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The maximum reverse voltage two diode series (PC817 , 1n4007)

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azadfalah

azadfalah

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Hello friends
I hope you are fine


i want sense 110V with 1n4007 resistor=45KΩ PC817 These are series

In the half-wave reverse PC817 burns?

Thanks
 

There is a risk of the PIV being exceeded because of leakage through the 1N4007.

A better method is to connect the 1N4007 acrooss the diode in the PC817 but reversed - cathode of 1N4007 to pin 1, anode of 1N4007 to pin 2. Keep the 45K resistor in series with them to limit the current.

That lets the PC817 conduct on one half cycle and the 1N4007 protects it on the other half cycle. The voltage across the PC817 will be at most 1.2V in one direction and 0.7V in the other.

Brian.
 
Thanks for the answer
I realized But i have Parallel Capacitor for Reducing ripple in your method my Capacitor discharge
for keep Capacitor i should use tow diode true ?
What do you think about my new method ?

aa.jpg
 

Hi,

Now a completely different situation. You could have posted the schematic with your first post...

110V AC is from about -150V to about +150V (peak).

During pusituve cycle the capacitor charges to about +150V.
At negative cycle the AC voltage goes down to -150V....while the capacitor may stay near +150V,
Therefore you have to expect -300V across the diode.

The capacitor voltage will always be positive. --> no negative voltage at the optocoupler.
The resistor needs to be calculated for peak voltage = 150V.

Klaus
 
I think capacitor Does not allow the reverse voltage And somehow neutralize negative voltage Was approved?
 

Hi,

in your circuit the diode does not allow reverse current flow...

Instead .. a polarized electrolytic capacitor (like that used in your schematic) will be killed by reverse voltage.

*******
The circuit D1, RN1 is useless... it does not influence capacitor current, capacitor voltage, optpcoupöer current.

Do you know why it is shwon here? Or do you know it´s purpose?

****

You don´t say, but I assume it is a 60Hz application.
Then the capacitor is charged to about 155V.
It is discharged by the optocoupler current.
The dominant load is the resistor R5.
It builds a time constant of about 100ms.
The time from "end of charge" to next "start of charge" is about 15ms. (estimated)
Within this 15ms the voltage at the capacitor drops (about constant) to about 135V.
Then within the next about 1.5ms it is charged again to about 155V

Klaus
 

What do you think about this?

ss.jpg


It works?
 

Hi,

Why?...What is it good for?

In post#4 I clearely said:
"The capacitor voltage will always be positive. --> no negative voltage at the optocoupler."

Klaus
 

If the capacitor is fully discharged At the moment the signal Contact
If the wave is negative
Opto burn.
 

I'm confused too.

The schematic doesn't show where the 110V is applied across so I assume it's the points marked v110 and both the points (linked together) marked C110.

That makes it a half wave rectifier, with C1 wired backwards and D95 doing absolutely nothing. The voltage across the PC817 diode will never be more than about 1.2V and it will never pass more than (110 * sqrt(2)) / 47K = 3.3mA so it will be quite safe. The capacitor C1 will probably explode though !

Brian.
 

better to have line input to full diode bridge to 47K then to 0.1 -1uF cap in parallel with LED and cathode back to neutral.
Collector output should go into Schmitt trigger with small RC delay to avoid glitches during " brown out"

capiche?
 

Thanks friends
But there is a problem I can not use of capacitor in Input / Capacitors (470nf 180v) are very big and ugly :cry: (we have 7 terminals)
So I am using reverse diode (method mr betwixt in post two ) And I want reduced ripple After Opto

I should use capacitors after OPTO After opto have a Pullup Resistor (10K or 6.8K) and a 74hc573 I want reduced ripple
But I do not want to change the cut-off time
Please help in choosing capacitors after OPTO (Voltage = 5.1V)
 

It seems the easiest way to do this (subject to the delays already mentioned) is:

AC ---> resistor ----> bridge rectifier ----> opto input

opto output (with pull-up and SMALL capacitor to ground) ---> 74HC573

The 74HC573 needs negligible input current so the pull-up can be quite high in value and the capacitor quite small with only a >5V rating.
The bridge rectifier can be small signal diodes (1N4148 for example). The LED in the optocoupler will flash at twice AC frequency but the RC network at it's output will filter that back to near DC levels. I think for safety, I would use a resistor in each leg of the AC rather than relying on only one.

Brian.
 

Brian your solution might be better with a brown-out threshold adjustment and hysteresis, depending on what the user wants. blackout or brownout detection?
 

betwixt said:
It seems the easiest way to do this (subject to the delays already mentioned) is:

AC ---> resistor ----> bridge rectifier ----> opto input

opto output (with pull-up and SMALL capacitor to ground) ---> 74HC573

The 74HC573 needs negligible input current so the pull-up can be quite high in value and the capacitor quite small with only a >5V rating.
The bridge rectifier can be small signal diodes (1N4148 for example). The LED in the optocoupler will flash at twice AC frequency but the RC network at it's output will filter that back to near DC levels. I think for safety, I would use a resistor in each leg of the AC rather than relying on only one.

Brian.
Click to expand...


we should Read 7 Contact 110V ac and can't use bridge (The circuit was crowded)

AC ---> resistor ----> reverse Counteractive Diode ----> opto input

opto output with pull-up (10k) and SMALL capacitor(????) to +5 ---> 74HC573

I want capacitance if Current required 74HC573 is 1ua and pull-up current is 510ua in (5.1v)
.011mSeconde (AC=50hz)/9,980K = i want 1.1uf capacitor

But I'm not sure Current required 74HC573 is 1ua For low level input ?
Here Say Yes : https://www.edaboard.com/threads/358098/#post1533290

Tomorrow I'll try the circuit in action

ss.jpg


Do you think it works?
 
Last edited:

azadfalah said:
we should Read 7 Contact 110V ac and can't use bridge (The circuit was crowded)

AC ---> resistor ----> reverse Counteractive Diode ----> opto input

opto output with pull-up (10k) and SMALL capacitor(????) to +5 ---> 74HC573

I want capacitance if Current required 74HC573 is 1ua and pull-up current is 510ua in (5.1v)
.011mSeconde (AC=50hz)/9,980K = i want 1.1uf capacitor

But I'm not sure Current required 74HC573 is 1ua For low level input ?
Here Say Yes : https://www.edaboard.com/threads/358098/#post1533290

Tomorrow I'll try the circuit in action

View attachment 131544
Click to expand...


Hi,
Opto have current tolerance? I think not please help
i want to have an integrated signal
 

The schematic is correct except C1 should go to ground, as though it was connected between pins 4 and 3 of the PC817. The idea is it charges up slowly through the 10K resistor but discharges quickly through the optocoupler. The discharge happens once per AC cycle when the opto LED lights up so when AC is present it keeps the capacitor mostly discharged. If the AC stops, so does the discharge so the voltage across C1 rises high enough for the HC573 to read it as logic high.

The opto transistor current will be zero when the LED is not lit (except for a tiny leakage current) and 5/10K = 0.5mA plus capacitor current when the LED is on. Those are good values. The input to the HC573 will be almost zero, it is a CMOS device and it needs far less than 1uA for high and low level input currents.

Brian.
 
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