Making sense of a transistor circuit

Discussion in 'General Electronics Chat' started by Alemarius Nexus, Mar 31, 2015.

  1. Alemarius Nexus

    Thread Starter New Member

    Mar 31, 2015
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    Hi,

    I've recently started to learn a bit about transistors, and I'm currently fiddling around with some test circuits in LTspice. One of them looks like this:
    ltspice1.png

    I've removed some wires for testing, so there are some unconnected components (R11 and R12), and there are some low-valued dummy resistors for measuring current. I figured most of the circuit isn't really that important to my question here. Also, the wire is deliberately interrupted at VV. The interesting thing for me here is the PNP transistor Q2, whose currents are plotted as the four resistances in the center are stepped. The emitter and collector currents are measured going into the transistor, the base current is measured going out of the transistor. I can't make sense out of these measurements. They say that current leaves the emitter, but also leaves the base. If they are opposite direction, it can't be an emitter-base-current, can it? Where else does the base current come from? And why does the transistor not open-circuit when there's no emitter-base-current? The magnitude and direction of the base current seems to imply that it comes from the collector, but why does it do that (especially if there's no emitter-base-current)?

    What I wanted to achieve here is that when the wire at VV is left open, both Q2 and Q3 go open-circuit too, so that the voltage between the four resistors in the center comes entirely from the upper half of the circuit, cutting off the lower half (it's a simulation of a control circuit for a 4-terminal resistive touch screen).


    Any ideas?
     
  2. Lectraplayer

    Member

    Jan 2, 2015
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    The current does split within the BJT (common bipolar junction transistor, like you have) but the base current is teeny tiny. The base to emitter current ratio is what your beta is. For example, a beta of 200 means your emitter current is 200 times what your base current is.
     
  3. Alemarius Nexus

    Thread Starter New Member

    Mar 31, 2015
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    0
    Ok, but from what I understood, the emitter current splits into emitter-base and emitter-collector current (conventional flow for a PNP transistor. For NPN the base-emitter and collector-emitter currents "fuse" into one big emitter current), so the emitter current should always have greater magnitude than either of the two other currents. In my example, this is not the case: The emitter current is ~70uA leaving the transistor, the collector current is ~315uA entering the transistor, and the base current is ~245uA leaving the transistor, so the emitter current is actually lower than the base current.

    Also, thinking about beta: If beta is the ratio of emitter current to base current, that would mean I'd have beta = -70uA/245uA, which is negative because emitter and base currents have opposite direction. How can that be negative?
     
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    I have no comment on your circuit, but I can show you how to get more out of LTSpice.

    44.gif

    Look a my simple example. Note that I am able to ask LTSpice to sweep the base current from 1nA to 1mA in logarithmic steps, and use that as the X-axis of the plot, using the directive .dc dec I1 1n 1m 5 . I choose to make the sweep logarithmic to show what happens over a wide range of base currents...

    I can plot the collector current Ic(Q1) and the emitter current Ie(Q1) (without putting in any dummy resistors to monitor the current). These quantities are automatically saved in the .raw file when the simulator runs. You can add them to any plot just by pointing to the collector, base, emitter pin on the transistor symbol. Note that the cursor automatically changes to an "Amprobe" when you hover over the pin.

    Note that the light blue trace falls perfectly on top of the dark blue trace, meaning that at the resolution of this plot, the emitter current and collector current are for all practical purposes equal. Note that the y-axis for the Ic(Q1) and Ie(Q1) plots is the right axis in Amps, which I also chose to plot on a log scale.

    I also added the base voltage V(b). I could do that because I named the node "b". It is plotted on its own axis on the right side in mV. LTSpice auto-ranged the plot from ~400mV to ~1V.

    Just to show you another trick, I also plot an expression Ic(Q1)/I1 = Ic(Q1)/Ib(Q1), which when I went to school, we called β. Notice that LTSpice knows that this is dimensionless, so it plots it on its own scale on the left side, showing that the β of this transistor is close to 100 for a large range of currents.

    There are many other features that the novice or casual user of LTSpice wont discover for a while...
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Are you sure that you have Q2 hooked up correctly. I think that you might have the emitter and collect swapped. This would mean that you are running it in reverse mode and current would enter the collector and then split into the current coming out of the base and the current coming out of the emitter. Most BJTs will function this way, but the properties are generally pretty poor.


    It would really help if you drew your schematic according to the conventional norm which has current flowing from top to bottom in a circuit and your power rails and center-tapped common properly stacked.
     
  6. ScottWang

    Moderator

    Aug 23, 2012
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    Are you sure the wires of e of Q3 and e of Q4 connected to the output of op amp is correct?
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    As best as I can tell, the output of the opamp is establishing the "ground" for the circuit midway between the supply rails. So the emitters are simply grounded.
     
  8. ScottWang

    Moderator

    Aug 23, 2012
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    Oh, yes, your are right, he created a virtual ground.
    Why I missed the input voltage and just focus on the voltage follower, and I also found that you already answer him and why you not point out that point, it must be the weather drive me loosing my mind to thinking deeper and the temperature changing from 16 to 31 C degree up and down during these days, from Winter suddenly became the summer, and the hard time is coming ... :(
     
    Last edited: Apr 1, 2015
  9. ramancini8

    Member

    Jul 18, 2012
    442
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    Junk LTSPICE because it is to advanced for you. Buy a simple book that contains some analog circuits with analysis, study the book, and build the circuits, then compare theory to actual. You will learn much faster that way.
     
  10. Alemarius Nexus

    Thread Starter New Member

    Mar 31, 2015
    3
    0
    I think it's connected the way I wanted it. I want Q2 to conduct when the wire at VV is connected. In this case, the emitter is at 3.3V, and the collector (usually) lower than that. Base is hard-wired to ground. When VV is open-circuited, I want Q2 to open-circuit. In this case, the voltage at Q2's collector is indeed higher than the voltage at it's base, but I thought unless current flows from the emitter to the base, the voltage between emitter and collector doesn't matter, because the transistor open-circuits anyway?

    What I'm trying to do is find the resistance ratios R1/R2 and R3/R4 in two steps: First only Q2 and Q3 are open-circuits (only the top half of the circuit matters, I can measure R3/R4), then only Q1 and Q4 are open-circuits (only the bottom half matters, I can measure R1/R2). The resistors form a voltage divider, so I can measure the ratio by looking at the voltage at the center wire (in reality, I can't measure at the center wire, only at L/R/T/B).


    Exactly. I apply 3.3 volts to the L and T wires, and 1.65 volts to the R and B wires. I can't use 0V for R/B, because there is also a case where the center wire is not connected, in which case the cutoff part of the circuit is pulled-down to 0V by R11 and R12 (which are not connected in this picture). I use 1.65 volts as "ground" so that this case is different from e.g. R2 being 0 ohms.


    I will eventually build up this circuit on a breadboard, but I find it easier to debug and play around with it in LTspice first. As for getting a book: I might do that, but I currently have a project which is not that analog heavy that I want to finish first. Then I might go through the analog stuff a bit more in-depth.


    Thanks for your help!
     
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