[SOLVED] Behavior of a Transistor

[bhoke]

Hi to all,

I have a question about the behavior of a transistor.




A voltage source with 50Ω impedance generates voltage. Waveform of Vin is shown above(it varies between -2 and 4 Volts), and the total circuit is shown below.



Positive level of Vin is being changed, while negative level is being kept same. As a result, figures below show the voltage waveforms of Vc and V1.




Orange line shows the Vc(5V/div), Purple line shows V1(2V/div). Can you explain the behavior of the transistor in this circuit?

 

[FvM]

In brief, you see the effect of storage delay in saturated switching.

 

[bhoke]

In brief, you see the effect of storage delay in saturated switching.

Thank you for your answer, but I have one more question.

Well, why does storage delay change as positive voltage level changes?

 

[FvM]

Base charge increases with excessive base current, discharge current is kept constant.

 

[WimRFP]

This is normal behavior of a bipolar junction transistor. It is also present in any PN junction. When you supply more current then you need to almost saturate the transistor, some of the current is stored in the form of base charge.

During your negative half of the pulse, this charge has to be removed before Vbe drops to a level where the transistor goes to off-state. The charge removal take time. This charge is not because of depletion layer capacitance (as used in varicap diodes), it really depends on current.

In the transiistor model this effect is modelled with the reverse transit time parameter. This value is 50...200 higher then the forward transit time parameter.

 

[bhoke]

Thank you very much. I got it.

 

[WimRFP]

Deep saturation is sometimes avoided by placing a schottky diode parallel to the BC junction (baker clamp). Excess base current turns on the schottky diode (as the voltage drop is around 0.4V instead of 0.7V for the BC junction). This bypasses the BC junction and the current adds to the collector current.