Transistor Unstabilized Circuits

Discussion in 'Homework Help' started by howartthou, May 27, 2009.

  1. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hi All

    Must admit I am finding transistor circuits pretty difficult to grasp right now so I would really appreciate some help here.

    Attached are 3 pictures:

    1. Circuit diagram of unstablized transistor circuit.
    2. Close-up graphic of a "heating resistor" (whatever that is?).
    3. Photo of my attempt at wiring the circuit on a trainer breadboard.

    My questions:

    1. Have I wired the trainer breadboard correctly?

    2. Can someone please explain how I am meant to wire in the "heating resistor" and what role it plays in the circuit?

    3. Please describe the electron flow.

    My attemp:

    1. See the attached photo of the wired circuit on the trainer breadboard. I did that.

    2. Really confused with the "heating resistor". Is it meant to touch the transistor to keep it "cool"? Why is it wired with solder? Is that so you can bend it over to touch the transistor?

    3. Electron flow:

    Starting at yellow -10V the electron flow goes through R1, then collector, then base, then R2, then 2 on 1k POT, then from POT to who knows? Stuck here. Help.

    At the same time the electron flow goes through the Emitter, then to Ground, then 1 on 1k POT, then from POT to who knows?

    It seems that when the flow goes into either 1 or 2 on the 1k POT it then flows back to -10V??? Confused here, how would it go back to where it started? Negative is not attracted to negative, but it looks that way...

    And I would guess the "heating resistor" would be wired to the red +15v and to ground, and is there to "cool" the transistor when it is bent over to touch it?

    I know there are 3 questions here so I am really hoping to get 3 answers. Any help appreciated.
     
  2. Audioguru

    New Member

    Dec 20, 2007
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    When the pot is adjusted correctly it applies a voltage and current to the base-emitter junction of the transistor which causes the transistor to conduct a little current from collector to emitter. The collector-emitter current flows in the collector resistor R1 which causes a voltage across it.

    Applying 15V to the 220 ohm resistor causes it to heat with 1.02W of heat.
    When the hot resistor touches the transistor and heats the transistor then the transistor conducts more current that can be measured by the voltage increasing across the collector resistor.

    It looks like almost everything is wired incorrectly. The potentiometer will smoke and burn. The very old transistor appears to have its collector and emitter shorted together.
    The 220 ohm "heater" is not powered.
     
  3. PRS

    Well-Known Member

    Aug 24, 2008
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    1. I think the board is wired correctly.
    2. The heating resistor isn't wired in at all. Insert it into two unused holes right next to the transistor. Connect one side of the resistor to ground and another wire to 15 volts. But until you're ready to experiment with heat vs current, keep the 15 volt wire in a neutral hole.
    3. As per Audioguru's description.

    It appears to me that you are first to examine the behavior of the base current vs the collector current using the pot. Then set the pot for, say 1 mA of collector current and turn on the heating resistor. (The latter isn't a special resistor. It was chosen so that you could safely dissipate 1 Watt through it to make it hot and act as a heater). Now watch the collector current rise with temperature.
     
  4. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Thanks so much to both of you. Together you have really helped clear things up. I have so much more to do and working through this one will give me the basics I need to go on. Thanks again :)

    I will try things out and see if I have more questions soon. Thanks!
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    It is difficult to see since the wires are all black but I think the pot is wired correctly.
    The collector and emitter of the transistor are shorted together and are both connected to ground.
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    You should have a variable voltage into the 100KΩ resistor, to bias the transistor base through a range.

    I can't make out the transistor, though this isn't a big deal. Never did like that case style, though it was common for older radios and whatnot.

    They are trying to demonstrate thermal run-away, a common problem with transistors of the BJT variaty. As the transistor heats up it conducts more. In a poorly designed circuit this can cause it to heat up even more, which causes it to conduct even more, until something frys.
     
  7. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Audioguru

    Just to clarify the photo:

    The germanium tranisitor has the Collector at the bottom and the Emitter at the top and the base on the left.

    Staring at the bottom of the tranistor is the Collector wire that is connected to R1 (1kohm) and then to the yellow negative terminal.

    The wiring on the breadboard does not match the layout diagram but I believe is wired correctly.

    For example, in the circuit diagram layout the emitter is on the bottom and the collector on the top, but on the breadboard its the other way round. This does not mean its wired incorrectly, just that the physical layout does not match the diagram but is still correctly wired.

    Try starting from the photo at the yellow negative 10 v. Follow the wire to R1 then to Collector which is at the bottom of the transistor and keep going from there...

    Sorry the photo layout does not match the diagram, but it doesn't have to, it just needs to be hooked up correctly as you would know.

    Anyway, please stay tuned as I have more questions...appreciate all the feedback sof far, thanks.
     
  8. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Bill, this experiment requires "the 1Kohm POT to be adjusted so that you have 5 V accross the collector resistor R1 which will set the original bias current at 5mA."

    It shows that when the transistor is heated the leakage current increases from 5 mA to about 9.8mA which is a change of 96% in the bias current.

    This is meant to show the circuit has no stabilization circuitry.

    Must admit I don't know exactly what I am talking about here but I understand the idea.

    I am still not clear on exactly what "bias" current means? What is being biased by what?

    Also not sure how the POT applies a voltage as in Audiogurus first answer?

    I thought the POT was just a variable resistor? So yes adjusting it would vary the V and C but the experiment requires it to be adjusted so that you have 5V accross the collector resistor R1.

    So where I am confused is that the circuit diagram shows a -10V terminal thats yellow then the ground terminal thats black. So does the mean ground is acting as the positive terminal in the circuit????? If so, I thought ground would always be negative as red is positive and black is negative. Really really confused with this one...help!

    I find the trainer breadboard confusing as I don't fully understand it and this circuit diagram doesn't show a - and a + terminal but a - and ground. Its ground that I find confusing. Does it have a split personality? Can ground act as either + or - depending on how the circuit is wired???
     
    Last edited: May 28, 2009
  9. PRS

    Well-Known Member

    Aug 24, 2008
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    If you set the pot to exactly 5 volts you'll get (5-.2)/100K = 0.05mA, approximately. If then the current gain of your transistor (beta) is 100 the collector will have 5 mA, approximately. I think your instructor wants you to adjust the pot for exactly 5mA at the collector and then monitor the current as you heat the resistor. The current will go up with temperature.

    This is due to the physics of the device.

    ic=Is*[e^(vbe/Vt)]

    where Vt is the thermal voltage and note the exponential! Also, Is, the saturation current, doubles for every 5 C rise in temp. I'll tell you more about these so-called constants if you want. But you'll find them in your book and it will tell you why all this is so.

    The purpose of the experiment seems to be to impress on you the need for temperature compensation in machines that go through extreems of temperatue such as Air Force jets radar equipment. You can't just use the standard amplifier designs that assume constant temperature.
     
  10. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hi Paul

    I appreciate the insight but believe or not I am still having trouble with some of the basics - like "ground" in the circuit. This really is hindering my progess.

    Could you please read the quote in this message and answer the question.

    I would also appreciate a description of current flow from starting at the yellow negative terminal because thats where current flow starts.
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    Ground is a common reference point, and is completely arbitrary. In this case you have a power supply that provides 2 voltages, variable, up to 15 volts. They would be refered to as ± (Plus and Minus).

    For your application ground is the minus power side of the power supply.

    If you put the negative lead of a voltmeter on ground, the plus power supply will read positive voltage, while the negative power supply will read negative, as measure from ground. Electronics will make heavy use of this for many things, op amps come to mind.
     
    Last edited: May 30, 2009
  12. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hi Bill

    Understood what you just said, even tested it, yes ground is arbitary. Thanks.

    But I am still confused about the quote above. How can the minus power side be ground? There are 2 terminal in the circuit, one is yellow and the other is black. One of them has to be negative and the other positive, surely?

    So how can you say the minus power side is ground? Are you excluding the black terminal when you say that?

    I know I am missing the obvious here
     
  13. Wendy

    Moderator

    Mar 24, 2008
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    Part of what you are missing is that Ground is a definition as much as anything. If you were going to earth the power supply or chassis, ground is what it would be.

    When designing electronics I have a battery. I need to declare one side or another as ground, as a reference for other voltages. All the other voltages I will measure will be defined from this one spot. It also simplifies drawing schematics, since I get to use a simple little symbol to designate this one spot.

    The original concept of ground is rooted from AC power, then RF communications got involved. At this point they were talking about the same thing, but from completely different points of views, and were interested in different properties. From early radio it got involved with modern electronics, and transmuted even more, where ground no longer means earthed in this application, but the voltage reference by which all others are measured.

    Since we are talking definitions, and that can vary a little depending on what part of the world you live in, what part of the world are you at?

    Some other takes on the subject...

    http://en.wikipedia.org/wiki/Ground

    http://en.wikipedia.org/wiki/Ground_and_neutral

    http://en.wikipedia.org/wiki/Ground_(electricity)

    http://en.wiktionary.org/wiki/ground
     
  14. PRS

    Well-Known Member

    Aug 24, 2008
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    Bill explained it all very well. Ground is a reference point. A voltage somewhere else in the circuit can be either negetive or positive with respect to the ground. You're not the only one to be confused on this point. And for this reason you've got to be careful when working on radios and tvs. You're oscilloscope ground lead is tied into earth ground through the third prong of the plug. Attaching that ground to a TVs ground is a sure way to short something out!
     
  15. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hi Bill

    Thanks for the links to up read on ground. I am from Melbourne, Australia. Over here we have a green and yellow wire that leads to earth, which as far as I know is just a copper pipe embedded in the ground.

    When you mentioned a battery has one side declared as ground thats usually the positive isn't it?

    I believe a car battery has the positive cable connected to the chassis - is that right?

    As I understand it, we need earthed circuits in houses so excess voltage in the case of a short has somewhere to go, and blow a fuse along the way.

    But if I take a simple 12v battery, a globe and wire it up, the circuit won't be grounded will it? So where is the reference point in this circuit?.
     
  16. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hello Paul

    So where is the gound in a TV connect to?

    As you can see ground still confuses me. Where is the gound in a simple circuit that has a 12v batter and a globe. There isn't one is there? So where is the reference point?
     
  17. PRS

    Well-Known Member

    Aug 24, 2008
    989
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    The designer determines ground from the outset. After that he provides the voltages needed by the devices in the circuit. Take a simple circuit, for example. A battery, an led and a resistor all in series. You could declare the battery's negative as ground (the usual case) and the positive terminal as +9 volts. Or, if you want to go the other way, you can declare the possitive battery terminal ground and the negative -9 volts. In doing so there is no reorientation of the diode, just a change of names.
     
  18. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Hi I can't tell really good from the picture of your breadboard but the wire coming off of the 1k resistor at the bottom of picture is it connected to the middle lead going directly to ground. I don't see it on the right side terminal hole it looks like it is in the terminal line that you emitter lead is in.???

    I'll keep looking I mught delete my reply if I see something different.

    I still don't see where your connected to a third lead on the transistor...

    Looks like both leads are in terminal 17.

    Or is the back lead in term.18??
     
  19. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Thank you Paul! This makes perfect sense and has been confusing for a while. But would you please explain how you would represent a "ground earth" in this same circuit? The "ground earth" could be a chassis of a computer or an actual Earth wire on a power point. So if negative is usually ground (the reference), how would you show "ground earth" in the same circuit of a diagram? There is only one symbol for ground. Still confused but getting there...explain this and that should just about do it :)

    Hobbyist, I can only guess you have not zoomed the picture yet?

    The wire at f 33 coming from R1 goes into the Collector terminal of the transistor at f 17. Then, as per the circuit diagram, there is wire coming out of the Emitter side of the tranisitor at j 17 that connects to ground.

    So the transistor teminals are collector at g 17, emitter at i 17 and the base at h 15.

    The emitter is obscured by the body of the transistor but it can be determined if you zoom close enough.

    The circuit diagram on the other hand, shows collector on the top and emitter on the bottom, so I transistor in my circuit is placed oppositly to the one in the diagram. I just had a nasty thought...I could have assigned the emitter and collector incorrectly!

    There is a dot on the transistor that is supposed to indentify the collector terminal so thats why I think collector is at the bottom and wired it accordingly.

    But if the collector is actually at the top, as per the circuit diagram, then I have screwed up!

    I have attached photos of this old germanium transistor (2N2431 also known as B324). Its obviously marked in some ancient symbology because I can't determine which is collector and which is emitter?

    Would be good if you can tell me if you think the circuit is wired correctly or not because I still haven't turned it on yet...I think you may have picked up that the transistor is upside down based on how I have wired it!
     
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  20. hobbyist

    Distinguished Member

    Aug 10, 2008
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    The reason why I question it is because the breadboards have 5 common holes per strrip for each number column and the letters designate each hole per column strip.
    the connection of your 1k re. appears to be in the same column strip as the ground wire.

    Isn't g17 and i17 on the same term. strip?

    So the transistor teminals are collector at g 17, emitter at i 17 and the base at h 15.

    Transistor is probably shorting out from collector to emitter...

    So current will probably flow from neg. through 1k re. to ground bypassing the transistor altogether?
     
    Last edited: May 31, 2009
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