This emitter follower circuit was built and the output was measured on an oscill

Question

This emitter follower circuit was built and the output was measured on an oscilloscope. At 1 V and 1kHz, the output wave form is negatively clipped. Can someone explain how this negative clipping occurs? I'm confused as to whether it is the transistor going into saturation or cut-off that is causing the clipping. Also, why do bumps start to appear below ground when the input amplitude is greater than 1 V? And lastly, when V_EE is connected to V= - 15V instead of ground, why does a full sine wave appear? Thank you.

Explanation / Answer

Solution:

a)

Ib = ( Vin - 0.7 )/270

when, Vin= 1V

Ib = 1.1mA

Now, applying KVL from input to Vee we get,

Vin - 0.7 - Vout = 270 x Ib

Vout = Vin - 1

so When Vin 1

Vout = 1-1 = 0 V

Now we consider the voltage Vbc and Vec,

Vbc = 1-15 = -14 V

and Vec =Vout - Vcc = Vout - 15 v= -ve because Vout never be greater than supply

hence both base-collector and emitter collector are reverse biased and Vout going to be 0V hence this transistor is in cutoff region.

b) When input starts increasing Vout will be more than 0 but still in cutoff region initially which leads to appear bumps below ground level.

c) When Vee = -15

The output voltage is -

Vout + 15 = 3.3 K x (Ic +Ib)

Vout =  3.3 K x (Ic +Ib) - 15

So now Vce = 15 - Vout = 30 - 3.3 K x (Ic +Ib) which is +ve and operates in a reverse active region

hence, full sine wave will appear having phase change of 1800  

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