Output BJT current buffer

Discussion in 'The Projects Forum' started by mk1, Dec 8, 2009.

  1. mk1

    Thread Starter New Member

    Dec 8, 2009
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    Hi all. some help would really be appreciates with this problem I am having!

    I have designed a common emitter amplifier using a BJT. Running at 10MHz. I am taking a sine wave from a signal generator as the input (can be up to 10v pk-pk, but currently am taking 2v pk-pk as small input means i dont have such a large voltage at the emitter. i have an emitter bias resistor and emitter degeneration to limit the gain (I need 40v pk-pk output).

    However, I need more current than this can supply so I wanted to connect a common collector to the output. This reduces the output of the CE by about 10V though. Does anyone know what I am doing wrong or what I can do to fix this?

    Also, am I best to use a cascode amplifier? If so, I have no idea where to start so any guidance would be great!! Are there any pros/cons of the two approaches?

    I literally just need the 40v sine wave, but have smaller signal generators available. I am using BJTs because I struggled to find high frequency, high voltage op amps.

    Any feedback? Thanks a lot in advance!
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    Post your circuit.
     
  3. mk1

    Thread Starter New Member

    Dec 8, 2009
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    Ok. Hopefully this has uploaded successfully. It is just a screen grab from Proteus. When connecting the CC, I would connect the left hand side of C5 to the CE BJT collector. Is this clear enough?
     
  4. mik3

    Senior Member

    Feb 4, 2008
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    Increase the value of C5. If the problem is not solved increase the values of R6 and R9.
     
  5. mk1

    Thread Starter New Member

    Dec 8, 2009
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    Thanks for looking over it!

    I have increased C5 to 1uF. It was there before, must have changed it to try something! Sorry about that! Increasing R6, R9 does not help much. Increasing R7 as well helps to bring the voltage up a little, however it is still a lot smaller than when the CC is not connected. This is with 1M resistors. Smaller gives less benefit. I am a little worried about increasing these any more, since I am trying to draw plenty of current from the CC amp. (input impedance of target is shown on right (8Ω + 33pF)).

    Is there anything else I can try?
     
  6. mik3

    Senior Member

    Feb 4, 2008
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    What is the difference after you changed C5?
     
  7. mk1

    Thread Starter New Member

    Dec 8, 2009
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    changing from 135pF to 100nF did almost nothing. the graph looks slightly more linear. the amplitude is the same.
     
  8. mik3

    Senior Member

    Feb 4, 2008
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    You said that it is the output of the CE that is reduced, didn't you?

    If this is the case then the output impedance of the CE is high and drops much voltage because the input impedance of the CC is low due to R6 and R9 (draws a significant current). Redesign the CE as to have a lower output impedance or increase the values of R6 and R9.
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    What is the impedance of your load?
     
  10. mk1

    Thread Starter New Member

    Dec 8, 2009
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    Yes, the output of the CE drops. The problem is that the gain of the CE is dependent on the collector resistance, so it cannot be reduced by too much. And increasing all CC resistors to over 1MΩ does not help. I have read several places that following a CE with a CC is common practice, but nowhere gives examples of how to match them.

    Some guides seem to suggest a cascode, but again I really don't know anything about choosing values or how it works..

    The load impedance is 8Ω in series with 33pF (as shown on the right of the schematic). I am currently testing without this even connected, as shown.
     
  11. CDRIVE

    Senior Member

    Jul 1, 2008
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    If this amp is being designed to feed standard coax (50 Ohms) then the output impedance of this amp should also be 50 Ohms. As Ron asked, what is the impedance of the load in your spice model?
     
  12. Ron H

    AAC Fanatic!

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    He said he is driving 33pF in series with 8 ohms. I wonder what sort of real-world load this represents?
     
  13. mk1

    Thread Starter New Member

    Dec 8, 2009
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    Well the CE alone has been tested without (or with very large) load. I was hoping to use the CC to drive the smaller load (shown). When driving the CC with a signal source component, the smaller load does not adversely affect it. It is solely the connection of the CC to the CE that drops the voltage, with or without the 8Ω and 33pF.

    Does this answer the question? Sorry if not.. I hope I understood!
     
  14. mk1

    Thread Starter New Member

    Dec 8, 2009
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    It is an input to an EMCCD sensor! The high voltage, electron multiplying bit..
     
  15. CDRIVE

    Senior Member

    Jul 1, 2008
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    A 10 MHz speaker?:D:rolleyes:
     
  16. Ron H

    AAC Fanatic!

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    I'm guessing night vision.:D
     
  17. mik3

    Senior Member

    Feb 4, 2008
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    Test the CE amplifier with a 600R resistor connected between the collector and ground to see the difference. You can remove the coupling capacitor and the bias resistors for the CC and connect it directly to the output of the CE. This is not a good practise since any variations of the quiescent point of the CE will affect the quiescent point of the CC but it will increase the input impedance of the CC a lot. Try it and tell us the results.
     
  18. hobbyist

    Distinguished Member

    Aug 10, 2008
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    To recap what your saying thus far.

    1. You apply the appropriate input signal to the CE stage alone with NO load, and you get proper volt. output, is this correct?

    2. You apply proper input signal to the CC stage with the proper load impedance, and you get proper voltoutput. Is that correct?

    Since the CE stage cannot drive the low impeance load, than you are using the CC stage for impedance matching. Is that correct?

    So when you couple the two together you lose the original strage gain of the CE, and thus the output of the complete amp at the CC stage goes down. Is that correct?

    If all this is correct, than the CC stage is loading the CE stage and there is improper matching of the stages.

    You need to redesign the CE stage using a collector resistor that will be at least 10 x smaller than toal impedance looking into CC input circuit. Thats including the emitter complex impedance as well as the divider impedance.

    It may also require you to redesign the CC stage to give high enough values for your CE stage to work with.

    I'm not saying anything different than what has already been suggested, from the other posters.

    Just trying to make it more clearer, to understand.

    Hope this helps.
     
    Last edited: Dec 9, 2009
  19. CDRIVE

    Senior Member

    Jul 1, 2008
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    Does this amp have to be operated class A as shown?
     
  20. CDRIVE

    Senior Member

    Jul 1, 2008
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    Running this as class A is really inefficient but this is what I roughed out. The output is just starting to clip at the input level shown.
     
    Last edited: Dec 11, 2009
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