What is the blue waveform? And the yellow one? Current? Voltage?
If there is power involved, it might be a good idea to decouple the supply voltage of the IC from the inductor & transistor.

 

What is the blue waveform? And the yellow one? Current? Voltage?
If there is power involved, it might be a good idea to decouple the supply voltage of the IC from the inductor & transistor.

Both are voltage waveforms. Blue one of the Timer555 and yellow of transistor. Yes it is de-coupled practically .

 

What are the traces?

Both traces are voltage traces.

 

Both traces are voltage traces.

Measured where?

 

@MikeML One side of the probe in between the collector and one side of the coil, and the other probe to ground.
No i haven't destroyed it as the max voltage MPSA42 can resist is 300V. Okay thanks for doing that.

 

You have destroyed the transistor by running it without the snubber diode.

Why start a completely new thread on a long-running topic? I have asked the moderator to merge it...

 

The open-circuit inductive kick as the transistor turns off can reach thousands of volts. By leaving out the snubber, you are causing the transistor to clamp at its breakdown voltage. There is finite probability that you will blow-up the transistor on any given time you turn it off (without the snubber).

 

@MikeML @BillB3857
fT :- 1. (MPSA42= 50MHz ; conditions (Ic=10mA, Vce=20V, f=100MHz)
2. ( BC 547 = 300MHz ; (Ic=10mA, Vce=5V,f=100MHz)
O/p Capacitance :- MPSA42 = 3pF . ( Vcb=20V, Ie=0 , f= 1MHz)
BC547 = 3.5 to 6pF. ( Vcb=10V, Ie=0 , f= 1MHz)
Can you explain me how does these parameters effect on the working of the transistor??

 

The open-circuit inductive kick as the transistor turns off can reach thousands of volts. By leaving out the snubber, you are causing the transistor to clamp at its breakdown voltage. There is finite probability that you will blow-up the transistor on any given time you turn it off (without the snubber).

Isn't the diode preventing the overvoltage in this case? Like in the typical double-pulse testing?
http://www.compoundsemiconductor.ne...d Semiconductor/July/Industry Sic PE/Fig4.jpg

What if high dV/dt triggers the transistor again when switching off resulting in an oscillation? In that case the transistor should still be not destroyed.

 

To demonstrate the existence of the back emf pulse in the basic switching circuit shown in the OP, with the available analogue CRO could you not just make the pulse more visible by putting a capacitor in series with the snubber diode. Maybe a bleed resistor across it so it will be operational again after a few seconds. I haven't tried this or worked out the values but I would try .1uf and 100K. It may come in under budget...Apologies in advance if this is a rubbish idea but no doubt someone will say !
Oh...and scope across the capacitor of course.

 

fT :- 1. (MPSA42= 50MHz ; conditions (Ic=10mA, Vce=20V, f=100MHz)
2. ( BC 547 = 300MHz ; (Ic=10mA, Vce=5V,f=100MHz)
O/p Capacitance :- MPSA42 = 3pF . ( Vcb=20V, Ie=0 , f= 1MHz)
BC547 = 3.5 to 6pF. ( Vcb=10V, Ie=0 , f= 1MHz)

Can you explain me how does these parameters effect on the working of the transistor??

 

fT=unity gain bandwidth (i.e. nominal Freq at which device gain=1 under the specified conditions)

O/p Capacitance = output (i.e. collector) capacitance (tested as per the stated conditions)

fT is important when considering desired response of the application

The output capacitance specification gives you a 'starting point' when designing the device-to-load matching network...

Best regards
HP

 

fT=unity gain bandwidth (i.e. nominal Freq at which device gain=1 under the specified conditions)

O/p Capacitance = output (i.e. collector) capacitance (tested as per the stated conditions)

Best regards
HP

how does these parameters effect on the working of the transistor??

 

how does these parameters effect on the working of the transistor??

Please re-read my previous reply -- I've amended it with further info

 

Limiting frequency Ft is defined three capacities (Cbe, Ccb and Cd_be=ie*tt/(K*T/e) T=300-->(K*T/e)=25.8mV)
Cbe=F(cje,Veb), Cbc=F(cjc,Vec).
I not comprehensible your purpose and I can not it is correct to answer the question.
Bordodynov.

 

Simply add a zener diode in series with the diode. In your schematic, at JP1. Pin 2 to positive of zener, and Pin 1 to Negative of Zener.
The clamping action is now at a voltage of zener+diode, instead of diode. You should be able to see the back emf.
In fact, doing this increases the life of the relay. Keep the zener voltage as high as your power supply.

 

M

Limiting frequency Ft is defined three capacities (Cbe, Ccb and Cd_be=ie*tt/(K*T/e) T=300-->(K*T/e)=25.8mV)
Cbe=F(cje,Veb), Cbc=F(cjc,Vec).
I not comprehensible your purpose and I can not it is correct to answer the question.
Bordodynov.

My purpose is that i was actually analyzing a waveform of the voltage of the transistor where i observed that when the transistor (MPSA42) should turn off , it turns off only for a moment and again shows some small swings , which i think is due to these parameters . BC547 shows proper results, it turns off for the entire expected period. i would attach the waveform for getting a clear idea.

 

What is the load of the transistor? Fluctuations indicate the presence of inductance. Capacitance of the transistor and the inductor to form a resonant circuit. It should be borne in mind that the inductance can be real and spurious. I had an experience when the output of the repeater to the two transistors (PNP + NPN) was a surge voltage of up to two supply voltages. Then the reason was the long power wires. Bypass power pins condenser cleared up emissions. To reduce Fluctuations shunt power stage capacity. Use a wide and thick wires.
Bordodynov.

 

What is the load of the transistor? Fluctuations indicate the presence of inductance. Capacitance of the transistor and the inductor to form a resonant circuit. It should be borne in mind that the inductance can be real and spurious. I had an experience when the output of the repeater to the two transistors (PNP + NPN) was a surge voltage of up to two supply voltages. Then the reason was the long power wires. Bypass power pins condenser cleared up emissions. To reduce Fluctuations shunt power stage capacity. Use a wide and thick wires.
Bordodynov.

A 12V dc motor is the load of the transistor.

 

A 12V dc motor is the load of the transistor.

For the stated application (motor driver) fT and interelectrode capacitance are not important characteristics --- Ic (maximum collector current), hfe or 'beta' (forward transadmitance [essentionaly current gain]) and maximum interelectrode 'voltages' will be the significant considerations...

Best regards
HP