Zerocross test issue

Thread Starter

Benengineer

Joined Feb 6, 2016
133
Here's my take on a zero crossing circuit that can meet the timing requirements while minimizing power without using a transformer.
It uses a capacitive voltage divider (C1 and C2) to losslessly reduce the voltage and rectifies that to get ≈6v to 14V DC to operate a LM393 zero crossing detector that drives the opto coupler.
C2 is a film or ceramic capacitor rated for at least 500V.
C1 and C3 should be rated for at least 20V.
Capacitor C1 becomes slight reversed biased on the negative half-cycle as determined by the clamp diode D2 but that voltage isn't high enough to damage the capacitor if it's electrolytic (which can generally tolerate up to 1.5V reverse bias).
The LTspice simulation shows the opto output zero crossing rise-time occurs within <30μs of the input zero crossing (top plot).
The main power consuming component is R3 which should have a 1W or greater rating.
The 1GΩ to ground (R5) at the output is just for simulation purposes. In the real circuit there would be no deliberate connection between the main's ground and the output ground.

View attachment 105700
Here is the capture I need. Please see the below:
upload_2016-5-9_15-38-4.png
This is the zoom out graph when input line voltage (sine wave,yellow) crosses zero, the relay( blue ) crosses 1 voltage.
upload_2016-5-9_15-40-24.png
This is the zoom in graph, you can see delay time

Is this circuit to realize this function, isn't it?

Thanks
 

Thread Starter

Benengineer

Joined Feb 6, 2016
133
Is your circuit identical to mine?
Your output signal looks inverted.
I don't change it. But I found that your simulation output result is little different from me. My output is when line voltage crosses the zero, the relay voltage is falling down crossing 1 volts.
 

crutschow

Joined Mar 14, 2008
25,108
Since your signal is inverted from mine, then your circuit is not the same.
My circuit has the output going high when the input goes high.
If you want similar results to mine then you need to use the same circuit.
Post your circuit.
 

Thread Starter

Benengineer

Joined Feb 6, 2016
133
Since your signal is inverted from mine, then your circuit is not the same.
My circuit has the output going high when the input goes high.
If you want similar results to mine then you need to use the same circuit.
Post your circuit.
My circuit is at the beginning in this thread. Please read the first post.
Thanks
 

crutschow

Joined Mar 14, 2008
25,108
I did read your post.
My design was to improve the response time of the circuit, which I understood was not satisfactory.
It switches within 30μs of 0V input.
Isn't that what you want?
 

Thread Starter

Benengineer

Joined Feb 6, 2016
133
Here's my take on a zero crossing circuit that can meet the timing requirements while minimizing power without using a transformer.
It uses a capacitive voltage divider (C1 and C2) to losslessly reduce the voltage and rectifies that to get ≈6v to 14V DC to operate a LM393 zero crossing detector that drives the opto coupler.
C2 is a film or ceramic capacitor rated for at least 500V.
C1 and C3 should be rated for at least 20V.
Capacitor C1 becomes slight reversed biased on the negative half-cycle as determined by the clamp diode D2 but that voltage isn't high enough to damage the capacitor if it's electrolytic (which can generally tolerate up to 1.5V reverse bias).
The LTspice simulation shows the opto output zero crossing rise-time occurs within <30μs of the input zero crossing (top plot).
The main power consuming component is R3 which should have a 1W or greater rating.
The 1GΩ to ground (R5) at the output is just for simulation purposes. In the real circuit there would be no deliberate connection between the main's ground and the output ground.

View attachment 105700
Crutschow,

Are you sure C1 and C2 are voltage divider? If they are a divider, C1 and C2 should series, not parallel. Please correct me if I am wrong.
Thanks,
 

crutschow

Joined Mar 14, 2008
25,108
Crutschow,

Are you sure C1 and C2 are voltage divider? If they are a divider, C1 and C2 should series, not parallel. Please correct me if I am wrong.
Yes, I'm sure C1 and C2 are in series, grasshopper. :)
You are confusing being drawn next to each other in a parallel fashion with being electrically in parallel.
There is no relation between the two.
They just aren't drawn in an axial straight line as you are typically used to seeing for a voltage divider.
You have to look at the connections.
To be electrically in parallel the capacitors would need to be tied together at both ends and they aren't.
One goes to the voltage input and one goes to ground with the other ends tied together, as for any proper series voltage divider.
Okay?
 

Thread Starter

Benengineer

Joined Feb 6, 2016
133
Here's my take on a zero crossing circuit that can meet the timing requirements while minimizing power without using a transformer.
It uses a capacitive voltage divider (C1 and C2) to losslessly reduce the voltage and rectifies that to get ≈6v to 14V DC to operate a LM393 zero crossing detector that drives the opto coupler.
C2 is a film or ceramic capacitor rated for at least 500V.
C1 and C3 should be rated for at least 20V.
Capacitor C1 becomes slight reversed biased on the negative half-cycle as determined by the clamp diode D2 but that voltage isn't high enough to damage the capacitor if it's electrolytic (which can generally tolerate up to 1.5V reverse bias).
The LTspice simulation shows the opto output zero crossing rise-time occurs within <30μs of the input zero crossing (top plot).
The main power consuming component is R3 which should have a 1W or greater rating.
The 1GΩ to ground (R5) at the output is just for simulation purposes. In the real circuit there would be no deliberate connection between the main's ground and the output ground.

View attachment 105700
Ok, thank your guys. Your guys are maters.

Here is what I got:

Xc2 = 1/2πfc2 = 1/2π*60*0.33x10^-6 = 8038.128Ω

Xc1 = /2πfc1 = 1/2π*60*10x10^-6 = 265.258Ω

Input voltage between 120VAC to 277VAC

Voltage divider V = 265.258/(8038.128+265.258)x120x √2 = 5.43V

Or V = 265.258/(8038.128+265.258)x277x√2 = 12.54V

For U2 output voltage is (-3k/100k) * 120x √2 = -5.091V or (-3k/100k) * 277x √2 = -11.752V

Am I right?

Since 4N25 forward current is 60mA and forward voltage is 1.3V (in the datasheet), we choose one input as our example; we can figure out 5.43V-1.2V(for 1N4148)=4.23V; then 4.23V – (1.3V-5.091V)= 8.021V;

Current flow through diode is 8.021/3k = 0.003A, which much less than 60mA. Is this Ok? Please let me know if I am right. Since I am a beginner, please forgive my bothering your guys. Basically, I want to know more design idea from your example.

Thanks
 

crutschow

Joined Mar 14, 2008
25,108
Your calculations are okay as far as they go.
You didn't include the effect of the clamp diode D2, which clamps the negative voltage at the divider to about -0.7V.
This means the unloaded peak output of the divider is nearly equally to the peak-to-peak voltage of the AC or double what you calculated.
The output load current reduces that voltage due to the impedance of the capacitive divider so the V+ voltage is about 6.5V for 120Vac input as shown in my first simulation.

The 60mA forward current for the 4N25 is the maximum current, not the operating current.
It will operate at any current down to zero.
So a few mA is just fine as an operating current.

Your calculation for the LED input current also didn't include the 1.3V forward drop of the input diode which subtracts from the voltage drop across the resistor.
 
Last edited:

Thread Starter

Benengineer

Joined Feb 6, 2016
133
Your calculations are okay as far as they go.
You didn't include the effect of the clamp diode D2, which clamps the negative voltage at the divider to about -0.7V.
This means the unloaded peak output of the divider is nearly equally to the peak-to-peak voltage of the AC or double what you calculated.
The output load current reduces that voltage due to the impedance of the capacitive divider so the V+ voltage is about 6.5V for 120Vac input as shown in my first simulation.

The 60mA forward current for the 4N25 is the maximum current, not the operating current.
It will operate at any current down to zero.
So a few mA is just fine as an operating current.

Your calculation for the LED input current also didn't include the 1.3V forward drop of the input diode which subtracts from the voltage drop across the resistor.
If I count D2, the current is about 50nA. It is very small. How do I apply for it?
What do you mean I didn't include the 1.3V forward drop of the input diode which subtracts from the voltage drop across the resistor 3k?
Is it 5.23V - 1.3V = 3.93V? I already know the current is about 60mA. Please let me know.
The other issue is that 0.33uF is so big size and not cost efficiency. Can we use high value resistors to made voltage divider?
Thanks,
Ben
 
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