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| university:courses:electronics:text:chapter-9 [02 May 2019 23:46] – [9.7.2 BJT Version DC Biasing techniques] Sergiy Gavrylenko | university:courses:electronics:text:chapter-9 [07 Oct 2020 16:37] (current) – [AT1 Diode bias generation] Doug Mercer |
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| The impedance R<sub>E</sub> reduces the overall transconductance g<sub>m</sub> of the circuit by a factor of g<sub>m</sub>R<sub>E</sub> + 1, which makes the voltage gain: | The impedance R<sub>E</sub> reduces the overall transconductance g<sub>m</sub> of the circuit by a factor of g<sub>m</sub>R<sub>E</sub> + 1, which makes the voltage gain: |
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| | {{ :university:courses:electronics:text:chptr9-e17.png?290 |}} |
| {{ :university:courses:electronics:text:chptr9-e17.png?300 |}} | |
| (when g<sub>m</sub>R<sub>E</sub> >> 1) | (when g<sub>m</sub>R<sub>E</sub> >> 1) |
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| Going back to our earlier biasing example, figure 9.5.1, values for I<sub>C</sub> = 2mA, R<sub>L</sub> = 3.4KΩ and R<sub>E</sub> = 3KΩ to calculate the small signal gain we first find g<sub>m</sub> = I<sub>C</sub>/V<sub>T</sub> = 2mA/25mV = 0.08. Using our formula for A<sub>V</sub>: | Going back to our earlier biasing example, figure 9.5.1, values for I<sub>C</sub> = 2mA, R<sub>L</sub> = 3.4KΩ and R<sub>E</sub> = 3KΩ to calculate the small signal gain we first find g<sub>m</sub> = I<sub>C</sub>/V<sub>T</sub> = 2mA/25mV = 0.08. Using our formula for A<sub>V</sub>: |
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| {{ :university:courses:electronics:text:chptr9-e18.png?400 |}} | {{ :university:courses:electronics:text:chptr9-e18.png?300 |}} |
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| ====9.5.2 Small signal input impedance with emitter/source degeneration==== | ====9.5.2 Small signal input impedance with emitter/source degeneration==== |
| Using our earlier biasing exercise in figure 9.5.1 as an example but splitting the 3KΩ R<sub>E</sub> into two resistors as in figure 9.5.4 with R<sub>E1</sub>= 1KΩ and R<sub>E2</sub> = 2KΩ with C<sub>1</sub> = 1uF we can recalculate the small signal gain for high frequencies, where C<sub>1</sub> effectively shorts out R<sub>E2</sub>, to be: | Using our earlier biasing exercise in figure 9.5.1 as an example but splitting the 3KΩ R<sub>E</sub> into two resistors as in figure 9.5.4 with R<sub>E1</sub>= 1KΩ and R<sub>E2</sub> = 2KΩ with C<sub>1</sub> = 1uF we can recalculate the small signal gain for high frequencies, where C<sub>1</sub> effectively shorts out R<sub>E2</sub>, to be: |
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| {{ :university:courses:electronics:text:chptr9-e20.png?200 |}} | {{ :university:courses:electronics:text:chptr9-e20.png?300 |}} |
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| The addition of by-pass capacitor C<sub>1</sub>, however, modifies the low frequency response of the circuit. We know from our two gain calculations that the DC gain of the circuit is -1.13 and the gain increases to -3.36 for high frequencies. We can therefore assume that the frequency response consists of a relatively low frequency zero followed by a somewhat higher frequency pole. The formulas for the zero and pole are as follows: | The addition of by-pass capacitor C<sub>1</sub>, however, modifies the low frequency response of the circuit. We know from our two gain calculations that the DC gain of the circuit is -1.13 and the gain increases to -3.36 for high frequencies. We can therefore assume that the frequency response consists of a relatively low frequency zero followed by a somewhat higher frequency pole. The formulas for the zero and pole are as follows: |
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| <m> F_Z = 1/(2πR_E2 C_1) </m> | <m>F_Z = 1/(2 pi R_E2 C_1)</m> |
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| <m> F_P = 1/(2R'_E C_1) </m> | <m>F_P = 1/(2R prime _E C_1)</m> |
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| where R’<sub>E</sub>= R<sub>E2</sub> || (R<sub>E1</sub> + r<sub>e</sub>) | where R’<sub>E</sub>= R<sub>E2</sub> || (R<sub>E1</sub> + r<sub>e</sub>) |
| <WRAP centeralign > Figure AT1.2 Inserting R<sub>2</sub>increases the input resistance </WRAP> | <WRAP centeralign > Figure AT1.2 Inserting R<sub>2</sub>increases the input resistance </WRAP> |
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| ====AT2 Bootstrapping for Higher Input Impedance==== | |
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| Depending on your component choices and signal source, the circuit in figure AT2.1 may load the source so that the input signal is noticeably attenuated when connected to the circuit. That is, at signal frequencies, the input impedance of the circuit may be low compared to the output impedance of the signal source, and so dissipation in the signal source causes attenuation of the signal entering the circuit. To ensure the current into the base of the transistor is negligible, the biasing network must have a relatively low equivalent resistance at DC when looking out of the base. However, there is a clever method we can use to raise the impedance of the network at signal frequencies when looking out of the capacitor. By bootstrapping some of the transistor’s output signal back into the input, we can make the input impedance (at signal frequencies) very large (i.e., approximately R<sub>E</sub>, the input impedance of the transistor). Consider the modified circuit in figure AT2.1. | |
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| Instead of Equation (3.3) and Equation (3.4), assume that | |
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| (R1||R2) + R<sub>B</sub> ≪ βR<sub>E</sub> and β ≈ 100 and C ≥ 1 / 2πf (R1 || R2 + R<sub>B</sub>). (A.1) | |
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| Otherwise, components can be chosen exactly as before. The bootstrapping capacitor C<sub>B</sub> must be very large so that it looks like a short circuit to signal frequencies. Theoretically, the resistor R<sub>B</sub> can be chosen arbitrarily. As long as Equation (A.1) can be met, a high choice of R<sub>B</sub> (e.g., R<sub>B</sub> > 1 kΩ) is a good idea. The signal at the transistor’s emitter follows the signal at its base. At signal frequencies, C<sub>B</sub> acts like a short circuit, and so both ends of R<sub>B</sub> see the same potential. Hence, R<sub>B</sub> carries no current at signal frequencies. Thus, the R1–R2 divider cannot load the input source because no current from the source makes its way across R<sub>B</sub> (i.e., R<sub>B</sub> ≈ ∞ at signal frequencies). The current through R1–R2 that would normally come from the source comes from the output instead. This method is called bootstrapping because we use the circuit’s own output to reduce current required from the input. | |
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| **Return to [[university:courses:electronics:text:chapter-8|Previous Chapter]]** | **Return to [[university:courses:electronics:text:chapter-8|Previous Chapter]]** |
