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  1. #1
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    Differential pair with dynamic resistance

    Hi,
    I need to calculate and draw asymptotes on the oscilloscope.
    Click image for larger version. 

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    The tested system looks as follows:
    Click image for larger version. 

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    One change - Rc = 20k Ohms.
    I know that for voltage Ub1 in range for +-0.1V (approximately) transistor T2 is in the active mode. And i notice that Uwy1 is constant and this voltage is 5V. So transistor T1 is in cut-off mode. But i don't know how T2 transistor works when Ub1 voltage is less than -0.1V and when Ub1 voltage is greater than 0.1V. Does it work in saturation mode? Could you explain me?
    Thanks for the help in advance!

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  2. #2
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    Dominik Przyborowski's Avatar
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    Re: Differential pair with dynamic resistance

    Basically 3Vt (≈80mV) is an input differential voltage range of a BJT diff pair (you can check math analysis in Grey, Hurst, Meyer book "Analysis and design...").
    If Vin_diff is higher than this 3Vt value one guy is cut-off, while second is saturated.

    However, If your Re is 2.5kΩ, Rc is 20kΩ, input common mode is on ground, VEE is -5.7V and VCC is 12V, then your amplifier is in trouble.
    Only left guy might be in active-linear mode, right is always squeezed by large Rc resistor.
    So, for input voltage below ground, output is going to be ca Vcc-(Vcc-Vee-Vcesat)Rc/(Rc+Re), while in opposite situation, output is pulled up by Rc to Vcc.

    Also, your question is not fully clear. Attached scope screenshot is in time domain and I believe it is showing full range response.
    So, asymptotes of what you need?



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  3. #3
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    Re: Differential pair with dynamic resistance

    T2 is not saturated with Rc=2k, only with 20k.



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    Re: Differential pair with dynamic resistance

    I forgot to mention that the system is powered by a current source IE = 1.5mA. And now I noticed that I sent the wrong layout. The following is correct:
    Click image for larger version. 

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    Re: Differential pair with dynamic resistance

    Okay, it's correct electronic circuit: Click image for larger version. 

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    and it's my assumptions and calculations:
    T1 is cut-off and T2 is saturated then output voltage Uwy2 is approximately -1V.
    My calculation:
    Ucc-Ic*Rc+UBC2=0
    UBC2-voltage base-collector on transistor T2
    UBC2=0.5V because UCE=UCEsat=0.2V and UBE=0.7V
    Then Ic=0.275mA
    Next Uwy2=UCC-Ic*Rc=-0.5V
    Is it OK? My calculations do not differ much from the oscilloscope, but I don't know if I do it correctly...



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  6. #6
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    Re: Differential pair with dynamic resistance

    You new schematic doesn't even give a value for Rc. How could we answer?

    As previously mentioned, saturation occurs above a certain value of Rc. Usually, saturation would be avoided. But it's not clear waht you want to achieve.



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