Hello,

I have a question regarding a basic npn tranisistor circuit.

I got the NJT4031N in the following circuit and everything seems to be fine when Vin is at around 1V (Vbe(on) according to datasheet). However, when it falls below 0.7V instead of turn off, it starts to get quite hot.
Can someone explain why this happens?

Thank you!

 

everything seems to be fine when Vin is at around 1V (Vbe(on) according to datasheet). However, when it falls below 0.7V instead of turn off, it starts to get quite hot.

Without having more information, e.g. collector voltage, I'd say the transistor is coming out of saturation and its power dissipation is increasing.

Do you have a current limiting resistor on the base?

 

... Looking at the datasheet, one specification (max limits) states that the maximum emitter-base voltage is 6.0 vdc. Another specification, two or three paragraphs further on, states that the max base-emitter on voltage is only 1.0 v. So pending further clarification, is it possible that the transistor has been damaged by slightly exceedingi that particular limit? If you have a different part to experiment with, try not to exceed or approach the V b-e limit.

 

Hello,

Have a look here to use the transistor as a switch:
https://electrosome.com/transistor-as-a-switch/

Bertus

 

If the input is around 0.6-0.7V the transistor can be partially on, causing it to heat up.
To turn it off, Vin should be <0.4V.

Below is the LTspice simulation of the circuit with the input going from 0.4V to 0.9V.
The transistor power dissipation (yellow trace) reaches a maximum of 2.7W at an input of 0.68V (blue trace).
At that point the collector current is 0.32A (red trace) and the collector voltage is 8.4V, 1/2 V2 (green trace).

 

Guys thanks for the replies.

I was able to get it working but to be honest I am a bit confused with the datasheet of the component.

I tried to work with the OFF characteristics and giving it 6V (with a current limit resistor) everything is perfect. However, until now I was trying to use it having in mind the ON characteristics (that's why the ~1V input).

But I am still not sure what the designers mean by OFF and ON characteristics.. Any ideas?

 

The 1V is the base to emitter voltage at the highest allowable base current. It is not the recommended operating point.

It is best to think of the bipolar transistor in terms of base amd collector currents instead of voltages.

For switching applications, the base current should typically be 1/10 of the collector current.

Bob

 

As Bob stated, you apply about 1/10th the collector current to the base to fully turn the transistor on.
You apply zero base current (and voltage) to fully turn it off.;
Look at the conditions in the data sheet for testing the collector-emitter saturation voltage (transistor fully on) .

 

Hello,

Looking at the graph that @crutschow showed, it is no wonder that the transistor gets hot.
Take a look at the thermal conditions for the transistor:



As for the saturation conditions that @crutschow mentioned, have a look at the following table:



Bertus