What is the function of D1 in this circuit?

[dnyberg2]

The attached circuit is not my design but I need to understand it completely.
This is a circuit that supplies power to a sensor.
When S1 is closed the power is removed from RL.
What is the purpose of D1?
In General, what happens if R1 were to short?
I know the base would be subjected to the max avail current of about 1 mA,
but would that ruin the BJT in this config or could it lead to the death of the BJT eventually?
In general, what is the failure mode of a BJT if the base is over driven?
Can one expect the collector & emitter to eventually short?
In this case Q1 is just a switch either in or not in saturation right?
So R4 and R2 rule how much current can get to the sensor even if shorted or does R1 play in that role? In my mind there is no voltage divider on the base so the base gets the full supply voltage. The CE junction is either full on or full off and delivery of current is subject to the datasheet limits of the BJT yes?
Thanks in advanced.

 

[schip666!]

As a SWAG, I think the diode may cause the base of the transistor to be biased up by one diode drop (~.7v) above the emitter voltage... The transistor will not conduct until the base is _more_ than a diode drop above the emitter so this could lower the threshold where the circuit switches on.

What the whole circuit does, and why, is a little mysterious to me though...

 

[dnyberg2]

As a SWAG, I think the diode may cause the base of the transistor to be biased up by one diode drop (~.7v) above the emitter voltage... The transistor will not conduct until the base is _more_ than a diode drop above the emitter so this could lower the threshold where the circuit switches on.

What the whole circuit does, and why, is a little mysterious to me though...

Thanks for the insight! Sound feasible to me! This tiny CKT is nothing more than a PWR switch for a sensor. All it does it turn a sensor on and off / supply current or not. Thanks again. i hope someone else knows more about the other questions on failure...

 

[dnyberg2]

As a SWAG, I think the diode may cause the base of the transistor to be biased up by one diode drop (~.7v) above the emitter voltage... The transistor will not conduct until the base is _more_ than a diode drop above the emitter so this could lower the threshold where the circuit switches on.

What the whole circuit does, and why, is a little mysterious to me though...

Although, now that I think about it some more, that's a Schottky diode with .325 forward voltage drop...

 

[berkeman]

The attached circuit is not my design but I need to understand it completely.
This is a circuit that supplies power to a sensor.
When S1 is closed the power is removed from RL.
What is the purpose of D1?
In General, what happens if R1 were to short?
I know the base would be subjected to the max avail current of about 1 mA,
but would that ruin the BJT in this config or could it lead to the death of the BJT eventually?
In general, what is the failure mode of a BJT if the base is over driven?
Can one expect the collector & emitter to eventually short?
In this case Q1 is just a switch either in or not in saturation right?
So R4 and R2 rule how much current can get to the sensor even if shorted or does R1 play in that role? In my mind there is no voltage divider on the base so the base gets the full supply voltage. The CE junction is either full on or full off and delivery of current is subject to the datasheet limits of the BJT yes?
Thanks in advanced.

It looks like it protects the BE junction when the switch is closed. The reverse BE breakdown is not usually very high. Seems like a regular diode would be fine there too -- maybe they are using the Schottky for its speed and not its lower forward voltage drop?

 

[dnyberg2]

It looks like it protects the BE junction when the switch is closed. The reverse BE breakdown is not usually very high. Seems like a regular diode would be fine there too -- maybe they are using the Schottky for its speed and not its lower forward voltage drop?

That sounds reasonable...
Thank you for your input.

 

[dnyberg2]

It looks like it protects the BE junction when the switch is closed. The reverse BE breakdown is not usually very high. Seems like a regular diode would be fine there too -- maybe they are using the Schottky for its speed and not its lower forward voltage drop?

Care to hazard a guess as to what might happen if the diode were to short?
This is an FMEA in case the crowd was not aware...

 

[berkeman]

Care to hazard a guess as to what might happen if the diode were to short?
This is an FMEA in case the crowd was not aware...

Then the transistor wouldn't work so great.

(FMEA = Failure Modes & Effects Analysis)

 

[yungman]

It is a protection diode. It is a schottky diode that prevent the reverse of the base to emitter voltage. It is common to put a diode like this to prevent damage to the transistor or circuits.

 

[yungman]

Care to hazard a guess as to what might happen if the diode were to short?
This is an FMEA in case the crowd was not aware...

All these years, never seen one fail unless you have arcing of high voltage. Much more often failure because the protection diode is not there.

Look at the application circuits of the good old voltage regulator 7805 etc. They suggest a diode go from output back to input in case the input get shorted by other part of the circuits. The output filtering cap will hold +5V ( 7805) and immediately putting 5 volt from output to input. The diode is used to prevent the voltage at the output more than 0.7V higher than the input to prevent damage...And I never see people put in that diode and the regulator do fail because of that.

For circuits in the high voltage enviroment, I use TVS (transorbs) instead of the diode because they are faster.

 

[Bob S]

When the switch S1 is opened, capacitor C1 charges through R2 and R4 thru Q1. When the switch closes, without the diode, C1 would discharge through R3 and RL (time constant about 1.6 seconds). With the diode, C1 (and the voltage across RL) would discharge through R2 and D1 about 100 times faster. This is probably the main reason for the diode.

Voltages and currents in Q1 are very low, and there does not appear to be any significant failure mode problem. The lowest Q1 voltage limit is the emitter base reverse voltage (about 5 volts). If R1 shorted, the base would be tied to the collector. This is called a transdiode connection, and has been used for emulating an ideal diode V-I curve. Because V_CEsat is about 0.5 volts less than V_BE, most the current will still flow through the collector. The base current is rated in excess of 5 mA (see Q1 datasheet).

Bob S