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Potential difference in a simple circuit with transistor

  1. Jul 30, 2015 #1
    Hello everyone..I have some difficulties in understanding the potential difference in a circuit with a transistor. Kindly refer to the picture provided. I understood that potential difference across Ry can be obtained from the formula in the picture. The potential difference across Rx plus Ry will be the same as V because I picture it as being parallel to the battery. Please correct me if i'm wrong. But what I would really like to know is how to find potential difference of R1 and R2? Is there any rules that I have to follow? One more thing is it necessary to put R1 in that circuit? I hope someone can answer my questions because I really like to know the answer. http://i.imgur.com/5eI0VkD.jpg
     
    Last edited by a moderator: Aug 3, 2015
  2. jcsd
  3. Jul 30, 2015 #2

    CWatters

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    That is correct if you make the assumption that the base current (in R1) is small compared to the current through Rx and Ry. You could make Rx and Ry small compared to R1 so this assumption is valid.

    For R1.... The base -> emitter of a transistor behaves like a diode so the base voltage will be about 0.7V. You know the voltage at the other end of R1 so you can calculate the voltage difference by subtraction.

    For R2.... This is harder. If you know the voltage across R1 you can calculate the current flowing through R1 into the base (eg the base current). If you know the gain (HFE) of the transistor you can work out the collector current by multiplying the base current by HFE. Once you know the collector current you can work out the voltage drop across R2.

    There are a few complications...

    There is a limit on the voltage drop across R2. The collector voltage cannot fall below the saturation voltage of the transistor (perhaps 0.3V). So the drop across R2 cannot be greater than the supply voltage - 0.3V.

    The gain of a transistor HFE isn't very well characterised. In other words it varies from transistor to transistor due to manufacturing tolerance. A typical small transistor might have a gain that is specified to be..

    min 20
    typ 100
    max 200

    This means the voltage on R2 might vary depending on how good the transistor is. In digital switching circuits it might not be a problem (you could just assume the minimum value). In analogue circuit this problem is typically avoided by including feedback in the design but that's a topic for later when you understand the basics.
     
  4. Aug 2, 2015 #3
    Thanks for the reply CWatters. So from your explanation I conclude that there are no fixed values for potential difference across R1 and R2. It depends on the current I guess? R1 depends on base current and R2 depends on emitter current? Can i use formula V=IR to calculate the potential difference in this case?

    VR1=Ibase x R1

    VR2=Icollecter x R2
     
  5. Aug 3, 2015 #4

    CWatters

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    No that's not really what I meant. I'm struggling to answer the question because I don't know the context to the circuit. Where did it come from? What do you think it does? It has no inputs or outputs, nothing is changing so why shouldn't the voltage on R1 and R2 be fixed?

    It looks like it might be part of a linear amplifier and I assumed that in my earlier description. In that case it would be normal to choose values for the supply voltage Vcc, the collector and base current and work back to values for R1 and R2.
     
  6. Aug 3, 2015 #5

    CWatters

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    The short answer is yes. Although normally when you design a circuit like this those voltages are known and you are trying to calculate what R1 and R2 should be.
     
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