# Current Sensing Switch

I have built the circuit depicted below in order to start a 10A 220V dust collector(Load 2) when I run my 20A 220V table saw (Load 1). So far, I have not been able to activate the relay to power load 2. The voltage output from the rectifier (as measured at the relay) is only 0.5VDC and the solid state relay does not trigger with less than 2.5VDC. I have measured the voltage potential between the secondary of the 1:100 CST and a common line to be 13VAC. I have checked the bridge rectifier with my multimeter and both legs show 0.56mV between AC legs and DC. I also built this circuit with a 1:500 CST but only produced 1.4VDC from the rectifier. Thoughts? simulate this circuit – Schematic created using CircuitLab relay data sheet https://cdn.sparkfun.com/datasheets/Components/General/SSR40DA.pdf CST data sheet https://www.coilcraft.com/senhitr.cfm

• (1) Please edit to include a link to the datasheets for the CT and SSR. (2) Can you confirm that the main load is run through the primary of the CT? (3) You have Load 2's neutral connected to the live of Load 1. It should be connected directly to neutral. Hit the edit link below your question ... Apr 3, 2020 at 11:59

The CS4100V-01L is not good enough for your needs in my opinion: - Below 2 kHz the response of the CT will fall-off and, by 50 Hz you'll barely get more than a few hundred milli-volts from it with "several" amps flowing through the main conductor. See also note 3. It's telling you what a typical burden will be like and you are a long way from that scenario.

But, if you want the math here goes... The secondary inductance is quoted as being 20 mH. This allows us to estimate the primary inductance by dividing secondary inductance by the square of the turns ratio hence, primary inductance is 20 mH / 10,000 = 2 uH. Given that you have no burden resistor and, assuming your load is extremely light, means we can use the impedance of the primary at 50 Hz to convert current into primary voltage. In other words, there is no reflected impedance back to the primary.

It's operating as a voltage transformer now. So 2 uH at 50 Hz is an impedance of 0.628 milli ohms reactive. If the primary inductance had 10 amps flowing through it, the primary voltage would be 6.283 mV RMS. Given that it's a step up device, the output voltage would be 100 times higher at 0.628 volts RMS and not enough to overcome the forward volt drop of the bridge.

I also built this circuit with a 1:500 CST but only produced 1.4VDC from the rectifier. Thoughts?

For a similar sort of CT tech, with 500:1 ratio, the secondary inductance might rise 25 times higher. This is because inductance is proportional to the square of turns. So now, the primary projected impedance will still be 2 uH but, you'll get more secondary voltage (5 times more). So, with 10 amps in the primary you'll get a secondary of 3.14 volts RMS. Less 1.4 volts for the bridge rectifier and you might see 3 volts DC. Figure 1. From Coilcraft datasheet.

Assuming that you are using a Fotek SSR-25DA then the trigger voltage is > 4 V and is listed as 7.5 mA at 12 V. This is equivalent to a 12/7.5m = 1.6 kΩ. The datasheet is recommending a terminating resistance, RT, of 2.9 Ω - and that's at a frequency range of 2 - 1000 kHz when you're operating at 50 Hz.

The second problem is that you have selected a zero-cross switching SSR. That means that it won't turn on unless the control voltage is on during or very close to the zero-cross voltage on the mains. Your circuit doesn't do that as the DC is pulsed with the mains so that the control voltage is zero just when you need it to be high. For more on this see my article Opto-triacs, solid-state relays (SSR), zero-cross and how they work.

I think you're going to have to increase the complexity of your control circuit somewhat. Maybe a fat capacitor on the output of the bridge might work.

Just use one of these to then drive an external contactor / relay...