## peter_pan 3 years ago How can I calculate the output impedance of an emitter follower (see the last post for details) and of a transformer? How matching impedances helps us increasing power transfer? Help!

1. emicho

the input impedance: The impedance when you are looking from the input of a device to know the value of the all load from this point " the input" and viceversa for the output impedance which is the one when you are looking from the output " here the emitter" For the follower emitter : $Rin =RB+Beta*(Hib+RL//RE)$ where RB= R1// R2 bias resistors $Rout=[RE//hib]+RB divBeta$

2. nick67

@emicho uhm... I'm not sure about Rout formula; could you better explain how did you obtain it ?

3. emicho

|dw:1342867918740:dw| well if we look at the equivalence circuit and if we include the source resistor"for more details" we will see That Ro is equal to Re//( hib + (Rs//Rb)/Beta ).. that's as i guess :)

4. emicho

ahhh i made a mistake at the first answer :D Re is in parallel with all not only Rl " which we ignore " :D thanks for ur note :)

5. nick67

thank you for explaining (but... that flying current generator betaIB... :D )

6. emicho

hehe but the type of ground you like ^^

7. peter_pan

Sorry, I'm back. I think I understood the first part of emicho's answer, but I'd like to understand the details of the calculations too. I'll write what I know and I hope somebody can help me the parts that put me in trouble. Input and output impedances are defined as follows: |dw:1343404015654:dw| $Z_{IN}={{V_{IN}} \over {I_{IN}}}$ I have connected the circuit A to a battery with e.m.f. = V_IN. Now, with reference to the following figure:|dw:1343404811562:dw| by definition we have: $Z_{out}={{V_{out}} \over I_{shortcircuit}}$ where I_shortcircuit is the current that would pass if I short-circuited the output terminals.

8. peter_pan

Let's now onsider an emitter follower (see file). We want to show that an emitter follower can match impedances. I've understood why $Z_{IN}=(\beta+1)Z_{LOAD}$, where beta is the current gain of the transistor. What about the output impedance? $Z_{out}={V_E \over I_E}={{V_B - 0.6 V} \over {I_B + I_C}}={{V_B - 0.6 V} \over {I_B(\beta+1)}}\approx {V_B \over I_B(1+ \beta)}$ Now, the final result should be: $R_{SOURCE}/(\beta +1)$ My question is: why is $R_{SOURCE}={V_B \over I_B}$ and not -as I would expect- $R_{SOURCE}={{V_B-V_{IN}} \over I_B} ?$ I'll take the opportunity to ask two more questions: 1) how would you calculate the output impedance of a transformer? |dw:1343406894395:dw| 2) What does impedance matching have to do with power transfer?

9. nick67

@peter_pan 0) as for internal source resistance, most books evaluate this parameter shutting down the source, that is Vin=0 (like you should do when using a multimeter to measure resistances in a circuit); other books evaluate Rsource setting Vin at a constant value (d.c.) and making all calculations on the variable part of signals (a.c.), in this second case again variable part of Vin is zero. 1) you should shut down the source and, using a sinusoidal variable frequency generator at the output, you should measure the ratio Vout(f)/Iout(f) at each frequency, reporting obtained values in a graph, that is Zout(f). 2) when a voltage source of value Vs, with internal resistance Rs, has a load with resistance RL, the voltage on the load is: $V _{L}=V _{S} R _{L}/\left( R _{S}+R _{L} \right)$To evaluate the power on the load we use the formula: $P _{L}=V _{L}^{2}/R _{L}$Now let's use a relation between internal source resistance and the load resistance, introducing the m factor: $R _{L}=mR _{S}$If we rewrite previous equation with substitutions we obtain: $P _{L}=V _{S}^{2}mR _{S}/\left( R _{S}+mR _{S} \right)^{2}$After a few manipulations, we obtain: $P _{L}=V _{S}^{2}/\left[ R _{S}\left( m+2+1/m \right) \right]$The way to maximize PL is to minimize the content in brackets: $m+2+1/m$Let's evaluate the first derivative: $(m+2+1/m)\prime=1-1/m ^{2}$To obtain null value we have to set m = 1 (excluding m = -1 that is a nonsense) If we evaluate the second derivative we obtain: $(m+2+1/m)\prime \prime=2/m ^{3}$This confirms that m=1 gives a minimum to the expression in brackets. So, m=1 is a value that gives maximum power to the load. If m=1 then$R _{L}=R _{S}$

10. peter_pan

@nick67: thanks for your answer! 0) Ok, I've got it; 1) I don't understand how this method matches with the definition of output impedance I know, which is $Z_{out}={{V_{AB|open circuit}} \over {I_{shortcircuit}}}$ where V_AB is the voltage between the output terminals of the circuit (the secondary circuit) when it is open (no loads or sources between A and B) and I_shortcircuit is the current that would pass if we shortcircuited A and B! (This is the definition I was given in class!) 2) I understood your reasoning; now, if we want to match impedances, as far as I know, it should be $Z_{IN}>>Z_{out}$. For example if we have a load, Z_IN=RL and Z_out be the impedance of the parts of the circuit which are schematized by the series of the voltage source and its internal resistance. In this case Z_out=RS. Therefore, shouldn't it be RL>>RS? But, if so, this is not RL=RS! So we are not maximizing the power given to the load! Where am I wrong?

11. nick67

@peter_pan 2) let me first answer on this point; when: $R _{S}<<R _{L}$you have maximum voltage on the load; when: $R _{S}>>R _{L}$you have maximum current on the load; as you know power on the load is: $P _{L}=V _{L}timesI _{L}$with a maximum in a situation in middle between first two seen before, i.e. when$R _{S}=R _{L}$This concept is largely adopted in transmission lines, where you have to match (set equal) impedances in order to obtain maximum power transfer to the load that also avoids signal reflections on the line (echoes)

12. nick67

as for point 1) your definition seems related to measuring the output impedance "seen" from the primary circuit (from left), not the output impedance "seen" from the secondary circuit (from right)...

13. peter_pan

As for point 1) I'm a little confused, because I used the same definition when I calculated the output impedance of the emitter follower... As for point 2), in conclusion which of the following ones is the impedance matching condition? RS<<RL (a) or RS=RL (b)? As an example, the emitter follower, which is used for matching impedances, has $Z_{IN}=(\beta+1)Z_{LOAD}>>Z_{out}={Z_{SOURCE} \over (\beta +1)}$ since $\beta \approx 100$ This is why I was convinced that the matching impedance condition was (a)!

14. nick67

@peter_pan May be there is a little confusion about language (I'm not English language mother tongue) when we say that two cascaded systems are "matched", we indicate that their impedances allow maximum power transfer, i.e. the output resistance of the previous one is equal to the input resistance of the second one (e.g. when a video signal amplifier has to be connected to a coaxial cable, you have to match impedances to 75 Ohm, without loosing power and avoiding reflections). With the emitter follower, instead, you have a "buffer" stage, that is a stage that have a very high input resistance and a very low output resistance. This condition ensures maximum voltage transfer from the source to the load (whatever is its value). In other words, combining the real source (with internal resistance >> 0 ) with the "buffer" you obtain kind of an ideal source (with very little internal resistance ). This way, no matter how much current the load requires, the "ideal" source can tolerate the job without decreasing its voltage level (you may use this solution, for instance, to "bufferize" a temperature sensor that has an high output resistance, allowing its output to be delivered to any other circuit, no matter how much higher is the other circuit input resistance, without loosing output voltage amplitude).

15. peter_pan

Ok, you have convinced me. I think I was wrong about the concept of impedance matching and I was wrong about the fact that an emitter follower matches impedances too. Actually, it allows us to change impedances, so that the signal doesn't change when you connect a load (yes, a "buffer"!). Right? Thank you very much!

16. nick67

@peter_pan you're welcome, and again sorry for any language misunderstanding