Hi all!
I am relatively new in electronics so I appreciate any helps from you all!
I will try my best to explain my design, please pardon me if there are a lot of error in it.

Overall, I am trying to design : An infrared sensor when detected something , will go through a comparator LM393, then its output to the input of 74HC04 HEX inverter IC. And in turn drive a motor.

I have grounded all the unused input of 74hc04.
It is a 5Vdc relay and 5v DC motor.

My problem is upon connecting to 6v power supply, it works. However when I supply it with a 9V battery that goes through LM7806 voltage regulator. The relay made that fast switching noise continuously and the motor wouldn't start.

What have I done wrong?? Been trying to figure it out for the last two days..
Thanks again in advance to all who reply!

 

Most likely the 9 volt battery cannot deliver the required current. What type of battery and show how the regulator is wired.

 

Hello,

It sounds as if the battery can not provide the current needed.
The voltage likely drops when the motor wants to start.

Bertus

 

Speaking solely on the relay, they have a rated voltage. Exceeding that voltage by a little usually won't hurt it but it will run hotter. Remember Ohm's Law: I = E/R. If your coil has a resistance of 100Ω and you power it with 5V then 5/100 = 0.005A and 0.025W (Watts is EI (or Volts times Current)). If you power that same relay with 10V then 10/100 = 0.01A and 0.1W (or 100mW), four times the heat.

Depending on the relay, it will need a minimum of current (consult the data sheet for the relay). If it needs 250mA and the battery is only capable of 200mA then the relay will not pull in. Whereas your 6V battery may easily supply 2A or more.

As you can see from my examples above, running a 5V relay on 9V is almost four times the heat, which will shorten the life of the relay.

 

The 74HC04 rated load is 25mA max per pin without the output voltage dropping a lot. How much current does the relay draw? A BJT would be better than that IC for switching the relay.

 

It sounds as if the battery can not provide the current needed.
The voltage likely drops when the motor wants to start.

I like this thought. However, if the motor doesn't start until the relay pulls in, BUT the motor draws more current than the battery can support in way of motor and relay, then the relay, which did click in will click out. Upon shutting down the motor the relay will click in again. The result will be a chattering relay. While it's a good point to be made, it may be evident of the battery simply not having enough power to pull the relay in - in the first place.

 

I think Tony name nailed it.

How much current does the motor take, stall current if you know? Alternately, what is the resistance f the motor winding?

Bob

 

What is the relay coil resistance and dropout voltage? With a 6V supply voltage, 74HC04 is spec'ed to provide 5.2mA with a the output dropping as much as half a volt.



74AC04 will sink or source 24mA, but the output voltage could drop more than 0.6V.

 

I think alex_t has it right.
The 74hc04 cannot drive the relay

 

I think alex_t has it right.
The 74hc04 cannot drive the relay

How does that explain why it works when driven by a power supply?

Bob

 

If it works on the 6 volt supply then the 74HC04 is not the problem. However all 6 inverters can be paralleled for more output.

 

My money is on the battery.

 

Posts #3 and #6 explain exactly what is going on.

Bob

 

An electric motor is like a piece of wire when it is stalled and when it starts running. Then its current is very high and is much more than a little 9V battery can produce.

A new Name Brand 9V alkaline battery can produce 8V at 250mV for a few minutes. A Super Heavy Duty 9V battery from China can produce 5V at 100mA for a minute.

An LM7806 needs a minimum input of 8V.

 

The 74HC04 rated load is 25mA max per pin without the output voltage dropping a lot.

Actually, it is 25 mA max. per pin before the part is destroyed. Throughout the datasheet, the output is not chatacterized above 6 mA.

For a more beefy output, use something in the AC series. The output stage is characterized up to 20 mA source and sink. One or more sections in parallel should do the job, but so would a single 2N7000.

OTOH the LM393 output might be able to drive the relay directly. Its low output voltage at 10 mA is les than 1 V, way better than an HC low output.

ak

 

All we know of the relay is that it's a relay. It could be one of those ice cube sized 4PDT relays, or it could be one of those tiny, half the size of an AA battery size relays. The TS hasn't responded since the initial post. He/She may be busy with other things for the moment. What we need is a picture of the relay OR the model number so we can look up the specs and K•N•O•W what we're working with. Otherwise we're guessing and arguing about our guesses. Truth is - none of us know as yet what relay we're talking about. For all we know we could be talking about one of those industrial size relays - or a high amp contactor.

THIS is all we know so far about the relay: (and it isn't even drawn correctly)(though that's not important).

 

Another thing I just noticed: The question states a 6V voltage regulator.

When on a 6V POWER SUPPLY - everything works. But when powered through a 9V battery AND a 6V REGULATOR - we have issues. That in itself opens up new issues. A 9V battery ? ? ? Old? New? Enough current to drive the relay? Is the REGULATOR robust enough for the relay? (though that's not a likely issue - still - - - ).

 

How does that explain why it works when driven by a power supply?

Bob

Power supply has enough current but the inverter output does not.

 

For starters, if the circuit is wired exactly as shown, that hex inverter is not powered. That may be a problem. Next, the internal resistance of the 9 volt battery causes a serious voltage drop when the load is applied. If you put a voltmeter on it you will see that.. Finally, all of the parts will work quite well with a supply voltage of 9 volts. so you can bypass that regulator and remove it. That will leave you with only the battery internal resistance to contend with.

 

Hello,

Next to the internal resistance of the battery that @MisterBill2 mentioned, there is also the voltage drop of the regulator to take in account.

Bertus