# Electronic Circuits and Projects Forum

1. ## circuit to replace mechanical switch, using solid state relay design verifying needed

see attachment. i have designed the circuit in attachment 1 to interface with circuit in attachment 2.

it convert the two mechanical switches into electrical switches using solid state relay.

for one of the them the input voltage is 12vdc and the other is 18vac. when the input voltage is applied to any of the circuits the output SSR is activated and drives the signal to ground so the EZ-TAG controller know some one has pushed a button and takes action needed.

the led in the circuit is use just to verify that the circuit has been activate.

can some one verify the design and to make sure it is safe.

thanks for the help in advance.

2. http://www.vishay.com/docs/83821/lh1522ab.pdf
First!...you need a current limiting resistor in series with the SSR's LED.
Then simplify.
Second, I would just put the indicator LED in series with the SSR's LED, and scale the current limiting resistor accordingly.
Third, I would eliminate the regulator and capacitor in the 12VDC input version. The limiting resistor would be (12V-3.3V-1.2V)/10mA=750Ω...for 10% tolerance use 680Ω
Fourth, I would eliminate the regulator in the 18-24VAC input. Increase the voltage rating on the capacitor to ~50VDC, and change the limiting resistor to ((24V*1.414)-3.3V-1.2V))/10mA=2.9KΩ...for 10% use 2.7KΩ

Ken

3. thanks for the advice KMoffett,
i did put a current limiting resistor of 0Ω, as i thourght the currect and voltage was lower then the max the SSR LED could take.
also i included the regulators to give some protect if the voltage did vary.

i do like the way you have simplefied the design and will take a look at it tommorow and redraw another circuit diagram to see how it looks.

i was not sure and do you thing on the output of the full wave bridge rectifier is there need for a smoothing capacitor or not?

4. The LED in the SSR's input has no internal current limiting so you need a much higher resistance than "0Ω". It needs the same kind of current limiting as your indicator LEDs only for a 1.2VDC Vf. My resistor calculations allow for quite large variations in the input voltages, so I don't think the regulators would add any benefit. And yes, you do need a capacitor after the bridge rectifier to turn the 120HZ pulsating DC into a smoother DC. I think I would also add a 10K resistor across the filter capacitor to assure that it is fully discharged after the AC input is off.

Ken

5. Hi

What is the smallest size capacitor do you thing i can use on the output of the full wave bridge rectifier?

Thanks

6. How often does the 18-24VAC turn on and off...mSec?...seconds?...minutes?...?

Ken

7. Hi

It will turn on about 10 seconds max at a time.
It will only be for no more then 15 times a day max.

Thanks

8. Hi

There was a error in the orignal circuit and just found it out now the voltage should have been 12-18VAC
Sorry about that and will rework out the value from the posts before and update.

Thanks

9. Hi

The working outs are below and new circuit attacted. let me know what you think and also if the working outs are correct.

Indicator LED have a power consumpation of 20mA.

Limiting resistor on the 12-18VAC i work it as below:
((18V-3.3V-1.2V))/(10mA+20mA)=450Ω...for 10% use 390Ω

390Ω*(10mA+20mA)=11.7v
((11.7v)-3.3V-1.2V))/(10mA+20mA)=240Ω

Limiting resistor on the 12VDC i work it as below:
(12V-3.3V-1.2V)/(10mA+20mA)=250Ω...for 10% tolerance use 225Ω

225*(10mA+20mA)=6.75v
((6.75v)-3.3V-1.2V))/(10mA+20mA)=75Ω...

10. I don't know why you are adding currents...(10mA+20mA). The current in a series circuit...the limiting resistor, the LED and the SSR...is the same through all components.

For the 12-18VAC circuit: Use the maximum peak voltage: 18*1.414, subtract the forward voltage drops of the two LEDs: 3.3V and 1.2V, and use the current through the series circuit:10mA.
R=((18V*1.414)-3.3V-1.2V)/10mA=2095Ω... so use a 2.2KΩ resistor.

For the 12VDC circuit: Use the DC voltage: 12V, subtract the forward voltage drops of the two LEDs: 3.3V and 1.2V, and use the current through the series circuit: 10mA.
R=(12-3.3V-1.2V)/10mA=750Ω... so use a 680Ω or an 820Ω resistor. It's not critical.

For the capacitor in the AC circuit, I would try 10uF/50VDC electrolytic capacitor. (note the polarity)

And, I would change the 1.1K resistor across the capacitor to 10K.

Ken

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