What is the role of the transistor and diode in a soft start circuit? Announcing the arrival...
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What is the role of the transistor and diode in a soft start circuit?
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What is the role of the transistor and diode in a soft start circuit?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Soft Clipping with DiodeHow protective diode protects transistor from breakdown?What is the purpose of transistor in this circuit?What is the role of the diode in the circuit?Diode transistor circuit logicWhat's the role of the transistor in this circuit?Parallel Diode with AC voltage and DC voltageWhat is the use of the diode in this circuitLM317 Soft-start circuitESD diode with zener configuration is equivalent to a circuit containing diode, transistor and zener diode, so how this could be equivalent
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$begingroup$
Please can someone explain the purpose of both the transistor and diode in this soft start circuit
transistors circuit-analysis voltage-regulator diodes linear-regulator
edited 19 hours ago
Andy West
462317
asked 20 hours ago
Soubhagya Ranjan SahooSoubhagya Ranjan Sahoo
615
New contributor
Soubhagya Ranjan Sahoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Please can someone explain the purpose of both the transistor and diode in this soft start circuit
transistors circuit-analysis voltage-regulator diodes linear-regulator
edited 19 hours ago
Andy West
462317
asked 20 hours ago
Soubhagya Ranjan SahooSoubhagya Ranjan Sahoo
615
New contributor
Soubhagya Ranjan Sahoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
this exact circuit(some resistor values changed) along with a description of how it works appears in the datasheet for the LM317
$endgroup$
– IC_Eng
11 hours ago
add a comment |
$begingroup$
Please can someone explain the purpose of both the transistor and diode in this soft start circuit
transistors circuit-analysis voltage-regulator diodes linear-regulator
edited 19 hours ago
Andy West
462317
asked 20 hours ago
Soubhagya Ranjan SahooSoubhagya Ranjan Sahoo
615
New contributor
Soubhagya Ranjan Sahoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Please can someone explain the purpose of both the transistor and diode in this soft start circuit
transistors circuit-analysis voltage-regulator diodes linear-regulator
transistors circuit-analysis voltage-regulator diodes linear-regulator
edited 19 hours ago
Andy West
462317
asked 20 hours ago
Soubhagya Ranjan SahooSoubhagya Ranjan Sahoo
615
New contributor
Soubhagya Ranjan Sahoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 19 hours ago
Andy West
462317
asked 20 hours ago
Soubhagya Ranjan SahooSoubhagya Ranjan Sahoo
615
New contributor
Soubhagya Ranjan Sahoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 19 hours ago
Andy West
462317
edited 19 hours ago
Andy West
462317
edited 19 hours ago
Andy West
462317
462317
asked 20 hours ago
Soubhagya Ranjan SahooSoubhagya Ranjan Sahoo
615
New contributor
Soubhagya Ranjan Sahoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 20 hours ago
Soubhagya Ranjan SahooSoubhagya Ranjan Sahoo
615
asked 20 hours ago
Soubhagya Ranjan SahooSoubhagya Ranjan Sahoo
615
615
New contributor
Soubhagya Ranjan Sahoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Soubhagya Ranjan Sahoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Soubhagya Ranjan Sahoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
this exact circuit(some resistor values changed) along with a description of how it works appears in the datasheet for the LM317
$endgroup$
– IC_Eng
11 hours ago
add a comment |
3
$begingroup$
this exact circuit(some resistor values changed) along with a description of how it works appears in the datasheet for the LM317
$endgroup$
– IC_Eng
11 hours ago
3
3
$begingroup$
this exact circuit(some resistor values changed) along with a description of how it works appears in the datasheet for the LM317
$endgroup$
– IC_Eng
11 hours ago
$begingroup$
this exact circuit(some resistor values changed) along with a description of how it works appears in the datasheet for the LM317
$endgroup$
– IC_Eng
11 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The diode is there to discharge C2 through the bulb when the battery is disconnected.
Discharging C2 "resets" the soft start circuit. When C2 is discharged and the battery voltage is applied, the LM317 outputs some voltage at its output (pin 2) this pulls up the voltage at the emitter of the PNP transistor. Since C2 is discharged the PNP's base is still at 0 Volt (I'm assuming the battery's negative connection is ground, unfortunately there is no ground symbol drawn in this schematic).
So there will be some voltage between base and emitter of the PNP which will switch it on. That will limit the voltage at the emitter of the PNP to about 0.7 V.
The LM317 tries to maintain 1.25 V between its pins 1 (ADJ) and 2 (OUT) so the output voltage is now limited to about 0.7 V + 1.25 V = 1.95 V. As long as C2 is not charged.
However, R3 will charge C2 so the voltage across C2 will increase, the output voltage of the LM317 will increase with it. The PNP transistor behaves as a voltage buffer, it buffers (copies, with a 0.7V shift up due to Vbe) the voltage at C2 to the ADJ input (pin 1) of the LM317. The output voltage will then be about: Vout = 1.95 V + V(C2).
The charging of C2 stops when the normal output voltage (set by R1 and R2) is reached then the voltage at pin 1 of the LM317 will no longer increase. Then almost no current will flow through the PNP and C2 will be charged to the same voltage as ADJ pin of the LM317.
When the battery is disconnected C2 needs to be discharged quickly so that the circuit is ready for the next startup. This discharging is done by the diode. Without the diode C2 would have to discharge through R3 and the rest of the circuit. That will take a while since R3 has a high value. Through the diode, discharging is almost "immediate".
answered 19 hours ago
BimpelrekkieBimpelrekkie
51.7k246115
$endgroup$
$begingroup$
Just checking to see if I can understand further... Would the ratio between R1/R3 be determining the "rate" of the soft start? (as well as the capacitance of C2, but if that is fixed)
$endgroup$
– Stian Yttervik
18 hours ago
1
$begingroup$
@StianYttervik How fast the voltage is ramped up is only determined by R3 and C2. R1/R2 only sets the end level output voltage. If R1/R2 is changed such that Vout is increased then the ramp up time increases but not the "rate". Getting to the end voltage is done with the same speed (R3, C2) but it takes longer to get there (R1/R2).
$endgroup$
– Bimpelrekkie
16 hours ago
add a comment |
$begingroup$
At the beginning, C2 is not charged so the base of the transistor is at ground and the transistor is conducting (its resistance R is low). This means that the ratio R2/R that dominates the behavior of the LM317 here is high and the LM317 is almost not conducting. As C2 charges, the transistor is less and less conducting and the ratio R2/R becomes lower and lower, which causes the LM317 to conduct more and more. Finally, the transistor is not conducting and the behaviour of the LM317 is dominated by the ratio R2/R1, that fixes the final output voltage.
The diode may be here to protect the LM317 from some reverse current (but I don't see what current), or more probably to discharge C2 after turning off.
answered 19 hours ago
MikeTeXMikeTeX
709416
$endgroup$
$begingroup$
I was editing my answer when Bimplerekkie has posted his own first. Sorry for the repetition.
$endgroup$
– MikeTeX
19 hours ago
$begingroup$
That's no problem. Bimpeirekkie's is a bit easier to read. Use 2 x <Enter> for paragraph breaks.
$endgroup$
– Transistor
9 hours ago
add a comment |
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2 Answers
2
active
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2 Answers
2
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$begingroup$
The diode is there to discharge C2 through the bulb when the battery is disconnected.
Discharging C2 "resets" the soft start circuit. When C2 is discharged and the battery voltage is applied, the LM317 outputs some voltage at its output (pin 2) this pulls up the voltage at the emitter of the PNP transistor. Since C2 is discharged the PNP's base is still at 0 Volt (I'm assuming the battery's negative connection is ground, unfortunately there is no ground symbol drawn in this schematic).
So there will be some voltage between base and emitter of the PNP which will switch it on. That will limit the voltage at the emitter of the PNP to about 0.7 V.
The LM317 tries to maintain 1.25 V between its pins 1 (ADJ) and 2 (OUT) so the output voltage is now limited to about 0.7 V + 1.25 V = 1.95 V. As long as C2 is not charged.
However, R3 will charge C2 so the voltage across C2 will increase, the output voltage of the LM317 will increase with it. The PNP transistor behaves as a voltage buffer, it buffers (copies, with a 0.7V shift up due to Vbe) the voltage at C2 to the ADJ input (pin 1) of the LM317. The output voltage will then be about: Vout = 1.95 V + V(C2).
The charging of C2 stops when the normal output voltage (set by R1 and R2) is reached then the voltage at pin 1 of the LM317 will no longer increase. Then almost no current will flow through the PNP and C2 will be charged to the same voltage as ADJ pin of the LM317.
When the battery is disconnected C2 needs to be discharged quickly so that the circuit is ready for the next startup. This discharging is done by the diode. Without the diode C2 would have to discharge through R3 and the rest of the circuit. That will take a while since R3 has a high value. Through the diode, discharging is almost "immediate".
answered 19 hours ago
BimpelrekkieBimpelrekkie
51.7k246115
$endgroup$
$begingroup$
Just checking to see if I can understand further... Would the ratio between R1/R3 be determining the "rate" of the soft start? (as well as the capacitance of C2, but if that is fixed)
$endgroup$
– Stian Yttervik
18 hours ago
1
$begingroup$
@StianYttervik How fast the voltage is ramped up is only determined by R3 and C2. R1/R2 only sets the end level output voltage. If R1/R2 is changed such that Vout is increased then the ramp up time increases but not the "rate". Getting to the end voltage is done with the same speed (R3, C2) but it takes longer to get there (R1/R2).
$endgroup$
– Bimpelrekkie
16 hours ago
add a comment |
$begingroup$
The diode is there to discharge C2 through the bulb when the battery is disconnected.
Discharging C2 "resets" the soft start circuit. When C2 is discharged and the battery voltage is applied, the LM317 outputs some voltage at its output (pin 2) this pulls up the voltage at the emitter of the PNP transistor. Since C2 is discharged the PNP's base is still at 0 Volt (I'm assuming the battery's negative connection is ground, unfortunately there is no ground symbol drawn in this schematic).
So there will be some voltage between base and emitter of the PNP which will switch it on. That will limit the voltage at the emitter of the PNP to about 0.7 V.
The LM317 tries to maintain 1.25 V between its pins 1 (ADJ) and 2 (OUT) so the output voltage is now limited to about 0.7 V + 1.25 V = 1.95 V. As long as C2 is not charged.
However, R3 will charge C2 so the voltage across C2 will increase, the output voltage of the LM317 will increase with it. The PNP transistor behaves as a voltage buffer, it buffers (copies, with a 0.7V shift up due to Vbe) the voltage at C2 to the ADJ input (pin 1) of the LM317. The output voltage will then be about: Vout = 1.95 V + V(C2).
The charging of C2 stops when the normal output voltage (set by R1 and R2) is reached then the voltage at pin 1 of the LM317 will no longer increase. Then almost no current will flow through the PNP and C2 will be charged to the same voltage as ADJ pin of the LM317.
When the battery is disconnected C2 needs to be discharged quickly so that the circuit is ready for the next startup. This discharging is done by the diode. Without the diode C2 would have to discharge through R3 and the rest of the circuit. That will take a while since R3 has a high value. Through the diode, discharging is almost "immediate".
answered 19 hours ago
BimpelrekkieBimpelrekkie
51.7k246115
$endgroup$
$begingroup$
Just checking to see if I can understand further... Would the ratio between R1/R3 be determining the "rate" of the soft start? (as well as the capacitance of C2, but if that is fixed)
$endgroup$
– Stian Yttervik
18 hours ago
1
$begingroup$
@StianYttervik How fast the voltage is ramped up is only determined by R3 and C2. R1/R2 only sets the end level output voltage. If R1/R2 is changed such that Vout is increased then the ramp up time increases but not the "rate". Getting to the end voltage is done with the same speed (R3, C2) but it takes longer to get there (R1/R2).
$endgroup$
– Bimpelrekkie
16 hours ago
add a comment |
$begingroup$
The diode is there to discharge C2 through the bulb when the battery is disconnected.
Discharging C2 "resets" the soft start circuit. When C2 is discharged and the battery voltage is applied, the LM317 outputs some voltage at its output (pin 2) this pulls up the voltage at the emitter of the PNP transistor. Since C2 is discharged the PNP's base is still at 0 Volt (I'm assuming the battery's negative connection is ground, unfortunately there is no ground symbol drawn in this schematic).
So there will be some voltage between base and emitter of the PNP which will switch it on. That will limit the voltage at the emitter of the PNP to about 0.7 V.
The LM317 tries to maintain 1.25 V between its pins 1 (ADJ) and 2 (OUT) so the output voltage is now limited to about 0.7 V + 1.25 V = 1.95 V. As long as C2 is not charged.
However, R3 will charge C2 so the voltage across C2 will increase, the output voltage of the LM317 will increase with it. The PNP transistor behaves as a voltage buffer, it buffers (copies, with a 0.7V shift up due to Vbe) the voltage at C2 to the ADJ input (pin 1) of the LM317. The output voltage will then be about: Vout = 1.95 V + V(C2).
The charging of C2 stops when the normal output voltage (set by R1 and R2) is reached then the voltage at pin 1 of the LM317 will no longer increase. Then almost no current will flow through the PNP and C2 will be charged to the same voltage as ADJ pin of the LM317.
When the battery is disconnected C2 needs to be discharged quickly so that the circuit is ready for the next startup. This discharging is done by the diode. Without the diode C2 would have to discharge through R3 and the rest of the circuit. That will take a while since R3 has a high value. Through the diode, discharging is almost "immediate".
answered 19 hours ago
BimpelrekkieBimpelrekkie
51.7k246115
$endgroup$
The diode is there to discharge C2 through the bulb when the battery is disconnected.
Discharging C2 "resets" the soft start circuit. When C2 is discharged and the battery voltage is applied, the LM317 outputs some voltage at its output (pin 2) this pulls up the voltage at the emitter of the PNP transistor. Since C2 is discharged the PNP's base is still at 0 Volt (I'm assuming the battery's negative connection is ground, unfortunately there is no ground symbol drawn in this schematic).
So there will be some voltage between base and emitter of the PNP which will switch it on. That will limit the voltage at the emitter of the PNP to about 0.7 V.
The LM317 tries to maintain 1.25 V between its pins 1 (ADJ) and 2 (OUT) so the output voltage is now limited to about 0.7 V + 1.25 V = 1.95 V. As long as C2 is not charged.
However, R3 will charge C2 so the voltage across C2 will increase, the output voltage of the LM317 will increase with it. The PNP transistor behaves as a voltage buffer, it buffers (copies, with a 0.7V shift up due to Vbe) the voltage at C2 to the ADJ input (pin 1) of the LM317. The output voltage will then be about: Vout = 1.95 V + V(C2).
The charging of C2 stops when the normal output voltage (set by R1 and R2) is reached then the voltage at pin 1 of the LM317 will no longer increase. Then almost no current will flow through the PNP and C2 will be charged to the same voltage as ADJ pin of the LM317.
When the battery is disconnected C2 needs to be discharged quickly so that the circuit is ready for the next startup. This discharging is done by the diode. Without the diode C2 would have to discharge through R3 and the rest of the circuit. That will take a while since R3 has a high value. Through the diode, discharging is almost "immediate".
answered 19 hours ago
BimpelrekkieBimpelrekkie
51.7k246115
edited 9 hours ago
answered 19 hours ago
BimpelrekkieBimpelrekkie
51.7k246115
answered 19 hours ago
BimpelrekkieBimpelrekkie
51.7k246115
answered 19 hours ago
BimpelrekkieBimpelrekkie
51.7k246115
51.7k246115
$begingroup$
Just checking to see if I can understand further... Would the ratio between R1/R3 be determining the "rate" of the soft start? (as well as the capacitance of C2, but if that is fixed)
$endgroup$
– Stian Yttervik
18 hours ago
1
$begingroup$
@StianYttervik How fast the voltage is ramped up is only determined by R3 and C2. R1/R2 only sets the end level output voltage. If R1/R2 is changed such that Vout is increased then the ramp up time increases but not the "rate". Getting to the end voltage is done with the same speed (R3, C2) but it takes longer to get there (R1/R2).
$endgroup$
– Bimpelrekkie
16 hours ago
add a comment |
$begingroup$
Just checking to see if I can understand further... Would the ratio between R1/R3 be determining the "rate" of the soft start? (as well as the capacitance of C2, but if that is fixed)
$endgroup$
– Stian Yttervik
18 hours ago
1
$begingroup$
@StianYttervik How fast the voltage is ramped up is only determined by R3 and C2. R1/R2 only sets the end level output voltage. If R1/R2 is changed such that Vout is increased then the ramp up time increases but not the "rate". Getting to the end voltage is done with the same speed (R3, C2) but it takes longer to get there (R1/R2).
$endgroup$
– Bimpelrekkie
16 hours ago
$begingroup$
Just checking to see if I can understand further... Would the ratio between R1/R3 be determining the "rate" of the soft start? (as well as the capacitance of C2, but if that is fixed)
$endgroup$
– Stian Yttervik
18 hours ago
$begingroup$
Just checking to see if I can understand further... Would the ratio between R1/R3 be determining the "rate" of the soft start? (as well as the capacitance of C2, but if that is fixed)
$endgroup$
– Stian Yttervik
18 hours ago
1
1
$begingroup$
@StianYttervik How fast the voltage is ramped up is only determined by R3 and C2. R1/R2 only sets the end level output voltage. If R1/R2 is changed such that Vout is increased then the ramp up time increases but not the "rate". Getting to the end voltage is done with the same speed (R3, C2) but it takes longer to get there (R1/R2).
$endgroup$
– Bimpelrekkie
16 hours ago
$begingroup$
@StianYttervik How fast the voltage is ramped up is only determined by R3 and C2. R1/R2 only sets the end level output voltage. If R1/R2 is changed such that Vout is increased then the ramp up time increases but not the "rate". Getting to the end voltage is done with the same speed (R3, C2) but it takes longer to get there (R1/R2).
$endgroup$
– Bimpelrekkie
16 hours ago
add a comment |
$begingroup$
At the beginning, C2 is not charged so the base of the transistor is at ground and the transistor is conducting (its resistance R is low). This means that the ratio R2/R that dominates the behavior of the LM317 here is high and the LM317 is almost not conducting. As C2 charges, the transistor is less and less conducting and the ratio R2/R becomes lower and lower, which causes the LM317 to conduct more and more. Finally, the transistor is not conducting and the behaviour of the LM317 is dominated by the ratio R2/R1, that fixes the final output voltage.
The diode may be here to protect the LM317 from some reverse current (but I don't see what current), or more probably to discharge C2 after turning off.
answered 19 hours ago
MikeTeXMikeTeX
709416
$endgroup$
$begingroup$
I was editing my answer when Bimplerekkie has posted his own first. Sorry for the repetition.
$endgroup$
– MikeTeX
19 hours ago
$begingroup$
That's no problem. Bimpeirekkie's is a bit easier to read. Use 2 x <Enter> for paragraph breaks.
$endgroup$
– Transistor
9 hours ago
add a comment |
$begingroup$
At the beginning, C2 is not charged so the base of the transistor is at ground and the transistor is conducting (its resistance R is low). This means that the ratio R2/R that dominates the behavior of the LM317 here is high and the LM317 is almost not conducting. As C2 charges, the transistor is less and less conducting and the ratio R2/R becomes lower and lower, which causes the LM317 to conduct more and more. Finally, the transistor is not conducting and the behaviour of the LM317 is dominated by the ratio R2/R1, that fixes the final output voltage.
The diode may be here to protect the LM317 from some reverse current (but I don't see what current), or more probably to discharge C2 after turning off.
answered 19 hours ago
MikeTeXMikeTeX
709416
$endgroup$
$begingroup$
I was editing my answer when Bimplerekkie has posted his own first. Sorry for the repetition.
$endgroup$
– MikeTeX
19 hours ago
$begingroup$
That's no problem. Bimpeirekkie's is a bit easier to read. Use 2 x <Enter> for paragraph breaks.
$endgroup$
– Transistor
9 hours ago
add a comment |
$begingroup$
At the beginning, C2 is not charged so the base of the transistor is at ground and the transistor is conducting (its resistance R is low). This means that the ratio R2/R that dominates the behavior of the LM317 here is high and the LM317 is almost not conducting. As C2 charges, the transistor is less and less conducting and the ratio R2/R becomes lower and lower, which causes the LM317 to conduct more and more. Finally, the transistor is not conducting and the behaviour of the LM317 is dominated by the ratio R2/R1, that fixes the final output voltage.
The diode may be here to protect the LM317 from some reverse current (but I don't see what current), or more probably to discharge C2 after turning off.
answered 19 hours ago
MikeTeXMikeTeX
709416
$endgroup$
At the beginning, C2 is not charged so the base of the transistor is at ground and the transistor is conducting (its resistance R is low). This means that the ratio R2/R that dominates the behavior of the LM317 here is high and the LM317 is almost not conducting. As C2 charges, the transistor is less and less conducting and the ratio R2/R becomes lower and lower, which causes the LM317 to conduct more and more. Finally, the transistor is not conducting and the behaviour of the LM317 is dominated by the ratio R2/R1, that fixes the final output voltage.
The diode may be here to protect the LM317 from some reverse current (but I don't see what current), or more probably to discharge C2 after turning off.
answered 19 hours ago
MikeTeXMikeTeX
709416
answered 19 hours ago
MikeTeXMikeTeX
709416
answered 19 hours ago
MikeTeXMikeTeX
709416
answered 19 hours ago
MikeTeXMikeTeX
709416
709416
$begingroup$
I was editing my answer when Bimplerekkie has posted his own first. Sorry for the repetition.
$endgroup$
– MikeTeX
19 hours ago
$begingroup$
That's no problem. Bimpeirekkie's is a bit easier to read. Use 2 x <Enter> for paragraph breaks.
$endgroup$
– Transistor
9 hours ago
add a comment |
$begingroup$
I was editing my answer when Bimplerekkie has posted his own first. Sorry for the repetition.
$endgroup$
– MikeTeX
19 hours ago
$begingroup$
That's no problem. Bimpeirekkie's is a bit easier to read. Use 2 x <Enter> for paragraph breaks.
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– Transistor
9 hours ago
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I was editing my answer when Bimplerekkie has posted his own first. Sorry for the repetition.
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– MikeTeX
19 hours ago
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I was editing my answer when Bimplerekkie has posted his own first. Sorry for the repetition.
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– MikeTeX
19 hours ago
$begingroup$
That's no problem. Bimpeirekkie's is a bit easier to read. Use 2 x <Enter> for paragraph breaks.
$endgroup$
– Transistor
9 hours ago
$begingroup$
That's no problem. Bimpeirekkie's is a bit easier to read. Use 2 x <Enter> for paragraph breaks.
$endgroup$
– Transistor
9 hours ago
add a comment |
Soubhagya Ranjan Sahoo is a new contributor. Be nice, and check out our Code of Conduct.
Soubhagya Ranjan Sahoo is a new contributor. Be nice, and check out our Code of Conduct.
Soubhagya Ranjan Sahoo is a new contributor. Be nice, and check out our Code of Conduct.
Soubhagya Ranjan Sahoo is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
this exact circuit(some resistor values changed) along with a description of how it works appears in the datasheet for the LM317
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– IC_Eng
11 hours ago