Voltage Signal Circuit Zener

Look at the LM3914. This is a LED bar or dot driver, it contains a voltage reference and ten comparators. An indication of battery voltage could be given and, when the last LED is lit, a small amount of current could be taken and amplified to switch your relay.

 

You will need a resistor and zener diode (12V) to provide a power supply for the chip and a voltage divider to provide a sample of the measured output voltage.

When they say that the LED current is programmable, they mean it can be set. This is done with a single resistor. If you go for the dot mode, the power supply needs would be quite low.

You will need a transistor at the LED output and a high voltage MOSFET to drive a high voltage relay. I have a IRFR320 (400V, 1.7A) taken out of a dead compact fluorescent lamp, doubtless they can be bought. Another source may be computer switch mode power supplies.

Relays with high voltage DC coils seem to be rare but you could use a 230V AC relay with a series resistor.

Duke

 

The weather is nice here at the moment so I should be out in the garden.
The LM3914 should be supplied with about 12V with a resistor and zener.
The internal reference voltage is low but they provide a means of raising it with an internal op-amp and resistors. I would set it for something less than 12V, say 8V.
You then need a voltage divider to give say 7.5V at your required switching level.

I did not read your post properly and did not realise that you had a solid state relay. It is likely that this could be driven directly by the 3918 in place of a led or in series with a led.

If you ae serious, I could refresh my memory on the 3819 and do a few resistor calculations but I must go and mow the lawns now. Last time before winter?

Duke

 

I attach a diagram of what I thought could work. The voltage is divided by 50 to make 250V=5V at the chip. Making the reference 5V supplying the resistor chain and placing a 40k resistor to ground, gives a range of 4V to 5V which would indicate from 200V to 250V. Since there are 10 LEDs these would change every 5V. Select which LEDs you need to drive your relay.

A power supply is needed and I chose 12V. The feed resistor will need to be about 10W, quite a large power loss.

You could try the simple resistor- zener- relay at little cost but the zeners I have seen have a tolerance of about 5% so some try and mod would be necessary.

Duke

 

The current drawn is
10 mA LED (perhaps change scaling resistors)
10 mA relay
10 or 20mA zener depending on whether relay is energised
Total 30mA

6.8k at 223V = 33mA, 7.3W
zener 20mA at 12V = 240mW, use 0.5W
The 1M sense resistor dissipates 50mW only but I would go for 0.5W resistor to get better voltage capability.
You may want to adjust the resistors around the 3914 to get standard values.

The 7.3W loss in the feed resistor seems a lot. If you try the simple zener relay solution the loss will only be there when the voltage is high and power is available.

You may be able to reduce the power consumption by using a 12V switch mode power supply. Some modern wall warts are switch mode and presumably rectify the input AC to get DC to switch. The wart should then be able to run on DC. A norrnal transformer wall wart would go pop on DC.

 

This is unlikely to work since you are dividing down voltage difference with the voltage.
You could try seven 33V zeners (BZX79 0.5W) in series to drop most of the voltage without reducing the voltage difference greatly. The details I have seen is that they have a tolerance of 5% so it may be necessary to consider what the effect of this will be.
I would test the zeners and relay with an adjustable power supply.

I am not too clear on the spec of the relay or whether the voltage will be protected from going above 235V which could cause excess current. The current drain will be very low on low voltage and rise rapidly as voltage rises above the zener voltage.