This is what I have designed, and part tested with low voltages, a device to allow me to power up using a 12 V on-off switch and also provide some slow start (typical 100ms but we can easily alter this).
Please let me know what you think.
Please let me know what you think.
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
You are going to think I am nitpicking now.
Your circuit looks OK at a quick glance... perhaps connect a pulldown resistor between the base/emmiter of the darlington pair, or use a single FET (2N7000) and small value timing cap, high value resistor.
For a real challenge
,why not design, so that in the off state, it draws no current. That means perhaps using a small 3.6volt backup battery to ensure self starting. A 240volt coil relay switched with an opto triac would work well.
Don't underestimate surge currents either. Thick print on the PCB and heavy duty relays required for transformers over 200/300va.
Instead of all those resistors (are you sure they will survive... you don't need that much series resistance anyway) why not try a thermistor.
Your circuit looks OK at a quick glance... perhaps connect a pulldown resistor between the base/emmiter of the darlington pair, or use a single FET (2N7000) and small value timing cap, high value resistor.
For a real challenge
,why not design, so that in the off state, it draws no current. That means perhaps using a small 3.6volt backup battery to ensure self starting. A 240volt coil relay switched with an opto triac would work well.Don't underestimate surge currents either. Thick print on the PCB and heavy duty relays required for transformers over 200/300va.
Instead of all those resistors (are you sure they will survive... you don't need that much series resistance anyway) why not try a thermistor.
Well, I would say the same thing! Instead of all those resistors why not try a thermistor. I would use one in the hot and one in the neutral. all these before power transformer.
What is an Inrush Current Limiting Thermistor (Surge Limiter)?
Inrush Current Limiters are among the most common design options used in switching power supplies to prevent damage caused by inrush current surges. A thermistor is a thermally-sensitive resistor with a resistance that changes significantly and predictably as a result of temperature changes. The resistance of a Surge Limiting thermistor decreases as its temperature increases. As the inrush current limiter self-heats, the current begins to flow through it. Its resistance begins to drop and a relatively small current flow charges the capacitors in the power supply. After the capacitors in the power supply become charged, the self heated inrush current limiter offers little resistance in the circuit. So low that the voltage drop is an insignificant factor with respect to the total voltage drop of the circuit.
Some members do not feel the need for Thermistors and some do.it is all up-to the design and the project and of course the people.
What is an Inrush Current Limiting Thermistor (Surge Limiter)?
Inrush Current Limiters are among the most common design options used in switching power supplies to prevent damage caused by inrush current surges. A thermistor is a thermally-sensitive resistor with a resistance that changes significantly and predictably as a result of temperature changes. The resistance of a Surge Limiting thermistor decreases as its temperature increases. As the inrush current limiter self-heats, the current begins to flow through it. Its resistance begins to drop and a relatively small current flow charges the capacitors in the power supply. After the capacitors in the power supply become charged, the self heated inrush current limiter offers little resistance in the circuit. So low that the voltage drop is an insignificant factor with respect to the total voltage drop of the circuit.
Some members do not feel the need for Thermistors and some do.it is all up-to the design and the project and of course the people.
50r 25W is perfect for 300VA to 1000VA on UK voltage supplies.
With a 100ms delay they will never get warm. No matter how many times you start up your transformer in any hour.
Will it instant OFF on loss of mains power?
I still don't like the switching transistors following the slow rise of the RC time constant.
Will they pass through the conduction phase without overheating.
I'd prefer to see a fast/snap ON to reduce the time that the transistors spend with medium current and high voltage
With a 100ms delay they will never get warm. No matter how many times you start up your transformer in any hour.
Will it instant OFF on loss of mains power?
I still don't like the switching transistors following the slow rise of the RC time constant.
Will they pass through the conduction phase without overheating.
I'd prefer to see a fast/snap ON to reduce the time that the transistors spend with medium current and high voltage
wrong.lanchile said:
The resistance of the Thermistor modulates with changes in the amplifier current. This is easily measured seen on a scope monitoring the output voltage of the PSU and the output voltage of the amplifier fed by that PSU.
I posted results of the Krell Klone with quiescent current ~ 2.6A and even this ClassA amplifier showed this Thermistor effect.
AndrewT said:
Hmm , let me think. The 2pole relay will switch as soon as the capacitor feeding it gets depleted. I think it is like 300-400 Ohms and the cap is 1000uF, but the cap is also feeding the second relay, another 300-400 Ohms.
Will 400 Ohms each relay coil, and a 1000uF capacitor with no other load (for example a LED on the on/off switch) the relays will flip 200ms after loss of mains. Not sure which of the two will go first. If there is a LED or other current consumption on the 12 VDC then the relays will flip sooner, subject to their own speed of course, which is measured in 10s of ms.
But why does it matter if the circuit is an instant OFF on loss of mains?
AndrewT said:
I now see what you mean, very subtle point, and not something I had thought about, even though I did think or power/current requirements for the driver transistors and chose the BC639 which is 1A and 1W.
In the ordinary life of the circuitry, the transistor will go from OFF to ON in an instant (say 200ms), and then stay ON, it will not oscillate.
Assuming the worst, we are talking about 6 Volts into 300-400 R, that is 120mW so we should be OK.
Am I right?
Hi,
first, I would use low power 24V coil relays for the mains switches.
Second, I would use 30 to 36Vdc feed to these relays.
Third, I would run the 24V relays on ~18Vdc when pulled in, i.e. resistor in the feed to reduce running current.
Fourth, I would use a 555 as the timer, with a tiny bit of positive feedback to ensure a fast switching action for the transistor.
The instant OFF is to ensure the timer circuit resets the relays to the OFF condition ready for the next power up when the mains comes back on. The auto isolators that supply the mains (in the UK) retry to establish power after a second or two of delay from initial power down.
When the transistor is OFF the current is near zero and the voltage is Vsupply.
When the transistor is ON the current is the relay coil current and the voltage should be about Vcesat if the base current is sufficient.
For both these conditions the dissipated power is very low.
You now realise that the in between condition is the one that must be minimised to ensure reliability.
Finally, fitting that diode across the relay coil to catch the back emf pulse extends the dropout time delay. Using a Diode + Zener or Diode plus Resistor reduces the dropout delay. ESP has posted some details on his results.
What's the resistance of a 6V mains relay coil? I can't believe it only draws 17mA (350r).
first, I would use low power 24V coil relays for the mains switches.
Second, I would use 30 to 36Vdc feed to these relays.
Third, I would run the 24V relays on ~18Vdc when pulled in, i.e. resistor in the feed to reduce running current.
Fourth, I would use a 555 as the timer, with a tiny bit of positive feedback to ensure a fast switching action for the transistor.
The instant OFF is to ensure the timer circuit resets the relays to the OFF condition ready for the next power up when the mains comes back on. The auto isolators that supply the mains (in the UK) retry to establish power after a second or two of delay from initial power down.
When the transistor is OFF the current is near zero and the voltage is Vsupply.
When the transistor is ON the current is the relay coil current and the voltage should be about Vcesat if the base current is sufficient.
For both these conditions the dissipated power is very low.
You now realise that the in between condition is the one that must be minimised to ensure reliability.
Finally, fitting that diode across the relay coil to catch the back emf pulse extends the dropout time delay. Using a Diode + Zener or Diode plus Resistor reduces the dropout delay. ESP has posted some details on his results.
What's the resistance of a 6V mains relay coil? I can't believe it only draws 17mA (350r).
AndrewT said:
I used 12 Volt relays and a 3.5VA 9V transformer with 10000% regulation, so I am expecting 13 volts at least. I believe the coils can be driver with 85% of the rated 12 Volts.
If you think that I need to "overdrive" the coils to start with, I could use the two 9V windings in series and get more than 25 VDC at my disposal, and then use a small resistor in series with the transistors?
AndrewT said:
I have seen this technique used, but as the consumption is just 30mA, why do we care?
AndrewT said:
OK so we want the transistor to go ON, not to go "slowly" to ON. I will think a bit more about this.
Of course and that is why I calculated Vcc/2 * Ic for the worst condition (6V into the load from a 12V supply).AndrewT said:
AndrewT said:
What do you mean by "dropout delay"? Sorry for being thick.
AndrewT said:
It is a 12V coil, 400R, 30mA. I mentioned 6V as the worst condition in terms of power dissipation on the driver transistor.
Off topic and insulting posts deleted.
When a relay switches off, back-EMF is generated. Without the diode that goes into the circuit and does more or less damage there. The relay falls off immediately. With diode that energy is forced to circulate through the relay coil until it is used up. the relay remains active during that time. A fly-back diode always delays the deactivation time of a relay.akis said:
You can reduce that time by adding a Zener or a resistor (R =~Rrelay) in series with the diode.
Damn I had this saved for "further reading" but no time. Thanks for pointing out the problem. I will read some more.
An externally hosted image should be here but it was not working when we last tested it.
OK just so I understand.
With a simple diode as relay protection.
When we take power off the coil, it generates an opposite voltage. This opposite voltage would potentially blow something , so we put a diode P at the ground, N at the +Vcc side, which under normal operation is open circuit, but when the coil tries to push its energy back into the circuit the diode conducts and the current flows through it and through the coil, in other words we short the coil.
However this delays the release time of the coil because ( I suppose) it keeps current flowing. But the current is flowing in the opposite direction so the relay plunger should now be pulled completely to the other side, which is exactly what we want, so what's wrong with that????
We then add a resistor/zener at approximately coil voltage, eg in this case 12 Volts. The diode isolates the zener under normal operation. When the coil "discharges" then the diode opens and the zener acts as a auto-variable resistor, so that the voltage drop across it stays below its rated, in this case 12 Volts. If the initial energy of the coil produced, say 20 Volts, the flow would be restricted by the zener which would increase its own resistance (or burn since there is nothing to protect it from unlimited current ? ) and therefore the resistor/zener would restrict current flow, but now the current will simply choose to go through whatever it would have chosen to go before (and thus cause damage), OR in any case current will still flow and the coil will still remail "active" - what am I missing please?
With a simple diode as relay protection.
When we take power off the coil, it generates an opposite voltage. This opposite voltage would potentially blow something , so we put a diode P at the ground, N at the +Vcc side, which under normal operation is open circuit, but when the coil tries to push its energy back into the circuit the diode conducts and the current flows through it and through the coil, in other words we short the coil.
However this delays the release time of the coil because ( I suppose) it keeps current flowing. But the current is flowing in the opposite direction so the relay plunger should now be pulled completely to the other side, which is exactly what we want, so what's wrong with that????
We then add a resistor/zener at approximately coil voltage, eg in this case 12 Volts. The diode isolates the zener under normal operation. When the coil "discharges" then the diode opens and the zener acts as a auto-variable resistor, so that the voltage drop across it stays below its rated, in this case 12 Volts. If the initial energy of the coil produced, say 20 Volts, the flow would be restricted by the zener which would increase its own resistance (or burn since there is nothing to protect it from unlimited current ? ) and therefore the resistor/zener would restrict current flow, but now the current will simply choose to go through whatever it would have chosen to go before (and thus cause damage), OR in any case current will still flow and the coil will still remail "active" - what am I missing please?
Not in the opposite direction. Look at the flow direction through the diode.akis said:
The Zener works the other way round. It conducts, when the voltage is higher than the Zener voltage and blocks, when the voltage is below. Look at the drawing in the above post. The Zener is connected in the opposite direction.akis said:
The discharge goes through the Zener until it falls below 12 V. Then the Zener will block it and the relay deactivates. Without that Zener the relay would remain on until the voltage falls below its release voltage, which can be as low as 10 % of its nominal voltage, e. g. 1,2 V for a 12 V relay.
The Zener is protected from unlimited current by the relay's coil impedance. And the time, during which the current is high is very short.
Just try it out. It has been done successfully before.
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