If the arrows are the diodes, they are drawn in the wrong direction.

Ken

Haha, thanks, they are just there to indicate that they are outputs.

Thanks again, Phil.

 

Hi all, after some minor refinement I have finished with this circuit, which in comparison is prety much the same as one posted earlier on by another member, note to self, listen to that member more often haha.

anyway here is the schematic, If you spot a problem please let me know.

Thanks, Phil.

 

Ohh, and the reason for the components rated at 2W, im a bit of a worrier and would prefer to know for sure things arnt going to melt.

Thanks, Phil.

 

I can understand the 2W 100 ohm resistor since it takes only 14.14 volts to yield 2 watts. The 2W 10K and the 1W 47K resistors will take 141.4V and 216.8V respectively before their power ratings are exceeded.

Better safe than sorry.

hgmjr

 

I can understand the 2W 100 ohm resistor since it takes only 14.14 volts to yield 2 watts. The 2W 10K and the 1W 47K resistors will take 141.4V and 216.8V respectively before their power ratings are exceeded.

Better safe than sorry.

hgmjr

If the voltage were to raise to 14.14 volts what would happen to the resistor? atm it runs only slighly warm at 13.8V would it burn out or just run too hot?

Thanks, Phil.

 

Raise D2 to 9V, as there is no real advantage to 5V, and R1 will not have to dissipate as much heat. Put a 1K resistor between T1's emitter and T2's base. With both turned on, you are essentially shorting the voltage at D1 to ground.

Ken

 

Raise D2 to 9V, as there is no real advantage to 5V, and R1 will not have to dissipate as much heat. Put a 1K resistor between T1's emitter and T2's base. With both turned on, you are essentially shorting the voltage at D1 to ground.

Ken

The reason for 5V is to allow for the 2V excess voltage, this take it to 7V minimum operation. If I were to raise to 9V this means that the minimum operating voltage would be 11V, this would be too high for my application.

Ahh right yeh, will this 1K resistor not affect the operation of T2.
What power resister would be good for the 1K?

Thanks, Phil.

 

Your original spec was 12v to 30v. You then changed it to 11V to 28V. Does it really need to operate down to 7VDC?
All resistors except R1 can be 1/4W.
Have you tried R1 at 28V?
P=((28-5)*(28-5))/100=5.3W
That's 230mA. Do you know how much current the circuit will draw? R1 might be able to a much higher value, dissipating less heat, both in R1 and D1.

Ken

 

Your original spec was 12v to 30v. You then changed it to 11V to 28V. Does it really need to operate down to 7VDC?
All resistors except R1 can be 1/4W.
Have you tried R1 at 28V?
P=((28-5)*(28-5))/100=5.3W
That's 230mA. Do you know how much current the circuit will draw? R1 might be able to a much higher value, dissipating less heat, both in R1 and D1.

Ken

Hi Ken, thanks for your time.

Although I did originaly specify 11V to 30V The circuit returned unexpected results and this was due to the ignition line dropping as low as 8V briefly under load, so the specification was changed to 7/25 Volts.

I have now changed all resistors to 1/4Watt.

I havnt tried R1 at 25V's, because my calculations state that it will be way over powered and probably burn out.

The circuit draws approx 150mA highest when the capacitor is completely discharged and the realy is energised.

I tried changing R1 to a 200 ohm, 2 watt resistor but it still gets hot. I can scan/attach all of my calculations if you would like to see them but they are basic V, I, and R equations.

When you say dissipating less heat in D1, I assume you mean D2, the Zener diode. It runs cool, as it is a 2 watt diode, this is operating way below its capabilities.

Thanks, Phil.

 

The zener shunt regulator is wasting a lot of power in the dropping resistor to save on component count. Just the space taken up by a 4-5W resistor (what you should use) is a waste. The attached circuit will draw 5mA max and dissipates only 50-100mW in the regulator. R1/D1/D2 are there to protect the regulator from over-voltage and negative spikes. The MOSFET is a voltage controlled device, so doesn't require drive current. Since it uses a 5V regulator I switched the 2N7000 to a logic-level MOSFET.

Re-breadboarded the circuit with the IRL520. Chance the 4.7uF cap to 10uF for 2 seconds.

Ken

 

A diode across the relay coil would be a prudent addition.

 

Phil,
The attached is just a layout of my circuit for size comparison. Just the 5W resistor for your Zener version will be that big, and dissipate 50 times as much heat.

Ron,
I think the OR-ing diode D4 will protect the MOSFET from the relay's flyback voltage.

 

Ron,
I think the OR-ing diode D4 will protect the MOSFET from the relay's flyback voltage.

What does the ORing diode connect to? Did I miss something?

 

D4 connects to my monostable, and D5 connects to a unit that normally pulls the relay low after 2 seconds. See post #19.

Ken

 

Sorry to drag this one back up, Nice and easy question with regards to KMoffet's design:

What calculations did you use to optain the 2 second components as normal RC circuits that I've dealt with are in the opposite configuration, did you mearly inverse the results?

Ive increased the resistance to approx. 880kOhm, this increases the delay to approx. 3.4 Seconds, would It be better to change the capacitor instead of having a resistor this large.

Thanks, Phil

 

The switching time depends on the switching characteristics of the particular MOSFET and relay you're using. It was experimental in my case. You can change either one.

Ken