 (according to page 3 in the datasheet ScottWang provided) which means that the current output could be as high as 187ma (using your base current measure), but your circuit only draws ~55ma so your transistor is in saturation (i.e. it's acting as a switch either completely on or completely off).


    Soft saturation means you're driving your transistor with barely the needed current to get it saturated, i.e. if the β is 100 then the minimum current to drive a 100ma load is 1ma.

    Hard saturation is a term that means you're driving your transistor with more current than needed to ensure saturation, an usual figure is ten times the needed base current.
     
    Last edited: Jan 11, 2013
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  4. Brownout

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    Collector current if 55mA. So, collector voltage is:

    6.17V - 3.2V - .055/3*150 = 0.22V. Base voltage is .81V, and thus the CB junction is forward biased. Indeed, saturation has been acieved.
     
  5. ramancini8

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    Jul 18, 2012
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    You show the collector-emitter voltage to be 0.11V, if that is true the transistor is in saturation, hard saturation as you define it.
     
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  6. ErnieM

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    Apr 24, 2011
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    Well you have to use what you have, and having several meters can be a blessing... until you try to compare readings and don't know which one (if any!) is the correct one.

    Breaking the circuit to insert an ammeter can induce differences when you remove that meter, I prefer to either not use one or leave it in circuit while I take other measurements.

    But if I just have one "good" meter I'm not stuck either. (I have three... usually just use one).

    By knowing the exact resistance we can calculate the current... or "close enough" to make an informed decision. So first, use the best meter to measure the resistances of all your resistors: measure both ways to make sure there isn't a sneak path or you will have to remove the resistor to get it's value.

    Next, just use the one meter to make voltage measurements about the circuit. Then...

    Let's assume you did that, and the voltage measurements and resistances in your picture are accurate.

    R4 has 2.25 - .83 =1.45 volts across it, so the current is 1.45 / 460 = 3.15 mA

    R3 has 2.92 - .11 = 2.81 volts across it, so the current is 2.81 / 150 = 18.7 mA. Same (may hold, check the voltages!) is about what R1 and R2 have, so the collector current is 18.7 mA + 18.7 mA + 18.7 mA = 56.2 mA, or close to what you measures (always good to have a check!).

    Your transistor is operating with a gain of 56.2 mA / 3.15 mA or 17.8. To "force" saturation a gain of 10 is usually used to guarantee sat, but that isn't cast in stone. .11 volts C to E on Q1 isn't bad, it's just .11V * 56.2 mA = .00618 watts, so the transistor is coasting.

    In your first post you say your input gives you 3.3 volts, but your figure shows only 2.25 volts there. That make me curious why there is a difference.

    Hope this helps, seems you are on a good path.
     
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  7. tracecom

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    Ernie,

    Apparently you saw my last post and responded to it before I deleted it, and I apologize for that. What happened was that after I posted it, I discovered that I had a wiring error that made many of the readings wrong, so I deleted the post. I really appreciate your response and regret that I wasted some of your time. I am reassembling the circuit and will post again when I have (hopefully) finalized readings.

    Thanks.
     
    Last edited: Jan 11, 2013
  8. tracecom

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    OK, I have done that, and the final results are on the following schematic.

    [​IMG]

    Now, I am going to go back over everyone's comments and try to understand them all. Thanks for everyone's input and patience.
     
  9. tracecom

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    Using the data in the previous post and the guidance provided, here are my calculations.

    Base current
    E = I R
    = .000940A x 460Ω
    = .43V, which is close enough to the .45V I measured

    Collector current
    E = IR
    = (.0545A / 3) x 147
    = .0187 x 147
    = 2.67V, which is close enough to the 2.73V I measured

    Q1 gain
    β = Ic / Ib
    = 54.5mA / .940mA
    = 57.98 ≈ 58

    Q1 power dissipation
    P = EI
    = .17V x .0545A
    = .0093W

    So, that seems to take care of the math.

    When I measure the output from the PIR with no load, it measures 3.3V, which is what the manufacturer states. But, when I connect it to R4, it drops to 1.26V. I don't know why. It is powered from the same 6V lantern battery as Q1 and the LEDs.

    I am still a little hung up on what defines saturation of a BJT. Any additional elucidation would be appreciated.
     
    Last edited: Jan 11, 2013
  10. Brownout

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    That happens if the output resistance of the sensor is high, and you are observing a voltage drop across the sensor's output resistance. You can fix this by using an amplifier with a higher input resistance that is higher than the sensor's output resistance. Like a darlington, for example.


    The BJT will be in saturation if the C-B junction is forward biased. If if VCB >= -.6V, you have achieved saturation.

    Now I must go look up the word 'elucidation'
     
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  11. tracecom

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    So, my measured Vcb is -.64V, which technically is barely there?

    By your answer to my request, you apparently knew the word. :)
     
  12. Brownout

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    I wouldn't say barely. You won't get much more than .6V no matter how "much" into saturation you are. Think about a forward biased diode. What is the volage of a "strongly" forward biased diode?
     
  13. tracecom

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    There is a school of thought that says for saturation, base current should be at least 1/10 of collector current. That would mean that in the circuit in post 8, R4 should be 87Ω to give a base current of 5.4mA. That seems to be quite different from some opinions. What's a neophyte to do? :eek:
     
  14. ErnieM

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    Apr 24, 2011
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    Read about what and why others have done, listen to their reasoning, then follow it blindly.

    Or try things on your own, make your own assessment and be willing to accept the consequences of your actions.
     
  15. vrainom

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    Sep 8, 2011
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    Uhm, don't you mean E-B junction? Also, BJTs are current amplifiers, ~.6v is their forward biasing voltage but it raises with the base current.

    To make it clearer, a BJT is in saturation when any raise in its base current won't make the collector current raise as well. This is because the base current multiplied by the transistor current gain (or β) equals the collector current the transistor is able to provide, and if its load is consuming less than that amount of current, the transistor is saturated.
     
  16. Brownout

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    Nope. I mean C-B voltage, just as I stated.

    True enough, but for saturation criteria, it's C-B voltage that is critical.

    I don't really follow you. Saturation occurs when the collector-base junction is forward biased. This can happen when the load * collector current satisfies this:

    VCC - βIB*RC < VB - .6V.

    For an NPN transistor.
     
  17. vrainom

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    Sep 8, 2011
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    Oh yeah, sorry, it was me who didn't get your point about the collector voltage being lower than the base voltage for an NPN transistor in saturated state, but from my point of view, that's explaining the effect, not the cause.

    What I meant to say is that a transistor (npn or pnp) is saturated when the base current multiplied by the current gain is equal or bigger than the current the load is able to consume.
     
  18. tracecom

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    It seems that I continue to find conflicting information about BJT saturation, or maybe it's just that I don't understand what I am reading.

    Here is info from Sedra/Smith, Microelectronic Circuits, Fifth Edition. Is there a conflict here about the Vcb at which a BJT enters saturation, or is it my lack of understanding?
     
  19. Brownout

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    There's no conflict. When VCB is about -.4V, the BJT is entering saturation. By the time it reaches -.6V, it is fully in saturation.

    BTW, I noticed an error in my equation

    Of course, that should have been VCB <= -.6V.
     
  20. tracecom

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    @ Brownout,

    I wasn't really questioning your statement. I was confused because in one statement from the book cited, it says 0.4V and another says -0.4V.


    ETA: I guess the top statement must be a typo. (They left out the minus symbol.) The graph and the caption for the graph agree, so that must be the author's intention.
     
    Last edited: Jan 13, 2013
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