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BJT Transition Frequency for a High-Frequency Design

  1. Mar 5, 2012 #1
    Hey all,

    If I understand correctly, transition frequency is synonymous with the unity-gain bandwidth frequency. With that said, if you divide the gain at the operating frequency by the transition frequency, you should get the operating frequency. Please correct my logic if I am incorrect.

    The reason why I ask is because I am designing a system that uses HF signals (120 & 125 MHz), and I am in need of a BJT that can handle these sorts of HF signals. At this frequency, I need a current gain, hFE/β somewhere between 5-40 (not exactly sure incoming base current at this time). I'm assuming the gain would need to be somewhere around 5-10.

    Due to its availability and cost ($0.00), I have researched the 2N2222A NPN BJT. It has a transition frequency of 300 MHz. With my logic above, if I were to operate at 125 MHz, I would receive a current gain close to 2.5, well under what I need for my design. Is my logic correct here? Do I need to look for a BJT with a transition frequency closer to 1 GHz to obtain the required current gain I need?

    Thank you for your assistance,

    - Satchmo05

    P.S. I apologize for posting this twice, I realized it would be better for me to post this in the EE section, rather than General Engineering Design.
  2. jcsd
  3. Mar 6, 2012 #2
    I don't have the sure answer, I just join in the discussion.

    Rough estimate, if you want gain of 10 at 125MHz, you want [itex]f_T\;[/itex]=1.25GHz which is 10 time 125MHz. I have designed BJT amplifier circuits with comparable gain faster than 125MHz using 1GHz BJTs

    But in reality, I don't think it is nothing that simple. I think the [itex]f_T\;[/itex] is defined in certain input driving condition. There are other factors in the input driving characteristics that are very important. Don't quote me on this, but... It has a lot to do with input capacitance and Miller capacitance: [itex]C_{be} \;\hbox {and } C_{bc}\;[/itex] respectively. If your drive is 50Ω, you form an RC and there is a pole that relates to [itex]f_T\;[/itex]. But if your drive is very low impedance, it would be faster.

    Another factor is the transit time or something in the semi conductor which I am not familiar with, also play a role in the speed of the BJT.

    I know the s-parameter of RF BJT take into consideration of the input capacitance and Miller capacitance, that the reason you have [itex] S_{11} \;\hbox { and } S_{12}[/itex]. That's the reason you see from Smith Chart the input impedance actually become reactive and goes quite low.

    When you talk about gain, I assume is voltage gain. But if frequency is very high, the input impedance become quite low and driving become a problem. In RF, we mostly talk about power gain rather than voltage gain as it take power to drive the BJT to get output. In another word even if you have a voltage gain equal 10, but if your input impedance become so low that it takes a lot of power to even drive the input.

    Someone more familiar with BJT might be able to answer this.
    Last edited: Mar 6, 2012
  4. Mar 6, 2012 #3
    I surf around a little. [itex]f_T\;[/itex] is defined as frequency where [itex]\beta\;[/itex] falls to unity. It is the frequency current gain drops to unity.

    In another word, [itex]R_{in}= \beta R_E≈R_E\;[/itex] The input impedance beome low and the transistor kind of disappeared!!!!

    Here is a link:

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