I tried using a schottky, but its was not a reliable solution.Hi,
when powered ON for a while the voltage at C1 becomes saturated. Maybe at about 0.7V. It is limited by V_BE of Q1.
Now you want the capacitor to be discharged by D2. But it has a similar forward voltage than V_BE, thus the discharge process fails.
It mainly discharges via R2.
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There are several solutions.
1) A simple solution is to use a schottky diode in place of D2. This give not the perfect improvement.
Klaus
Yes, i Think i should try this, adding a 3.3V or 4.7V Zener Diode. where again 1N4007 will help me out for faster discharge, atleast till 0.7V2) One could be to charge C1 to a much higher voltage.
maybe add a zener between C1 and Q1
My problem is C1, because its the only capacitor which stays charge.Your problem is not C1, it’s C2. C1 will discharge through D2, R4 and R8, but the discharge path for C2 is through that big R2. You need a low resistance discharge path for C2. Does it really need to be 470uF?
No, C1 is not the the only capacitor that stays charged. Try running a simulation of this.I tried using a schottky, but its was not a reliable solution.
Yes, i Think i should try this, adding a 3.3V or 4.7V Zener Diode. where again 1N4007 will help me out for faster discharge, atleast till 0.7V
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My problem is C1, because its the only capacitor which stays charge.
C2 wont stay charged for longer time because R3, R4, R8 is the only path where C2 would get discharge easily, it wont cross R2. Ofcourse R4 and R8 makes 172 Ohms. and R2 itself is 560 Ohms. and even if luckily it crosses R2, RV1, which is sometimes 50K, would block the path.
i am still facing a problem with adding a zener diode between Q1 base and C1.Hi,
2) One could be to charge C1 to a much higher voltage.
maybe add a zener between C1 and Q1
Klaus
yes, i agree with you. but my problem is how can i discharge the C1 Cap easily as rapidly.No, C1 is not the the only capacitor that stays charged. Try running a simulation of this.
R3,R4, R8, C1 has a time constant of 2.2 seconds, thus, it will take about 10 seconds for C1 to completely discharge. The discharge path for C2 has a time constant of about 37milliseconds, but C1 is going to keep C2 charged.
And, yes, the diode drop of D2 will be a problem in that the base voltage of the transistor won't drop low enough.
Why did you add one? R5 already was there.have added 4K7 resistor to Base and Emitter (GND)
Yes, but with the 3V3 zener the threshod voltage at the capacitor is around 3.7V.as the main as put to OFF, the voltage from 0.72V to 0.60V goes down very slowly which again takes aroung 6 to 7 seconds
After looking at this again, I realize this circuit is not working anywhere near close to how you THINK it's working.i am still facing a problem with adding a zener diode between Q1 base and C1.
have added 4K7 resistor to Base and Emitter (GND). so here when the mains are put to ON, voltage level on Q1 base reaches to 0.72V MAX. but as soon as the main as put to OFF, the voltage from 0.72V to 0.60V goes down very slowly which again takes aroung 6 to 7 seconds, and below 0.60V it decreases rapily say within MAX 1 Seconds.
i Also tried adding 4K7 Resistor across C1, but in such cases C1 does not gets charged. i tried increasing the resistor value too. but does not makes sense.
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yes, i agree with you. but my problem is how can i discharge the C1 Cap easily as rapidly.
for C1 to charge C2, the only path which is available is through RV1, R2, and R3. but before R3, it will go through R4 and R8 Too. which has too dissipate C1 rapidly, but it does not.
It's a rectifier and filter.What is the purpose of D3 and C2?
I don't know how you come to your conclusion...I realize this circuit is not working anywhere near close to how you THINK it's working.
He thinks C1 is controlling the delay, when C2 is the dominant time constant.I don't know how you come to your conclusion...
Maybe the OP can run a simulation....
Klaus
Duh.. I missed that the input was AC.It's a rectifier and filter.
Run a simulation if you’re not convinced. C2 gets charged to 24 volts when power is applied. When power is removed C2 discharges slowly through R3 (4.7K).Hi,
Me too thinks that C1 is controlling the delay....
Klaus
I have to admit, I indeed missed R3.When power is removed C2 discharges slowly through R3 (4.7K).
Yes Sir,Run a simulation if you’re not convinced. C2 gets charged to 24 volts when power is applied. When power is removed C2 discharges slowly through R3 (4.7K).
But, it's even worse: I don't think there's enough voltage to ever drive the base of Q1 on. There's a voltage divider of R3,R4,R8,R2,RV1,R6. With RV1 at minimum, there's about 0.73V. As RV1 is increased, presumably to increase the on-time, the voltage available will drop well below Vbe.
I still don't believe it. First of all, with a 10uF cap, you're going to have a LOT of ripple on your 24VDC. Secondly, with RV1 at 50K, you'll have a maximum of about 400mV on the base of Q1. Are you sure your schematic shown in the original post reflects your actual circuit? Thirdly, the voltage at the base of Q1 should RISE, not fall.Yes Sir,
You were right. Actually C2 was the issue. with the same circuit i changed the value of C2 from 470uF to 10uF.
and R6 to 47K. Which made the system work perfect. Now the voltage at Q1 Base drops down rapidly within a millisecond when RV1 is at minimum and hardly 2 seconds when RV1 is at maximum.
Thanks.
Yes,I still don't believe it. First of all, with a 10uF cap, you're going to have a LOT of ripple on your 24VDC. Secondly, with RV1 at 50K, you'll have a maximum of about 400mV on the base of Q1. Are you sure your schematic shown in the original post reflects your actual circuit? Thirdly, the voltage at the base of Q1 should RISE, not fall.
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