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Simple Thevenin circuit help

  1. Apr 24, 2016 #1
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    Hi, I am struggling to simplify this into a thevenin circuit. Firtsly I am unsure how the zener diode works and why the R1 is not required.(so does it create a voltage drop across the whole loop and , how does it work exactly?) Also unsure if the resistors are in series or not. I think you are menat to determine if the same current is flowing through each resistor to know whether they are in series or not, so if you assume the battery is at the 6.2Vwould you say that R3 and R4 are in series which you add, then the three remaining are in parallel?
    The actual measured values are:
    R2=5.6KO
    R3=2.7KO
    R4=68KO
    R5=3.3KO
    Rth=4.67Ko
    Thanks
     

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  2. jcsd
  3. Apr 24, 2016 #2

    gneill

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    Hi nathsen. A zener diode provides a nearly constant potential drop across itself so long as it conducts a current greater than its threshold value (it varies from device to device, but generally just a few mA are required to put it into its normal operating conditions). When its operating in this region it behaves much like a fixed voltage source with a pretty low internal resistance (usually below 20 Ohms and typically in the single digits).

    You should see if you can find a V-I curve for a zener to take a look at so that you can get a feel for what it does.

    For your purposes treat it like a 6.2 V DC voltage source. It's resistance will be much smaller than all the other resistances in the circuit so its a good bet that it can be ignored.

    R1 and the actual voltage source are ignored because they can't be "seen" behind the zener's fixed potential and low resistance.
     
  4. Apr 25, 2016 #3
    Thanks for your help. So assuming it's a 6.2 V DC source, under my understanding I just simplify the remaining resistors now.
    Where am I going wrong? I'm thinking R3 and R4 are in series, then add the resultant in parallel to R2 using product over sum, then that resultant product rule over sum to R5?
     
  5. Apr 25, 2016 #4

    gneill

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    I think you're having trouble recognizing the parallel and series components. The circuit diagram is laid out in a manner that can make it tricky.

    Take several colored pens and trace the wires of each node with a different color (or use some other means to uniquely identify the wiring associated with separate nodes). Components that are in parallel will be connected to the same pair of colors. Remember, when you suppress the voltage source (zener) it's replaced by a wire, so both of its connections become part of the same node.
     
  6. Apr 26, 2016 #5
    Thanks very much for your reply. I've tried to follow your advice (see picture). Is that correct in theory? I am still not getting the value desired but is it through other means rather than how I'm simplifying the circuit!?
     

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  7. Apr 26, 2016 #6

    gneill

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    Remember that when the voltage source is suppressed it is replaced by a short circuit (wire). So your pink and green nodes should become all one color.
    upload_2016-4-26_16-5-54.png
     
  8. Apr 26, 2016 #7
    Ok so i've gone over the green with pink. Now I need to identify components connected to "same pair of colours". So actually Resistor 3 and 4 are both connected to pink and blue which means they are in parallel. And 2 and 5 are also in parallel? Sorry i'm getting a mental block over this!
     
  9. Apr 26, 2016 #8

    cnh1995

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    Last edited: Apr 26, 2016
  10. Apr 27, 2016 #9
  11. Apr 27, 2016 #10

    gneill

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    Sometimes redrawing the circuit in a slightly different way can yield insights; you pick up relationships that you missed before.

    If you move R2 and the zener to different locations in the same wire (so no topological changes) as shown below, then you should be able to see that the circuit on the left can be morphed in to the circuit on the right:

    upload_2016-4-27_21-37-54.png

    If you then suppress the zener as you would a voltage source, what do you get for the Thevenin resistance of the network?
     
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