#34 Emitter degeneration: less gain, easier input drive

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Start with the gain-versus-input tradeoff

   

   Tradeoff: smaller output swing and less input current; both outputs invert

   Symbolic model; same I_CQ established separately, not by adding R_E alone

   Shared scale within each gauge family; derive and replay

   Narration transcript

   Why would you deliberately make an amplifier produce less output? Watch two separately biased common emitter stages receive the same small input. The left emitter is at signal ground; the right has an external resistor with no bypass capacitor. Both collector outputs invert. But the right stage has a smaller output swing and takes less current from its input. That is our question: what do we gain by accepting less voltage gain? These are instruments showing a calculated signal model, not moving charges inside a transistor. Both stages have the same collector bias current, established separately. Adding a resistor alone would not preserve that operating point. Input instruments share a scale, both output instruments share a different scale, and the two base-current instruments share a third. We will derive the tradeoff and replay this experiment at the end.

2. 2. Start with the gain-versus-input tradeoff

   

   Input at base, output at collector; external R_E has no bypass

   Bias network omitted; separately establish forward-active Q

   Narration transcript

   Read the actual signal path. The input reaches the base through the first coupling capacitor. The collector resistor connects to the supply, and the second capacitor takes the output from the collector. The emitter now reaches ground through an external resistor. There is no bypass capacitor across it in this lesson. The original drawing omits the base bias network. As before, assume a valid forward active operating point has already been established; this signal diagram alone is not a complete buildable amplifier.

3. 3. The emitter is no longer signal ground

   

   Midband: coupling capacitors ≈ shorts; supply is signal ground

   This lesson takes r_o → ∞, unlike the prior example

   Emitter node can move across the unbypassed R_E

   Narration transcript

   We again use a midband small signal model. Coupling capacitors are approximated as signal short circuits, transistor capacitances are neglected, and the constant supply becomes signal ground because its small signal change is zero. In this particular source, transistor output resistance is omitted: we explicitly take r o as infinite. Do not carry the finite r o from the previous numerical example into this simplified circuit. The external emitter resistor remains, so its upper node can move with the signal.

4. 4. The emitter is no longer signal ground

   

   Transistor responds to base–emitter gap, not base height alone

   v_π=v_in−v_e; local negative feedback

   v_in=v_e+v_π simultaneously, not a delayed sequence

   Collector still inverts; feedback reduces junction change

   Narration transcript

   Think of the emitter as a floor that can rise under the base. The transistor responds to the gap between them, not to the base height alone. A positive base signal raises emitter current and the voltage across the emitter resistor. Subtract that emitter voltage from the base input to obtain the controlling junction voltage. This is local negative feedback, often called emitter degeneration. In the close-up, input equals emitter voltage plus junction voltage at every instant. We are explaining the relations in order, not claiming that the circuit waits between steps: all the displayed values solve the same instantaneous model. The output is still taken at the collector and still inverts. Feedback reduces the junction change; it does not erase the input.

5. 5. Follow both currents through the emitter resistor

   R_E carries i_e=i_b+i_c, not i_b alone

   $\displaystyle i_{c}=\beta i_{b}; i_{e}=\left(\beta +1\right)i_{b}$

   Narration transcript

   Now identify the currents in the hybrid pi model. Base current flows through r pi from base to emitter. The controlled source adds collector current into the emitter node. Therefore the external emitter resistor carries their sum, not base current alone. Inside this model collector current is beta times base current, so emitter current is beta plus one times base current. Keeping that plus one is important when distinguishing an exact relation in the chosen model from a high beta approximation.

6. 6. Follow both currents through the emitter resistor

   $\displaystyle v_{\mathrm{in}}=i_{b} r_{\pi }+\left(\beta +1\right)i_{b} R_{E}$

   R_in=r_π+(β+1)R_E; before bias-network shunting

   Narration transcript

   Walk from the base to ground. The input voltage equals the drop across r pi plus the drop across the external emitter resistor. The first drop is base current times r pi. The second is beta plus one times base current times emitter resistance. Factor out base current and divide input voltage by it. The input resistance is r pi plus beta plus one times emitter resistance. This is the resistance looking into the transistor base in the stated model, before adding any actual base bias network in parallel.

7. 7. Keep the minus sign and the assumptions

   v_o=−i_c R_C; more current lowers collector voltage

   A_v=−βR_C/(r_π+(β+1)R_E); finite β kept

   Narration transcript

   At the collector, current balance gives output voltage equal to negative collector current times collector resistance. The minus sign has the same origin as in the grounded emitter stage: more collector current creates more resistor drop, so collector voltage falls. Replace collector current with beta times base current and divide by the input voltage we just derived. The voltage gain is negative beta times collector resistance divided by r pi plus beta plus one times emitter resistance. No large beta approximation has been used in this step.

8. 8. Keep the minus sign and the assumptions

   $\displaystyle A_{v}=-\frac{g_{m} R_{C}}{1+g_{m} R_{E}\left(1+\frac{1}{\beta }\right)}$

   Large β:

   $\displaystyle A_{v}\approx -\frac{R_{C}}{\frac{1}{g_{m}}+R_{E}}$

   Original final box missed minus sign; algebra and physics require inversion

   Narration transcript

   There is another useful form. Since r pi equals beta divided by g m, divide the denominator by beta. Gain becomes negative g m times collector resistance divided by one plus g m times emitter resistance times one plus one over beta. The extra denominator shows the reduction caused by feedback. Only when beta is large may we neglect one over beta and write approximately negative collector resistance divided by one over g m plus emitter resistance. The final boxed formula in the original note omits a minus sign; the preceding algebra and the physics both require inversion.

9. 9. Keep the minus sign and the assumptions

   Only with large β AND g_m R_E≫1:

   $\displaystyle A_{v}\approx -\frac{R_{C}}{R_{E}}$

   Capital R_E external; small r_e=V_T/I_EQ intrinsic

   g_m=α/r_e; replacing α with 1 is an approximation

   Narration transcript

   If g m times emitter resistance is also much greater than one, the high beta expression approaches negative collector resistance divided by emitter resistance. The gain then depends mainly on a resistor ratio, but only under both stated limits. Also keep two similarly named quantities separate. Capital R E is the external resistor. Small r e is the intrinsic emitter resistance defined as thermal voltage divided by emitter bias current. The exact identity is g m equals alpha divided by small r e. Replacing it with one over small r e is another high beta approximation.

10. 10. Compare two stages at the same bias current

    Added example:

    $\displaystyle I_{\mathrm{CQ}}=1 \left[\mathrm{mA}\right]; \beta =100; V_{T}=26 \left[\mathrm{mV}\right]; R_{C}=2 k\Omega ; R_{E}=200 \Omega$

    $\displaystyle 1 \left[\mathrm{mA}\right]=0.001 A; 26 \left[\mathrm{mV}\right]=0.026 V; g_{m}=\frac{I_{\mathrm{CQ}}}{V_{T}}$

    g_m=0.038462 S≈38.462 mS; retain unrounded value

    Bias/headroom must be checked separately; supply voltage unspecified

    Narration transcript

    The source gives no numbers, so these are additional teaching values. Choose one milliamp of collector bias current, beta one hundred, thermal voltage twenty six millivolts, collector resistance two kilohms, and external emitter resistance two hundred ohms. First convert units: one milliamp is zero point zero zero one amp, and twenty six millivolts is zero point zero two six volt. Transconductance is collector bias current divided by thermal voltage. Dividing zero point zero zero one by zero point zero two six gives zero point zero three eight four six two siemens, or thirty eight point four six two millisiemens. A siemens is an amp per volt. Keep the unrounded value for the next step. The bias network is assumed adjusted for forward active operation; no supply voltage is given, so actual voltage headroom still needs a separate design check.

11. 11. Compare two stages at the same bias current

    $\displaystyle r_{\pi }=\frac{\beta }{g_{m}}=100\times \frac{0.026}{0.001}$

    $\displaystyle 100\times 0.026=2.6$

    $\displaystyle r_{\pi }=2600 \Omega =2.6 k\Omega$

    Junction resistance, not full input resistance; equal Q gives equal g_m and r_π

    Narration transcript

    Next find r pi, the incremental resistance from base to emitter. Divide beta by transconductance. With the exact transconductance from the previous step, this is one hundred times zero point zero two six divided by zero point zero zero one. First multiply one hundred by zero point zero two six: that gives two point six. Then divide by zero point zero zero one: the result is two thousand six hundred ohms. Dividing by one thousand converts this to two point six kilohms. This is the junction resistance, not yet the full resistance seen at the base. Because our comparison circuits have the same separately established collector bias current and beta, they have the same r pi and transconductance.

12. 12. Compare two stages at the same bias current

    $\displaystyle R_{\mathrm{in}}=r_{\pi }+\left(\beta +1\right)R_{E}$

    $\displaystyle \left(100+1\right)\times 200 \Omega =20200 \Omega$

    $\displaystyle R_{\mathrm{in}}=2600+20200=22800 \Omega =22.8 k\Omega$

    Grounded emitter: 2.6 kΩ; larger R_in means less i_b

    Narration transcript

    Now build the resistance seen at the base. We already know r pi is two thousand six hundred ohms. The emitter contribution is beta plus one times the external resistor. One hundred plus one gives one hundred one. Multiply one hundred one by two hundred ohms: that gives twenty thousand two hundred ohms. Add the junction resistance: two thousand six hundred plus twenty thousand two hundred equals twenty two thousand eight hundred ohms, or twenty two point eight kilohms. Without the external resistor, this extra term is zero and input resistance is just two point six kilohms. This larger input resistance means less base current for the same applied base voltage; it does not mean emitter current equals base current.

13. 13. Compare two stages at the same bias current

    $\displaystyle -\beta R_{C}=-100\times 2000 \Omega =-200000 \Omega$

    $\displaystyle A_{v}=-\frac{200000}{22800}\approx -\frac{8.772 V}{V}$

    Grounded emitter:

    $\displaystyle A_{v}\approx -\frac{76.923 V}{V}$

    Both negative: amplitude decreases, inversion remains

    Narration transcript

    Use the same input resistance to find voltage gain. The numerator is negative beta times collector resistance. Negative one hundred times two thousand gives negative two hundred thousand ohms. Divide that by twenty two thousand eight hundred ohms: gain is approximately negative eight point seven seven two volts per volt. Ohms cancel; the gain is a voltage ratio. For the grounded emitter comparison, keep the same numerator but divide by two thousand six hundred. That gives approximately negative seventy six point nine two three volts per volt. Both signs are negative, so neither circuit has stopped inverting. The resistor reduces output amplitude, not the number of inversions.

14. 14. Compare two stages at the same bias current

    

    $\displaystyle v_{\mathrm{in}}=1 \left[\mathrm{mV}\right]=0.001 V; i_{b}=\frac{v_{\mathrm{in}}}{R_{\mathrm{in}}}$

    i_b≈4.386×10⁽−8) A=43.86 nA

    v_π=i_b r_π≈0.114 [mV]; only part of the input

    Narration transcript

    Where does a one millivolt base signal go in the feedback stage? First calculate base current. One millivolt is zero point zero zero one volt. Divide by twenty two thousand eight hundred ohms: the result is about four point three eight six times ten to the minus eight amp. That is forty three point eight six nanoamps; nano means one billionth. Keep the unrounded current when multiplying by the two thousand six hundred ohm junction resistance. The junction voltage becomes about zero point zero zero zero one one four volt, or zero point one one four millivolt. Only a small part of the applied one millivolt reaches the controlling junction. Now we can calculate the rest instead of guessing it.

15. 15. Compare two stages at the same bias current

    

    $\displaystyle i_{e}=\left(\beta +1\right)i_{b}\approx 4.430 \left[\mathrm{μA}\right]$

    $\displaystyle v_{e}=i_{e} R_{E}\approx 0.886 \left[\mathrm{mV}\right]$

    $\displaystyle v_{\pi }+v_{e}\approx 0.114+0.886=1 \left[\mathrm{mV}\right]$

    $\displaystyle v_{o}\approx -8.772 \left[\mathrm{mV}\right]$

    Simultaneous incremental voltages, one shared scale

    Narration transcript

    The emitter carries beta plus one times base current. Multiply the unrounded forty three point eight six nanoamps by one hundred one: emitter current is about four point four three zero microamps. Multiplying that by two hundred ohms gives about zero point zero zero zero eight eight six volt, or zero point eight eight six millivolt at the emitter. Add the two voltage changes: zero point one one four plus zero point eight eight six equals one millivolt. That closes the input voltage loop. Finally multiply the known gain, negative eight point seven seven two, by the one millivolt input. The output change is negative eight point seven seven two millivolts. In the close-up the input, emitter and junction instruments share one scale and move simultaneously. These are changes around the bias point, not complete node voltages.

16. 16. Compare two stages at the same bias current

    $\displaystyle \frac{1}{g_{m}}=\frac{0.026}{0.001}=26 \Omega$

    $\displaystyle A_{v}\approx -\frac{2000}{26+200}=-8.850$

    Compare finite β: −8.772; 1/g_m is not exactly r_e

    Narration transcript

    Let us test the high beta approximation, not assume it is exact. One over transconductance is thermal voltage divided by collector current. Zero point zero two six divided by zero point zero zero one gives twenty six ohms. Add the external two hundred ohms: twenty six plus two hundred equals two hundred twenty six ohms. Divide negative two thousand by two hundred twenty six. The approximate gain is negative eight point eight five zero. Compare this with the finite beta value of negative eight point seven seven two. They are close, but the approximation is not an identity. Also, one over transconductance is not exactly the intrinsic emitter resistance: that uses emitter bias current rather than collector bias current.

17. 17. Compare two stages at the same bias current

    Resistor-ratio approximation:

    $\displaystyle -\frac{2000}{200}=-10$

    g_m R_E≈7.692; dropping 1 is still noticeable

    Large β and strong feedback are distinct limits; retain full model for precision

    Narration transcript

    Can we simplify even further to a resistor ratio? Divide negative two thousand ohms by two hundred ohms: the result is negative ten. Why is that farther from the full model? This approximation also drops the one in the feedback denominator. With our values, transconductance times emitter resistance is zero point zero three eight four six two times two hundred, approximately seven point six nine two. That exceeds one, but is not overwhelmingly greater. Dropping the one is therefore a noticeable change. A large beta justifies one approximation; strong feedback justifies another. Do not use the second merely because the first worked. Keep the full finite beta expression when precision matters.

18. 18. Include the source, load and model limits

    

    With r_o → ∞ and zero input:

    $\displaystyle R_{\mathrm{out}}=R_{C}$

    Finite r_o changes this model result and gain

    Narration transcript

    What resistance does a load see looking back into the collector? Set the input signal to zero and remove the external load. With infinite r o, this model has no path for collector voltage to feed back into the base or emitter, so the dependent current change is zero and output resistance is simply collector resistance. This conclusion is specific to the model. Restoring finite r o changes the output resistance and gain. Emitter degeneration does not make every real transistor amplifier's output resistance exactly equal to its collector resistor.

19. 19. Include the source, load and model limits

    Separate extension:

    $\displaystyle R_{L}=10 k\Omega ; R_{\mathrm{sig}}=1 k\Omega$

    $\displaystyle R_{C} R_{L}=20000000 \Omega ²; R_{C}+R_{L}=12000 \Omega$

    R_C ∥ R_L≈1666.667 Ω; unloaded R_out remains R_C

    Narration transcript

    Now change the experiment explicitly: add a ten kilohm collector load and a one kilohm source resistance. These were absent from the paired comparison. We first combine the collector resistor and the load in parallel. In ohms, their product is two thousand times ten thousand, which equals twenty million. Their sum is two thousand plus ten thousand, which equals twelve thousand. Divide the product by the sum: twenty million divided by twelve thousand is about one thousand six hundred sixty six point six six seven ohms. That is smaller than either resistor, as a parallel combination must be. Use this combined resistance for loaded gain; it does not change the definition of unloaded output resistance.

20. 20. Include the source, load and model limits

    Loaded stage gain=−β(R_C ∥ R_L)/R_in

    $\displaystyle A_{v}\approx -\frac{166666.667}{22800}\approx -\frac{7.310 V}{V}$

    Reference is base voltage, not source voltage

    Narration transcript

    First measure gain from the actual base input, after source resistance. The loaded stage gain is negative beta times the parallel collector resistance divided by transistor input resistance. Multiplying negative one hundred by the unrounded parallel resistance gives about negative one hundred sixty six thousand six hundred sixty six point six six seven ohms. Divide by twenty two thousand eight hundred ohms. Stage gain is approximately negative seven point three one zero. The magnitude is smaller than the unloaded eight point seven seven two because the collector load gives more current a path to signal ground. Source resistance has not yet been counted in this ratio: its denominator is the voltage at the base, not the generator voltage.

21. 21. Include the source, load and model limits

    

    Divider=22800/(22800+1000)

    $\displaystyle \frac{v_{\mathrm{in}}}{v_{\mathrm{sig}}}\approx 0.957983$

    Source gain≈−7.003 V/V

    Magnitude falls; include real bias shunt and name both gain voltages

    Narration transcript

    Finally include the source divider. Its denominator is twenty two thousand eight hundred plus one thousand, which equals twenty three thousand eight hundred ohms. Divide twenty two thousand eight hundred by that sum. The base receives about zero point nine five seven nine eight three of the source voltage. Multiply the unrounded loaded stage gain by this fraction. Gain measured from the source is approximately negative seven point zero zero three. Its magnitude must be smaller than the stage gain because this passive divider only reduces the input. If a real base bias network is present, include its signal resistance in parallel at the input and recalculate. Always name the two voltages in a gain ratio; source gain and stage gain are not interchangeable.

22. 22. Include the source, load and model limits

    Original unloaded comparison at 1 [mV] input: −76.923 versus −8.772 [mV] output

    Both invert; v_π≈0.114 [mV]; R_in grows 2.6 → 22.8 kΩ

    Answer: easier input drive and reduced g_m sensitivity, a deliberate tradeoff

    Gauge-family scales differ; check real bias, swing and frequency

    Narration transcript

    Return to the opening question and replay the original two-stage experiment, without the added source resistance or load. The same one millivolt input now makes the difference visible. In the grounded emitter stage the output peak change is about negative seventy six point nine two three millivolts. With the external two hundred ohm resistor, it is about negative eight point seven seven two millivolts. Both still invert. The emitter now follows most of the input, leaving only about zero point one one four millivolt across the controlling junction. The base-current instrument also moves less because input resistance rose from two point six to twenty two point eight kilohms. So why accept less gain? To make the input easier to drive and make the gain less sensitive to transistor transconductance, under our stated feedback model. This is a deliberate tradeoff, not a broken amplifier. The replay uses shared scales within each instrument family, not identical scales for every quantity. Check real bias, headroom and frequency limits before applying this signal model to hardware.

Source video: Electronics Basics #34 | Emitter Degeneration: Why Choose Less Gain? (18:47)