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Beta of a Transistor (both NPN and PNP), how would I do that?

  1. Nov 25, 2012 #1
    Hello, I am trying to design a circuit that can measure the BETA of a PNP and NPN transistor, I have an ammeter, and I have a design that makes IB going in 0.1mA, so that the current IC of the collector with a shifted decimal is the Beta we would like to measure. So I used a current source, and designed a circuit (a picture of the two circuits I designed is attached to my blog here: http://vantraveller.blogspot.ca/). However, I don't think the current source is correct.
    The meter I am making should be able to measure any NPN or PNP transistor, but with the current source I am using, sometimes the value of IB changes, which it shouldn't.
    I am thinking that maybe the current source is wrong. Is there a way to correct the current source? Is there an easier way to fix IB so that I do not need to use a current source? Thank you so much!
     
  2. jcsd
  3. Nov 25, 2012 #2

    davenn

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    double post

    one of them needs to be removed :)

    Dave
     
  4. Nov 25, 2012 #3
    oops sorry, new to this site, not sure if I posted in the right section. Been stuck on this problem for the past 20 hours... haha
     
  5. Nov 26, 2012 #4
    Just look at the first diagram. I have no idea what is U1 and U2. Looks like Q1 is the current source set up the base current of Q2.

    Problem is you use two 1n4148 to set up about 1.4V to bias Q1 and you have emitter resistor of 900Ω. The Ic of Q1≈0.7V/900≈0.78mA. That's a lot of current for the base of Q2.

    Current gain β=Ic/Ib. β>100 in most case. So you should use a lot lower current for base. Use something like 1uA to 20uA so the Ic of Q2 will be in range of 100uA to 2mA etc. You don't need U1 or U2 to do that, get rid of them. You just measure the voltage drop across the R2 and calculate the Ic of Q2.

    Now come the question of the current source to provide the Ib for Q1. The circuit you use is not accurate and it can drift. This is because you are matching the Vbe of the 1N4148 to Q1. But you can get the feel of it. All you have to do is to measure the voltage drop across R5, then calculate the Ib of Q1. You can assume the Ie=Ic for Q1 at this point and use it as base current. If you create the current source like this, you don't need R3. So get rid of U1, U2 and R3.

    To get `1 to 20uA, increase the resistance of R5 so 0.7V/R5=base current of Q2.

    Work on this first and post back.
     
  6. Nov 26, 2012 #5

    sophiecentaur

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    Re: Problem with creating a beta measurer of PNP and NPN transistors?

    It is not clear what does what in those circuits. If I wanted a defined constant current source, I would use the output from a collector. What have you done? Feed that into the base of the transistor under test and then measure the collector current from that. That would give you a readable value for Beta - directly, if you used a round figure for the input base current (e.g. 0.1mA as you suggest).
     
  7. Nov 26, 2012 #6

    berkeman

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    Done -- two threads merged.
     
  8. Nov 26, 2012 #7
    What Im confused about is, for a PNP transistor, because the current is coming out from the base, is it right to put a current source to keep the current to 0.1mA? If it is, and I have 1n4148 diodes and resistors to make it, how would I make a current source of of the Base current to keep the base current 0.1mA?
     
  9. Nov 26, 2012 #8
  10. Nov 26, 2012 #9
    I am looking back to your original first diagram, I was reading wrong. I don't even know how you make it work. You should draw the voltage in a more traditional way, say the most positive voltage at the top and ground or negative voltage at the bottom. Drawing battery in one direction and then cross from top to bottom make it very hard to read. I cannot make sense of your new drawing.

    I draw up a circuit to measure the β of PNP, it is close to what you try to do in the second diagram. Q2 is to set up the base current for Q1. I use +15V as positive voltage. You can make R1=5K and VR1 is a 10K trim pot so you get 0 to +10V adjustable at the base of Q2.

    The V1 is a voltmeter to measure the voltage across RE. The current through Q2 is V1/RE. Since the beta of Q2 is over 100, you can assume the collector current of Q2 is same as V1/RE. This is the base current of Q1.

    V2 measure the voltage across Rc which gives the collector current Ic1=V2/Rc. Make sure you use DVM that has very high input impedance.

    From the two meters, you get the base current of Q1 and the collector current of Q1. β=Ic1/Ib1.

    Now, lets set up the operating condition. Lets set the range of collector current of Q1. Say I want to test from 0.1mA to 10mA. So if I use 100Ω for Rc, I'll get 0.01 to 1V full range.

    Now to set up the base drive constant current source for Q1. Lets assume β=100 for Q1, so the base current to get 0.1 to 10mA collector current is 1uA to 100uA. This will be the adjustment range of the collector current of Q2.

    We know the emitter of Q2 vary from 0 to 9.3V( from input of 0 to 10V). We want 1uA to 100uA. So RE=9.3V/100uA=93K. You can use say 82.5K which is a standard value for 1% metal film resistor. It is not important to be exact as I am making assumption of the β of Q1 anyway. But with the value of RE in the ball park range, you measure the EXACT base current by measuring the voltage across RE and Ie=V1/RE.


    With that you get both the base current and the collector current of Q1 and you can find β.

    Now you can adjust to various base current and read the collector current and see beta change with base current.

    Notice you only get one point that the base-collector voltage of Q1 is -14.3V. But as you know, BJT has very high output resistance which means the collector current don't change much when varying the collector voltage. It will be very complicated to do measurement with various collector voltage. That's what a curve tracer is for.

    You can work out the circuit for the NPN transistor.
     

    Attached Files:

  11. Nov 27, 2012 #10

    sophiecentaur

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    I'm not sure that you appreciate the difference between a 'source of current' and a 'current source', which has a specific meaning.
    Any source of voltage, in series with a resistor, will 'supply current'.
    An ideal 'current source' will supply a chosen value of current into a wide range of loads. It has an infinite resistance (don't try to think practically about this as it's not any more achievable than an ideal voltage source, with zero resistance). This can, in its simplest form, be a 'high enough' value of resistor in series with a 'high enough voltage'; the current into a range of low resistances will not vary much. Mostly, it's achieved using the collector of a transistor as shown in the above post, which relies on the inherent feedback provided by the emitter resistor. For better control, an OP amp can be used, utilising the very high gain available to produce a very constant current value.
     
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