# Calculating emitter-follower resistances given load power

#### YoGMan

Joined Sep 20, 2017
72
My load power should be 25mW.This is what I have done to choose the resistance values:
1) I put Vcc= 18 V
2) I choose Ve= 9 V to get maximum swing
3)Vb= 9 V+ 0.6 V =9.6 V , from that I got the ratio R1/R2
4) The load will take around 20 mA peak value to dissipate 25mW , so I have biased my transistor to draw 25 mA at Q point
5) From Re=Ve/Ie = 9 V/ 25 mA , I got Re= 360 ohms
6) The capacitance were calculated assuming a low cutoff of 20 Hz
As a summary : R1= 1k , R2=1.143k , Re= 360, RL= 150 ohms.
However the simulations are not giving me my expected values what have I missed here ?View attachment 162221

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#### crutschow

Joined Mar 14, 2008
23,486
So how is the simulation not working?

You are likely missing the fact that the emitter follower can only drive current in the positive direction.
In the negative direction, RE is the only path for the load current so the negative large-signal impedance is much higher than the positive impedance, causing clipping of the negative output.
You can use a lower value for RE to reduce that difference, but that increases bias current and power dissipation of course.
Also, try a smaller input voltage.

Note that just because you calculate 53.18μF for a capacitor value, it makes little sense to use that in the simulation, since you can't buy such a value.
Use the nearest standard value (in this case, 56μF).

#### Bordodynov

Joined May 20, 2015
2,418
Re<Ve_min/20mA=300Ohm

#### anil kumar_1540361006

Joined Oct 24, 2018
1
My load power should be 25mW.This is what I have done to choose the resistance values:
1) I put Vcc= 18 V
2) I choose Ve= 9 V to get maximum swing
3)Vb= 9 V+ 0.6 V =9.6 V , from that I got the ratio R1/R2
4) The load will take around 20 mA peak value to dissipate 25mW , so I have biased my transistor to draw 25 mA at Q point
5) From Re=Ve/Ie = 9 V/ 25 mA , I got Re= 360 ohms
6) The capacitance were calculated assuming a low cutoff of 20 Hz
As a summary : R1= 1k , R2=1.143k , Re= 360, RL= 150 ohms.
However the simulations are not giving me my expected values what have I missed here ?View attachment 162221
hihuhiu

#### Picbuster

Joined Dec 2, 2013
979
My load power should be 25mW.This is what I have done to choose the resistance values:
1) I put Vcc= 18 V
2) I choose Ve= 9 V to get maximum swing
3)Vb= 9 V+ 0.6 V =9.6 V , from that I got the ratio R1/R2
4) The load will take around 20 mA peak value to dissipate 25mW , so I have biased my transistor to draw 25 mA at Q point
5) From Re=Ve/Ie = 9 V/ 25 mA , I got Re= 360 ohms
6) The capacitance were calculated assuming a low cutoff of 20 Hz
As a summary : R1= 1k , R2=1.143k , Re= 360, RL= 150 ohms.
However the simulations are not giving me my expected values what have I missed here ?View attachment 162221
First step calculate the DC situation voltage over RE. ( no AC connected)
V_RE should have a value allowing the AC to flow to RL if not recalculate the base resistors.
next calculate the current in the base as result of 2KHz 6V and C2
then calculate impedance Zout =RL+ C1 // E
Zout carries DC + AC component
calculate AC current or voltage related to RL.

Good luck

Picbuster

#### YoGMan

Joined Sep 20, 2017
72
The circuit is giving me around 84 mW instead of 25 mW but I will try to revise the values and keep you updated on how its working .Thanks for you reply cheers

#### YoGMan

Joined Sep 20, 2017
72
So how is the simulation not working?

You are likely missing the fact that the emitter follower can only drive current in the positive direction.
In the negative direction, RE is the only path for the load current so the negative large-signal impedance is much higher than the positive impedance, causing clipping of the negative output.
You can use a lower value for RE to reduce that difference, but that increases bias current and power dissipation of course.
Also, try a smaller input voltage.

Note that just because you calculate 53.18μF for a capacitor value, it makes little sense to use that in the simulation, since you can't buy such a value.
Use the nearest standard value (in this case, 56μF).
Hello friend thank you .The input voltage was too high , 2 Vrms is the maximum I should input and I now get expected 25mW maximum output power.I will re adjust the values to readily available ones

#### crutschow

Joined Mar 14, 2008
23,486
2 Vrms is the maximum I should input
My simulation with 2Vrms shows slight clipping on the negative peaks.
Reducing the value of RE or increasing the value of R2 will eliminate that.

#### YoGMan

Joined Sep 20, 2017
72
My simulation with 2Vrms shows slight clipping on the negative peaks.
Reducing the value of RE or increasing the value of R2 will eliminate that.
I did it and its working just fine now. I am designing two more voltage divider amplifier to connect to the emitter follower and only one problem I'm facing. The internal capacitances of the BJTs are fixed and to get a high cutoff frequency at around 5kHz require around 2M ohm .I wanted to know if I can add a small capacitor of around 100nF between the base and ground to bring the high cutoff at 5 kHz or is there any other way beside building a low pass filter at the output ?

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#### crutschow

Joined Mar 14, 2008
23,486
if I can add a small capacitor of around 100nF between the base and ground to bring the high cutoff at 5 kHz
You need to add an input series resistor to give an RC rolloff.
100Ω in series with 240nF to ground will give a -3dB rolloff at about 5kHz.