Is this circuit correct?

Is these circuit correct ? I made this for a 12V < 1A Solenoid switch. I want to know especially if the Bi-Color LED circuit is correct or not.

Is it better if I put a 1k resistor between pins 2 and 5 of Inverters after D2.

for 20mA LED's you are exceeding the rated current by 50%. Never use absolute. max. by design. Also you are exceeding reverse voltage on LEDs rated at -5V absolute max.

Where is the bi-colour? I see two complementary LEDs

I have a 3 pin common cathode LED. In proteus there is only 2 pin Bi-Color LED. So, I have used ordinary LED in the circuit to make Bi-Color LED but as only the yellow LED lights up brightly in Proteus model I have used them.

5V / 10 mA = 280E but for R2 I have used 330E. So, ILED is < 10 mA. Why you say it exceeds 20 mA ?

I am getting -2.74 V reverse voltage across LED.

I assume that the 74hc040 gets 5volt power.

The input of U1:A is connected to 12volts via the relay coil.
The input has an ESD protection diode to the 5volt supply, it conducts when the input is higher than 5volts.
When the transistor is off current will flow from +12v, through the relay coil and pull the input up to about 5.7volt.
Current into the input pin will exceed IC rating.
The voltage across the relay when it should be off will be about 6.3volts, instead of about zero as intended, at least until the inverter input fails. Maybe not low enough for the relay to release if it is already on.

One way of avoiding that would be to connect switch pullup to 5volts, connect switch to an inverter input and use an inverter output to drive the transistor base through a resistor.

LED connection should be O.K., but U1A will be damaged by exceeding the permitted input voltage range respectively input protection diode current limits.

Is this new circuit correct ?

Are you sure the relay voltage falls short of minimum dropout voltage?

I am not using relay. I am using Solenoid (12 V). There is no Solenoid Valve in Proteus so I used relay.

I will use one of these Solenoid Valves.

**broken link removed**

**broken link removed**


Here is the new circuit.

Are you sure the relay voltage falls short of minimum dropout voltage?

Click to expand...

You mean to say that even 3 or 4 volts will operate the relay ?

When relay is off I am getting 1.2V across relay coil which is 10 times less than its rating (12V).

Why nobody has answered ? Please somebody confirm that the last circuit is ok or not.

Try any of these options.

In the 2nd circuit, LED connected to collector of transistor doesn't light up. I can't use the first circuit you posted because it doesn't show the status of solenoid or transistor. If one of them fails still led will glow.

To design it properly, we need to know the resistance of the coil. Try using two different resistors for both the led's and make the resistance value smaller for the led connected to relay.

milan.rajik said:

I am not using relay. I am using Solenoid (12 V). There is no Solenoid Valve in Proteus so I used relay.


Here is the new circuit.



You mean to say that even 3 or 4 volts will operate the relay ?

When relay is off I am getting 1.2V across relay coil which is 10 times less than its rating (12V).

Click to expand...

There is no need to run 5-6 mA through your voltage divider to get the signal to the 74HC04. If you use 12k/ 8k2 for the divider it will be ok.

One of your valves is a 2W 12V type, and the nominal current in activated mode will be 166mA, through 72 ohms. The other valves current/ resistance/ wattage is not defined.

Just for the record, if the LED you use is a tripple type, you can't have more than one resistor in the in the common cathode side. If you need to reduce current more for some of the LEDs, you'll have to add a resistor in series with that LED's anode. I suppose you are aware of the colour mix when blending several colours in the same LED.

Since you always will have 2 LEDs lit at the same time, you should recalculate your resistors, or use anode resistors only to get a high enough voltage over the LED with the highest Vf.

Thank you Gorgon.

Thats what I mentioned that current rating is < 1 A. So, I think even BC337 is enough but as I don't know the current rating of another solenoid I have used TIP122 which is available here.

For led 10 mA is enough for me and it might result is less brightness for green color but it's not a problem as I am using water clear bi-color led. It is a bi-color and not tri color. The led is Red+Greeen. Reg indicated that solenoid is OFF and green, solenoid is ON. The other led in power supply is a 3 mm green 2 pin led which is for indicating 5V power for 74HC04. I have removed the NOT gate at the base of the TIP120/122 because the valve will be closed when no power and I need it to be open (solenoid ON) when float switch is open. So, without NOT gate at base of TIP12x the solenoid will be ON when float switch is open and solenoid turns OFF when float switch closes.

I am doing this system because in my country we don't get good quality ball valves and the water supplied by water department to houses will get wasted due to overflow as ball valves doesn't operate properly. Even when tank is full the water leaks through ball valves. For my system a battery will be used so that when there is no mains power the system will still work.


And yes, I will change the value of resistors in hardware.

I have modified Techgreeds 2nd circuit but the Circuit Wizard Simulator shows that reverse voltage across diodes are too high. Is this correct. How to make the circuit work without led getting killed. I am attaching simulation video which shows voltages and currents and also circuit wizard 1.5 file.

Edit: Problem solved. The circuit was asking for one more resistor to work that way. New video attached.

Use separate 1k resistors for the two LEDs, then they will not interact.

Connect the cathode to 0V and use two resistors in the anode circuits.

I have already done that Gorgon. My previous post was updated.

- - - Updated - - -

There is one problem. When the float switch closes then 120 mA current flows through 100E resistor and power dessipated is 1.44W and I have to use a 5W resistor.

- - - Updated - - -


Problem solved again. I used 2 X 200E 2W resisitors.

THe problem with simulators is they don't care how efficient a design is, nor show any real ESR in components such as the power source, nor show proper layout methods to reduce EMI. Dumping 100mA to turn off a transistor and bias a 10 mA LED would get rejected in my design team.

Can you think of a better way to use transistors or CMOS as low power switches to eliminate this?

Also you layout is crucial to how the solenoid coil current is circulated for inductive pickup and EMI disturbance reasons, and we are still waiting for the coil resistance.

Can you give the current rating for the contact switch and the bounce time and what would you expect to happen when the solenoid reacts to switch bounce?

You don't need to use a 100 ohm resistor, you could use the 1k resistor for the LED as a pullup, and reduce the 10k to 2k2 to keep the base current to around 1mA. This way you reduce the number of resistors too.
The switch is then in parallel with the LED.

I will use a 10 uF 16V across the switch which is a float switch. I need to get the solenoid only then I can provide details. It will take 10 to 15 days to reach me. There is no datasheet as I purchased it from eBay. Once I get the sensor I will provide all details after measuring.

@Gorgon

Your method worked. I used 1.1k instead of 1k and 2.2k instead of 10k.