Can I connect 7812 and 7805 to this circuit ?

[milan.rajik]

In my project I am using the battery charger L200 based circuit mentioned here.

**broken link removed**

I want to use the same circuit with 7812 and 7805 for power supply for additional circuits. So, can I connect 7812 to the output of 1N4001 in the L200 based circuit and can I also connect 7805 to the output of 7812 ?

I want to know if the 7812 and 7805 heats up or not. 7805 will power a PIC12F683 and 12V is used to drive a buzzer and a relay.

 

[betwixt]

I'm not sure why you want to use the output of a battery charger which has current limiting to source voltage to fixed regulators. It would make more sense to connect all the regulator inputs to the output of the bridge rectifier.

The 7812 certainly wouldn't work as you suggested because it needs about 15V at it's input pin to work correctly.

Chaining regulators where there is sufficient voltage overhead is possible but there is no real advantage to doing it. The total power dissipated as heat will be about the same. Nobody can tell you how hot the regulators will get without knowing the current your circuit draws through them. The PIC itself probably only needs one or two mA but loading on the pins can increase that to as much as 95mA and relays can take much more depending on their type. The power dissipated in the regulators is '(input voltage - output voltage)/output current' so when you know the current it should be easy to calculate the power. The result is in Watts.

Brian.

 

[milan.rajik]

Ok. I will connect 7812's input to output of bridge which will be 20V and I will connect input of 7805 to output of 7812. I will use good heatsinks for L200, 7812 and 7805. Actually the 7805 will be driving PIC12F, a 3mm LED @ 10 mA and base of transistor @ < 5 mA. The 7812 will drive a relay which draws 150 mA.

 

[kam1787]

Ok. The 7812 will drive a relay which draws 150 mA.

do connect a reverse-biased diode across the relay

 

[milan.rajik]

I am not actually driving a relay but I am driving a solenoid valve. Earlier I used a valve whose current rating was 150 mA @ 12V. Now I am using a better quality valve at its current rating is 2A. So, I will be using LT1083MK-12 to power the solenoid valve. Should I use LT1083MK-5 to get 5V for PIC circuit ? PIC circuit will not draw more than 100 mA.

I am asking this because I have a 8051 dev board and it has a 12V 500 mA adaptor and the board has a 7812 regulator with good heatsink. If I use 12V 500 mA adaptor with the board then everything is ok but if I use 12V 2A adaptor then within 15 to 20 seconds the 7812 heats up very much. I don't know the reason. I think it is due to high current adaptor.

That's why I am asking if it is ok to use a 7805 when input is 12V 3A. I know that only if the circuits (load) draws current then the regulator will deliver that much current but why in my dev board the regulator heats up if 2A adaptor is used ?

I am also going to use a 400V 6A diode in reverse across the solenoid coil which is rated at 12V 2 A.

It is for a Gas Leakage detection system where if LPG gas leakage is detected it will close the valve and shut of the gas flow. It is a NO type valve and hence I am using a battery so that if there is a power failure and also if there is a gas leakage then power is needed to close the valve. Battery will be 12V 10Ah Gel battery.

Finally I just want to know is it Ok to use LT1083MK-12 to power the solenoid and buzzer and LT1083MK-5 to power the PIC12F. Earlier there was an LED in PIC circuit which was used to display status that is the LED would blink once every time the while(1) loop got executed but now I removed it and hence 10 mA current consumption is also saved. I have used that pin to monitor mains voltage using ADC pin. If mains voltage goes below 180 V the PIC will turn OFF a relay and hence connect battery power to circuit and valve. If mains is ok then relay will be ON and battery out is cut off.

 

[betwixt]

Using a 7.5A regulator to supply a 12F683 is overkill to say the least. I would use a 5.1V Zener diode and a resistor, it will work just as well but be cheaper and much smaller. If you need to use a regulator, consider a 78L05 or similar.

The other point to watch is the maximum input voltage of the LT1083 which is 20V. The original circuit shows a transformer with 20V RMS secondary so I would expect something like 28V after the bridge rectifier.

The diode across the solenoid only has to withstand the operating voltage of the solenoid coil and the instantaneous current when it's power is removed. If your solenoid operates from 12V then all you theoretically need is a diode with 12V PIV and the current will be a brief spike of maybe a few hundred mA. I would suggest you use a 1N4001 or equivalent to save costs and space. Remember the diode only conducts for a few milliseconds after the solenoid voltage is removed, under normal operation it is reverse biased and passes no current at all.

Can you post a diagram of your proposed connections between the boards, PSU and solenoids please. It will make it clearer to see where your power demands really are.

Brian.

 

[FvM]

It is for a Gas Leakage detection system where if LPG gas leakage is detected it will close the valve and shut of the gas flow. It is a NO type valve and hence I am using a battery so that if there is a power failure and also if there is a gas leakage then power is needed to close the valve. Battery will be 12V 10Ah Gel battery.

Sounds like the least optimal problem solution...
Any reason why you are not using a NC valve as everybody would do?

 

[milan.rajik]

s my Circuit correct ?


In the actual circuit 7812 will be replaced by LT1083CP-12 (12V 7.5A out regulator)

J3 is for battery connection and a 12V 10Ah Gel battery will be used.

Initially when there is mains power detected by AN3 pin relay RL1 will turn on. It is a 5V relay. Both RL1 and PIC gets power from 7805 which is 1.5A type.

When this relay (RL1) turns ON battery Out is disconnected and system will be running from mains voltage.

If mains voltrage falls below 180 V then AN3 pin detects low voltage and relay RL1 turns OFF and connects battery out to RL2 (12V 2A solenoid valve) and also to input of 7805.

A diode D9 is used to prevent this battery voltage flow into 7812.

So, when there is low mains voltage 7812 will be providing 12V output and 7805 will be getting input from 7812 out and also battery out but battery out will not flow into 7812.

For mains 20V 3A secondary transformer is used.

whether system is running in mains or battery power PIC12F gets power from 7805 and buzzer and RL2 (12V 2A solenoid valve) gets 12V power.

A 400V 6A diode will be connected in reverse across the solenoid valve. Godd heatsinks will be used for all three regulators.

TIP122 may need a base current od 500 mA so that 2A current can flow in its collector. PIC will not be able to provide 500 mA base current for TIP122 and hence A 2N2222A is used which provides 500 mA base current for TIP122. 2N2222A base will be driven by PIC. Is this ok?

If Ic through valve and TIP122 collector is 3A then 2V drop accross Vce of TIP122 and valve is rated at 12V. So, is it ok to use a 12V regulator and provide supply to TIP collector circuit or should I use 7812 instead and power the buzzer from it and connect 20V out from bridge to TIP122 collector circuit with a high wattage resistor ? I may need to drop 8V 2A across that resistor


Edit: Ok. I will use 78L05 which gets power from LT1083 or battery out to power the PIC12F. What collector resistors should I use for 2N2222A and TIP122 ?
I will also replace 6A free wheeling diode with 1N4007.

@FvM

I can't use NC type valve because here power cuts are common and NC type valve closes if there is a power cut and I will have to use a battery to keep open the valve during these times. As solenoid draws 2 A current power consumption will be more.

So, I use NO type valve and only if gas leaks or if there is power cut with a gas leak then valve is operated and closed. Once valve is closed it remeins in that state and buzzer will be sounding. The leakage has to be rectified manually and then system has to be reset so that it continues functioning.

 

[FvM]

Is my Circuit correct ?

Mostly. Q2/Q4 will be destroyed if you don't add a current limiting resistor.

So, I use NO type valve and only if gas leaks or if there is power cut with a gas leak then valve is operated and closed. Once valve is closed it remeins in that state and buzzer will be sounding. The leakage has to be rectified manually and then system has to be reset so that it continues functioning.

I don't understand the reasoning behind. The intended function can be achieved with NC valve. But I don't want to argue about this point, just mention state-of-the-art. It's your decision to ignore it.

 

[Easyrider83]

Anodes of D5/D7 should be connected to ground.

 

[milan.rajik]

Yes I forget wiring from anodes of D4 and D5. I will add it.

@FvM

What value collector resistors should be used for Q2 and Q4 and what wattage ? What about Vin to LT1083 ? The datasheet says its input should be 25V or more. betwixt said there will be around 28V out of the bridge. So, can I input 28V into the LT1083 and take out 12 V ?

To get 2A in TIP122 what should be the base currents for 2N2222A and TIP122 ?

 

[FvM]

TIP122 is a darlington transistor with sufficient current gain. Simply omit Q4.

 

[betwixt]

I would also add a 100nF capacitor from the output of U3 to ground.

You do not show the VSS & VDD pins of the PIC but they should have capacitors across them as well, mounted close to the IC itself. I suggest 10uF and 100nF in parallel or you might find it does strange things when the relays operate.

Brian.

 

[milan.rajik]

Yes betwixt I will add 100 nF and 10 uF across VDD and VSS pins of PIC. Proteus doesn't show PIC power supply pins and hence I have not shown the related circuit. Yes I had added 100nF to output of LT1083 but later inserted that 1N4007 diode hence it went to the input side of 7805. I will add another 100 nF to output of LT1083.

@FvM

You said to use NC type valve. Will it not consume a lot of power as the valve has to be operated to keep it open all the time whether there is mains power or battery power ?

 

[FvM]

NC valve is a problem witI wash battery operation. I was following the basic idea that a safety system should be always fail-safe, go to safe state in case of power loss. There's no easy workaround, you can't have both long term battery operation and fail-safe behaviour.