I am making a circuit using PIC10f200 and i need to power the circuit sometimes with a capacitor. i can supply dc voltage upto 12v in the beginning but after 10sec the dc power will turn off for approximately 1.1sec but i still want my PIC10f200 to work during that time. Can i do that using a capacitor??
i will only be using one pin for output rest 5 pin's are left as is.

 

I am making a circuit using PIC10f200 and i need to power the circuit sometimes with a capacitor. i can supply dc voltage upto 12v in the beginning but after 10sec the dc power will turn off for approximately 1.1sec but i still want my PIC10f200 to work during that time. Can i do that using a capacitor??
i will only be using one pin for output rest 5 pin's are left as is.

You can do this:

C = I Δt/ΔV
Where C is the capacitance in farads, I is the current into the PIC, in amperes, Δt is the length of time the 12V supply will be low, and ΔV is the difference between the voltage C1 charges to when the 12V supply is high and the sum of the voltage drop across the regulator and the 5 volts into the PIC.

 

Hi Tapan,
the answer to your question is yes you can! But....
Consider the following befor you choose the capacitor.

You have to find-out how much power the PIC does use during the 1.1sec.
This will depend very much from what operations the pic is doing at the time.
For example, if the PIC is sourcing some outputs, say two LEDs, then the current consumption
could be as much as 20mA.

The speed of the the clock is also another factor to consider.
And finally you have to check the discharge curve of a capacitor.
As the capacitor discharges the voltage will drop, so the operating voltage of the PIC make a difference.

Regards

Francesco C

 

Hi Tapan,
the answer to your question is yes you can! But....
Consider the following befor you choose the capacitor.

You have to find-out how much power the PIC does use during the 1.1sec.
This will depend very much from what operations the pic is doing at the time.
For example, if the PIC is sourcing some outputs, say two LEDs, then the current consumption
could be as much as 20mA.

The speed of the the clock is also another factor to consider.
And finally you have to check the discharge curve of a capacitor.
As the capacitor discharges the voltage will drop, so the operating voltage of the PIC make a difference.

Regards

Francesco C

Thank you for answering my query.
The pic is required to give out a current of 4-10mA when it is high. The internal clock frequency is 4Mhz.
The function of the circuit is to restart the main power source after 1 sec. so basically i will just be using the timer of PIC. The delays are to be very precise hence i chose PIC because of it's low cost, size and simplicity.

 

You can do this:
View attachment 108148
C = I Δt/ΔV
Where C is the capacitance in farads, I is the current into the PIC, in amperes, Δt is the length of time the 12V supply will be low, and ΔV is the difference between the voltage C1 charges to when the 12V supply is high and the sum of the voltage drop across the regulator and the 5 volts into the PIC.

Thank you Sir,
The PIC works between 3v-5v properly. So what should i use as VCC for PIC. I believe 3v will provide a better solution as the current output of PIC at 3v is less than 5v. can u give me a rough estimate of the capacitor that i should use for this circuit. and in your opinion what will be the best main's power rating(12v or less).

 

Thank you for answering my query.
The pic is required to give out a current of 4-10mA when it is high. The internal clock frequency is 4Mhz.
The function of the circuit is to restart the main power source after 1 sec. so basically i will just be using the timer of PIC. The delays are to be very precise hence i chose PIC because of it's low cost, size and simplicity.

Hi tapan,
I had a loook at the datasheet of the PIC10f200 and you can power it from 2Vdc to 5.5Vdc.
If you could power it from the lower voltage of 3.3Vdc and you know that uses a max current of 10mA then you could do a sort of simulation using TLSPICE (free circuit simulator).

You just charge a capacitor at 12Vdc and starting at 2000uF, add a voltage regulator (3.3V) and a load of 10mA.
You can then see visually what is happening to the output and how long the voltage can be maintained.
If the voltage output lasts less than 1.1sec then increase the capacitance a bit, until you are satisfied with the results.

If you have never used LTSPICE please let us know, and one of us will do it for you.
I stongly recommend you use LTSPICE . It will give you a good approximation of your final design.

Regards

Francesco C

 

Hi tapan,
I had a loook at the datasheet of the PIC10f200 and you can power it from 2Vdc to 5.5Vdc.
If you could power it from the lower voltage of 3.3Vdc and you know that uses a max current of 10mA then you could do a sort of simulation using TLSPICE (free circuit simulator).

You just charge a capacitor at 12Vdc and starting at 2000uF, add a voltage regulator (3.3V) and a load of 10mA.
You can then see visually what is happening to the output and how long the voltage can be maintained.
If the voltage output lasts less than 1.1sec then increase the capacitance a bit, until you are satisfied with the results.

If you have never used LTSPICE please let us know, and one of us will do it for you.
I stongly recommend you use LTSPICE . It will give you a good approximation of your final design.

Regards

Francesco C

Thank you sir,
i have never used it but i will try and get back to you.

thanks

 

Thank you Sir,
The PIC works between 3v-5v properly. So what should i use as VCC for PIC. I believe 3v will provide a better solution as the current output of PIC at 3v is less than 5v. can u give me a rough estimate of the capacitor that i should use for this circuit. and in your opinion what will be the best main's power rating(12v or less).

I already posted a formula you can use to determine the size of the capacitor needed to hold up the voltage for any given time, but here's a design completely worked out for a 1.8 second delay.

You can change the output voltage by switching regulators or using an adjustable one, and you can change the drop-out time by changing the value of C3.

This work was done using LTspice, which is available, free, at http://www.linear.com/designtools/software/

 

I already posted a formula you can use to determine the size of the capacitor needed to hold up the voltage for any given time, but here's a design completely worked out for a 1.8 second delay.

You can change the output voltage by switching regulators or using an adjustable one, and you can change the drop-out time by changing the value of C3.

This work was done using LTspice, which is available, free, at http://www.linear.com/designtools/software/
View attachment 108154

Thank You SIr

 

Thank you sir,
i have never used it but i will try and get back to you.

thanks

Hi,
I had a go myself to solve your problem.
I have attached a LTSPICE file for you.
Just install LTSPICE and load the file 'Delay.asc' as attached.
You can change the capacitance and the voltage regulator to 5V to get different results.
Please remember that this is a simulation! A real circuit will behave a little different.
Also capacitors can be +-20% of the stated capacitance, and that will affect the discharge time.

Regards.

 

Hi,
I had a go myself to solve your problem.
I have attached a LTSPICE file for you.
Just install LTSPICE and load the file 'Delay.asc' as attached.
You can change the capacitance and the voltage regulator to 5V to get different results.
Please remember that this is a simulation! A real circuit will behave a little different.
Also capacitors can be +-20% of the stated capacitance, and that will affect the discharge time.

Regards.

Thanks a lot sir
i downloaded LTspice and will test your circuit.
Thanks alot for your help
People here are really helpful.