Does the vertical link between D2 and D3 'jump over' the horizontal link between R3 and D1, or is there an electrical connection between the two links?

 

What I think I am seeing is that the solenoid (relay coil) is energized for some short time period when the AC voltage is high enough to break-down zener diode D6 plus the forward drop of D4 and the base turn on voltage of Q1.
What is intended to happen after the contacts close is not clear.

 

Does the vertical link between D2 and D3 'jump over' the horizontal link between R3 and D1, or is there an electrical connection between the two links?

Yes there is a connection between them. D1 & D2 is connected with D3.

 

Yes there is a connection between them. D1 & D2 is connected with D3.

Yes, sol will be energized for short period.
I think the solenoid gets lower voltage until switch is closed and will remain energised till the power is present or any other interruption.

What was the purpose of C3, R4 and C5.

Or Is it like when C5 gets charged, it will make D5 reverse bias, Q1 will be off. Similarly after C3 gets charged, D3 is reverse bias, denergizing the solenoid..

Or how will the flow actually gets affected ?

 

THERE IS NO NEED FOR A PROTECTION DIODE ACROSS THE RELAY COIL!!! There is already an RC connected across the coil, and that would absorb any such impulse! BUT because the capacitor also discharges thru the coil when the supply voltage is removed, there is no instant reduction in current, and thus no big voltage pulse produced.
That is covered in the second semester basic electricity course.

 

There is already an RC connected across the coil

No. It's actually connected between the top end of the coil and ground, to supply the coil with a smoothish V+, it seems to me. There's nothing to protect the transistor from a switch-off spike.

 

No. It's actually connected between the top end of the coil and ground, to supply the coil with a smoothish V+, it seems to me. There's nothing to protect the transistor from a switch-off spike.

Except that the transistor does not switch off quickly because of the larger capacitor connected to the base.

 

THERE IS NO NEED FOR A PROTECTION DIODE ACROSS THE RELAY COIL!!! There is already an RC connected across the coil, and that would absorb any such impulse! BUT because the capacitor also discharges thru the coil when the supply voltage is removed, there is no instant reduction in current, and thus no big voltage pulse produced.
That is covered in the second semester basic electricity course.

Hi,

All cap's seems to imply that you are very sure of that part of the reply. How is it that you came to this conclusion? I would not want to guess at this just to save a 5 cent diode
The RC is not connected across the coil, and if the transistor turns off the cap will not be able to discharge through the coil.
Also, there is no capacitor on the base of the transistor it's past the diode and zener. It is a little interesting though that there is no base resistor to handle CB leakage current. I'll have to mention that.

 

Hey @MisterBill2 please refer to this circuit.

Hello again,

You might also want to add a base resistor to the transistor to handle the sometimes problematic collector to base leakage current. This tends to turn the transistor on.
The resistor would go from base to ground. 10k is usually good enough.

 

The base bias is provided by capacitor C5, and while it is not directly tied to the base, it is certainly going to reduce the speed of the transistor switching off, as it discharges thru the base.
In addition, this IS a circuit that would already be in operation, at least that is what I get from post #1. The TS was asking how it worked, rather than requesting some design assistance.

 

The base bias is provided by capacitor C5, and while it is not directly tied to the base, it is certainly going to reduce the speed of the transistor switching off, as it discharges thru the base.
In addition, this IS a circuit that would already be in operation, at least that is what I get from post #1. The TS was asking how it worked, rather than requesting some design assistance.

Point #1: Capacitor C5 turning transistor off slowly.
Point #2: Already in operation.

I don't see Point #2 as that significant unless maybe it's been running for 10 years. Even then it may be because of dumb luck.

For Point #1, you may be right, but then again it depends on how sharp the current cuts off because of the series zener and diode on the base. As the capacitor discharges, eventually the voltage gets lower than the Vz+Vd+Vbe voltage, and that means the transistor might turn off more abruptly when it gets to that point.
There is another point though, and that is if the transistor really does stay on as the cap discharges, then that would mean the current in the coil may be decreasing slowly, and that could mean the contacts open slowly, and that could mean significant contact arching.

Maybe a simulation would help, then we could check the possibility of the current in the coil decreasing too slowly. We'd really have to check this with an added back EMF diode too.

 

No. It's actually connected between the top end of the coil and ground, to supply the coil with a smoothish V+, it seems to me. There's nothing to protect the transistor from a switch-off spike.

I think this operation doesn't have any protection from switch off spike or antiparallel diode because of two reasons, one is that the frequency of operation is very low- 2-3 times a day. And second is, if solenoid have high inductance, back emf wouldn't play major role I guess.

 

The base bias is provided by capacitor C5, and while it is not directly tied to the base, it is certainly going to reduce the speed of the transistor switching off, as it discharges thru the base.
In addition, this IS a circuit that would already be in operation, at least that is what I get from post #1. The TS was asking how it worked, rather than requesting some design assistance.

Yes. This circuit is currently in use. But im not aware about since when. Yes i was asking about the working and flow of the processes.

 

Point #1: Capacitor C5 turning transistor off slowly.
Point #2: Already in operation.

I don't see Point #2 as that significant unless maybe it's been running for 10 years. Even then it may be because of dumb luck.

For Point #1, you may be right, but then again it depends on how sharp the current cuts off because of the series zener and diode on the base. As the capacitor discharges, eventually the voltage gets lower than the Vz+Vd+Vbe voltage, and that means the transistor might turn off more abruptly when it gets to that point.
There is another point though, and that is if the transistor really does stay on as the cap discharges, then that would mean the current in the coil may be decreasing slowly, and that could mean the contacts open slowly, and that could mean significant contact arching.

Maybe a simulation would help, then we could check the possibility of the current in the coil decreasing too slowly. We'd really have to check this with an added back EMF diode too.

When capacitor charges, the current (lb) is around 3.3 mA, when it discharges it is 2 mA. (Calculative)

Is requirement of base resistor can be omitted if I connect an npn in series with the existing npn? Emitter of 1st can be collector of 2nd, with common base.

Also I was trying to simulate in LTSpice but didn't get how to simulate Dpst relay. Could you provide any suggestion to this?

 

When capacitor charges, the current (lb) is around 3.3 mA, when it discharges it is 2 mA. (Calculative)

Is requirement of base resistor can be omitted if I connect an npn in series with the existing npn? Emitter of 1st can be collector of 2nd, with common base.

Also I was trying to simulate in LTSpice but didn't get how to simulate Dpst relay. Could you provide any suggestion to this?

Hi,

Why would you want to use a second transistor, is the gain of just one not enough?

You can look up how to use a switch (sometimes called just "SW") and then just use two of them. Trigger them both with the same pulse source that turns on after some time delay.

 

Hi,

Why would you want to use a second transistor, is the gain of just one not enough?

You can look up how to use a switch (sometimes called just "SW") and then just use two of them. Trigger them both with the same pulse source that turns on after some time delay.

No, not for gain improvement. Second transistor in series to divide voltage. can be helpful in stress management due to spikes?

 

With the slower "turn off" you don't get a spike.

 

You can look up how to use a switch (sometimes called just "SW") and then just use two of them. Trigger them both with the same pulse source that turns on after some time delay.

I tried. But these sw switches are Spst. How can I make it spdt?

 

No, not for gain improvement. Second transistor in series to divide voltage. can be helpful in stress management due to spikes?

Hi,

Not really, that's not the way to handle spikes. Clamping is the way it is usually done, or with a snubber circuit. You only need one transistor then and it's easier to deal with the switching of that one that way too.

You can also check to see how fast the transistor turns off. The large capacitor takes time to discharge, but the zener and series diode could cause the transistor to turn off slowly but then sharply after that. If it turns off sharply then you need to handle the spike.

Is this a one-off circuit or are you going into production or something with this? If it's for a large production run you may want to keep costs down, but for a one-off unit (or for hobby purposes) it doesn't matter much if you add one resistor or one diode.

 

I tried. But these sw switches are Spst. How can I make it spdt?

Hello again,

You said DPST before didn't you? But it doesn't really matter because you can make any type of simple switch.
It's good you know how to use the SPST switch already.

To make a SPDT switch you would just use two switches, but you would have one turn on at say 3v and the other turn off at 2v, something like that. So one turns on at 3v and the other turns off at 2v, or you could just invert the pulse so they both turn on at 3v (with a 5v pulse drive) but due to the inverter the second switch only turns on when the first switch turns off.
Sometimes you need a small "dead time" between the switches, which would mean it would be a break before make switch. In this case you can use two pulse sources, one set to turn on at say t=0.00 and off at t=0.99, and the other to turn on at t=1.00 and off at 1.99, and then the first one would turn back on at t=2.00, so only one switch is on at any time and there is 0.01 seconds between the on/off of the two.
To make a DPST switch (which is what you want I think) you just use two switches and set them both to turn on at 3v. Set the pulse to turn on sometime after startup.

Using these ideas you can make SPST, SPDT, DPST, DPDT, etc., even 3PDT, 4PDT, etc. To get multiple 'throws' you have to use more switches but have them turn on at different times using different pulse sources. For SP3T you'd have to use three switches but also three pulse sources. That's unless you already have logic state outputs that can drive them individually.