hi Beau,
The TS states his lamp takes 8 amps when running, that could be upto 5 times the current while lamp element reaches its operating temperature. ie: 40 Amps.

A poorly routed ground/0v line could easily develop a high negative spike that could take out his MCU.

E

It’s an HDI ( metal halide) lamp, so no “cold filament” inrush, but there is a nasty little circuit with a diac and an auto transformer making high voltage pulses until the arc strikes.

 

Using RC Filter or LC Filters for the Mcu?

 

Not disagreeing with you, but what would be the mechanism for damaging the chip with a badly routed ground connection?

Bob

looking at the picture in the original post, it is important how the dc-dc groudns are tied together, and they need to be for the circuit to work. In some configurations the voltage drop on the main current lines could add to the power supply voltage and overload the micro, or its i/o pins.

 

looking at the picture in the original post, it is important how the dc-dc groudns are tied together, and they need to be for the circuit to work. In some configurations the voltage drop on the main current lines could add to the power supply voltage and overload the micro, or its i/o pins.

Certainly having the load current cause a voltage drop on the return lines can affect the voltage of the processor, BUT not by increasing the voltage. It will drop the voltage. Only if the regulator ground voltage is separated from the processor common line can there be a problem.
So for any hope of the system ever performing correctly the connections to the supply must be totally separate.
And really, the five volt supply for the processor needs to have a large value capacitor connected across the processor board terminals so that when the supply gets yanked low by the spotlight inrush, it will not drop. AND , in that case, a diode connected across the regulator so that the output will not be more positive than the input. And the a forward diode between the battery and the input so that the load will not pull the input to the regulator down. Yes, it gets complicated, look at the regulator application notes for the explanation.

 

hi Beau,
The TS states his lamp takes 8 amps when running, that could be upto 5 times the current while lamp element reaches its operating temperature. ie: 40 Amps.

A poorly routed ground/0v line could easily develop a high negative spike that could take out his MCU.

E

Thank you for all the help. So what I am getting is that the ground from the main load needs to go straight to the power source?

 

Thank you for all the help. So what I am getting is that the ground from the main load needs to go straight to the power source?

The spike on the negative line will be positive, towards the positive source , and so it can not exceed 12 volts, at which time there might be a problem

 

hi Beau,
The TS states his lamp takes 8 amps when running, that could be upto 5 times the current while lamp element reaches its operating temperature. ie: 40 Amps.

A poorly routed ground/0v line could easily develop a high negative spike that could take out his MCU.

E

The TIP120 datasheet has a max collector current of 5 Amps? I am surprised it is handling 8A, it can handle 8 Amps 'pulse' continuous 5 Amps. What resistor value are you using from the micro to the Darlington to limit current on base?

 

The spike on the negative line will be positive, towards the positive source , and so it can not exceed 12 volts, at which time there might be a problem

TIP120 Collector to Emitter saturation voltage worse case is 4V. The device is only rated for 2W max power. At 8 Amps that is 32W or power! Of course that is worse case per the datasheet. Graphs show 1.25V typical at 5A at 8Amps per figure 2 of datasheet typical is 2V. Still 16W of power! You need an extremely large heat sink for the TIP120 and most likely a fan running over it to remove heat. I am surprised it is working at all being stressed beyond datasheet recommendations:

https://media.digikey.com/pdf/Data Sheets/ON Semiconductor PDFs/TIP120-22.pdf

 

TIP120 Collector to Emitter saturation voltage worse case is 4V. The device is only rated for 2W max power. At 8 Amps that is 32W or power! Of course that is worse case per the datasheet. Graphs show 1.25V typical at 5A at 8Amps per figure 2 of datasheet typical is 2V. Still 16W of power! You need an extremely large heat sink for the TIP120 and most likely a fan running over it to remove heat. I am surprised it is working at all being stressed beyond datasheet recommendations:

https://media.digikey.com/pdf/Data Sheets/ON Semiconductor PDFs/TIP120-22.pdf

If you go back and look at the circuit drawing in post #1, the transistor is only controlling the relay, which probably has 30amp contacts and a 200mA coil current. So the transistor probably does not even need a heat sink.

 

If you go back and look at the circuit drawing in post #1, the transistor is only controlling the relay, which probably has 30amp contacts and a 200mA coil current. So the transistor probably does not even need a heat sink.

Yes that’s correct

 

Yes that’s correct

That relay only draws 150 ma

 

If you go back and look at the circuit drawing in post #1, the transistor is only controlling the relay, which probably has 30amp contacts and a 200mA coil current. So the transistor probably does not even need a heat sink.

Oops! Sorry, my fault. It was late at night. I must have been sleeping. Never mind!

 

That relay only draws 150 ma

Yea, I should have simply gone to bed. Not awake, sorry.

 

I am starting to have some luck! thanks for all the suggestions. It gave me some ideas.
I did check the amps of my spotlight using a shunt and o scope and established a 40 amp current draw on start up.
So i have tidied up my power cables and earth to battery and its lots better./

 

I also put a ferrite choke on the main power cables.