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First of: The transistor doesn't work with a gate or like a MOSFET. A BJT that is used to switch (or in most of many other situations) will have a fixed voltage drop from its base to its emitter, if it's being used right. The current flowing into the base is amplified by the transistor by the factor mentioned in its datasheet to get the maximum current that can flow into the collector.

So, if you have +10.7V that goes into the base of the average single small signal BJT through 10kOhm, it will have a base voltage of about 0.7V, so across the resistor 10V will fall and the current into the transistor's base is 1mA. If it then has a hFe of 100 (small signal gain), it can pull a maximum of 100mA into its collector.

Then, why is there a capacitor?

This works as a little trick, first you assume the voltage across the capacitor C1 is 0V and the Q is also 0V, because it comes from a resting situation where nothing happened.

Then, the output Q of the flipflop goes up, and the capacitor still has 0V across it, because to change its voltage it needs to charge. So the transistor also sees the high voltage of +15V through R2. R2 will then source current into the base to allow a current to be drained through the LOAD.

But, because current flows through the capacitor now, it will charge up, the resistor R3 helps with that, pulling R2 to ground with an extra 22k, but remember that the current is mainly determined by the base voltage and R2, across the 22k only the base voltage falls, but it'll help getting the last bit of charge into the capacitor.

As the capacitor charges, it's voltage slowly rises, until at some point the voltage at R2 is so low that the transistor stops conducting.

So this is a way to make a hardware-fixed pulse length.

When the Q goes low again, the voltage between C1 and R2 will go to -15V (approximately) and it will be slowly charged up again through the resistors R2 and R3 and transistor as well (though the transistor now takes a back seat as it only conducts a few hundred nA or less). I would have preferred to see a diode between C1 and R2 to ground with a small series resistance, that quickly takes away the negative charge to best protect the BJT from negative voltages on its base, but...

EDIT: As Spehro points out in his answer, the R2 and R3 also protects from serious harm, because it brings the maximum negative peak down to within acceptable limits on the base, but a protection diode and 330ohm resistor both in 0603 package would not add significant footprint.

First of: The transistor doesn't work with a gate or like a MOSFET. A BJT that is used to switch (or in most of many other situations) will have a fixed voltage drop from its base to its emitter, if it's being used right. The current flowing into the base is amplified by the transistor by the factor mentioned in its datasheet to get the maximum current that can flow into the collector.

So, if you have +10.7V that goes into the base of the average single small signal BJT through 10kOhm, it will have a base voltage of about 0.7V, so across the resistor 10V will fall and the current into the transistor's base is 1mA. If it then has a hFe of 100 (small signal gain), it can pull a maximum of 100mA into its collector.

Then, why is there a capacitor?

This works as a little trick, first you assume the voltage across the capacitor C1 is 0V and the Q is also 0V, because it comes from a resting situation where nothing happened.

Then, the output Q of the flipflop goes up, and the capacitor still has 0V across it, because to change its voltage it needs to charge. So the transistor also sees the high voltage of +15V through R2. R2 will then source current into the base to allow a current to be drained through the LOAD.

But, because current flows through the capacitor now, it will charge up, the resistor R3 helps with that, pulling R2 to ground with an extra 22k, but remember that the current is mainly determined by the base voltage and R2, across the 22k only the base voltage falls, but it'll help getting the last bit of charge into the capacitor.

As the capacitor charges, it's voltage slowly rises, until at some point the voltage at R2 is so low that the transistor stops conducting.

So this is a way to make a hardware-fixed pulse length.

When the Q goes low again, the voltage between C1 and R2 will go to -15V (approximately) and it will be slowly charged up again through the resistors R2 and R3 and transistor as well. I would have preferred to see a diode between C1 and R2 to ground with a small series resistance, that quickly takes away the negative charge to best protect the BJT from negative voltages on its base, but...

First of: The transistor doesn't work with a gate or like a MOSFET. A BJT that is used to switch (or in most of many other situations) will have a fixed voltage drop from its base to its emitter, if it's being used right. The current flowing into the base is amplified by the transistor by the factor mentioned in its datasheet to get the maximum current that can flow into the collector.

So, if you have +10.7V that goes into the base of the average single small signal BJT through 10kOhm, it will have a base voltage of about 0.7V, so across the resistor 10V will fall and the current into the transistor's base is 1mA. If it then has a hFe of 100 (small signal gain), it can pull a maximum of 100mA into its collector.

Then, why is there a capacitor?

This works as a little trick, first you assume the voltage across the capacitor C1 is 0V and the Q is also 0V, because it comes from a resting situation where nothing happened.

Then, the output Q of the flipflop goes up, and the capacitor still has 0V across it, because to change its voltage it needs to charge. So the transistor also sees the high voltage of +15V through R2. R2 will then source current into the base to allow a current to be drained through the LOAD.

But, because current flows through the capacitor now, it will charge up, the resistor R3 helps with that, pulling R2 to ground with an extra 22k, but remember that the current is mainly determined by the base voltage and R2, across the 22k only the base voltage falls, but it'll help getting the last bit of charge into the capacitor.

As the capacitor charges, it's voltage slowly rises, until at some point the voltage at R2 is so low that the transistor stops conducting.

So this is a way to make a hardware-fixed pulse length.

When the Q goes low again, the voltage between C1 and R2 will go to -15V (approximately) and it will be slowly charged up again through the resistors R2 and R3 and transistor as well (though the transistor now takes a back seat as it only conducts a few hundred nA or less). I would have preferred to see a diode between C1 and R2 to ground with a small series resistance, that quickly takes away the negative charge to best protect the BJT from negative voltages on its base, but...

EDIT: As Spehro points out in his answer, the R2 and R3 also protects from serious harm, because it brings the maximum negative peak down to within acceptable limits on the base, but a protection diode and 330ohm resistor both in 0603 package would not add significant footprint.

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source | link

First of: The transistor doesn't work with a gate or like a MOSFET. A BJT that is used to switch (or in most of many other situations) will have a fixed voltage drop from its base to its emitter, if it's being used right. The current flowing into the base is amplified by the transistor by the factor mentioned in its datasheet to get the maximum current that can flow into the collector.

So, if you have +10.7V that goes into the base of the average single small signal BJT through 10kOhm, it will have a base voltage of about 0.7V, so across the resistor 10V will fall and the current into the transistor's base is 1mA. If it then has a hFe of 100 (small signal gain), it can pull a maximum of 100mA into its collector.

Then, why is there a capacitor?

This works as a little trick, first you assume the voltage across the capacitor C1 is 0V and the Q is also 0V, because it comes from a resting situation where nothing happened.

Then, the output Q of the flipflop goes up, and the capacitor still has 0V across it, because to change its voltage it needs to charge. So the transistor also sees the high voltage of +15V through R2. R2 will then source current into the base to allow a current to be drained through the LOAD.

But, because current flows through the capacitor now, it will charge up, the resistor R3 helps with that, pulling R2 to ground with an extra 22k, but remember that the current is mainly determined by the base voltage and R2, across the 22k only the base voltage falls, but it'll help getting the last bit of charge into the capacitor.

As the capacitor charges, it's voltage slowly rises, until at some point the voltage at R2 is so low that the transistor stops conducting.

So this is a way to make a hardware-fixed pulse length.

When the Q goes low again, the voltage between C1 and R2 will go to -15V (approximately) and it will be slowly charged up again through the resistors R2 and R3 and transistor as well. I would have preferred to see a diode between C1 and R2 to ground with a small series resistance, that quickly takes away the negative charge to best protect the BJT from negative voltages on its base, but...