555 driven relay causing issues with power loss

Thread Starter

electrichelmut

Joined Feb 17, 2021
5
So I have been creating a circuit (sic) to use a pulse sent from an automotive c/l unit, to send a timed pulse to operate a motor. With some help from another forum see here: https://www.briskoda.net/forums/topic/488951-555-timing-chip-help/ I have it working pretty much as it should, as many times as I want, when no load is placed on the relay(relay switches on off as it shoud)
But when a load is connected to the relay, and the pulse is sent, the c/l control unit goes quiet/stops working when the relay switches. The circuit does work, but just once, and because the control unit is now off and needs re activating, no further pulses can be sent. It is almost as if the control unit power is removed/bypassed or goes too low, when relay switches.
I sure there is something obvious I have missed here, but being a novice when it comes to electronics I'm not sure what.
If you read the posts on the other forum from last to first you will probably get an idea of what I am talking about, but it being a car based forum I think they have run out of steam over there.
Also I do know that my desgined circuit does have flaws(it's just how it evolved over time) but as seen as it works with no load, I can't be that far away, I hope.
 

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ElectricSpidey

Joined Dec 2, 2017
1,653
Do you have any noise/back emf suppression on the motor?

Do you have any de-coupling caps on the various chips/devices?

You can also get rid of the 100Ω resistor in the VCC path of the 555.

EDIT: see post #5.
 
Last edited:

Audioguru again

Joined Oct 21, 2019
3,172
The 555 should have a 10V zener diode and filter capacitor parallel with it and they are powered by the 100 ohm resistor.
The zener diode and capacitor prevent the 555 from being destroyed by common voltage spikes on a car's electrical system.
Of course the transistor should nave a series base resistor.
 

Thread Starter

electrichelmut

Joined Feb 17, 2021
5
So been back ironing out issues with it today, thought I would work through the posts one by one.
What happens to the battery voltage when there is a relay load?
And it would seem that my first issue was with power supply voltage(knackerd battery)
Voltage across battery terminals only 10volts with no load, then drops to 4.8volts with load, meaning not enough volts for the c/l unit
So with a different test battery, success. I was able to switch a test headlamp using the circuit in it's original form(not tryed it with the motor yet)

Now to the modifications I need to make to the original circuit
You can also get rid of the 100Ω resistor in the VCC path of the 555.
Easliy done

Of course the transistor should have a series base resistor
Ok, again pretty easy

Do you have any noise/back emf suppression on the motor?
NO, but as far as I understand it, It means adding a diode in series before the motor and a capacitor in parallel across the motors terminals. Is that correct?

Do you have any de-coupling caps on the various chips/devices?
Again No, Didn't even know what these were until I typed it into a search engine. Not a 100% sure where to place them in my circuit, I presume it goes like in Audioguru's modifications to my circuit

The 555 should have a 10V zener diode and filter capacitor parallel with it and they are powered by the 100 ohm resistor.
Is this just the same as the de-coupling capacitor except with a zener diode and reistor in series with it?

And finally does the whole circuit need a voltage regulator?

Thanks for your reply's





 

Audioguru again

Joined Oct 21, 2019
3,172
The filter capacitor is the decoupling capacitor discussed in the 555 datasheet.
The 555 already has a series 100 ohm resistor that can power the 555, the zener diode and the filter capacitor at the same time.
The zener diode is a voltage regulator that prevents car voltage spikes from destroying the 555.
 

Thread Starter

electrichelmut

Joined Feb 17, 2021
5
So the circuit would be modified as in the attatched diagram below, green being the modificationsgenb2modifpy - Copy - Copy.jpg
A few questions:
Is the zener diode the correct way around?
Would a 12V zener diode be more suited, it being a 12V battery?
Would this circuit work for back emf suppression from the motor?
It is ok that I am using different chasis ground locations isn't it?
 

sghioto

Joined Dec 31, 2017
2,360
But when a load is connected to the relay, and the pulse is sent, the c/l control unit goes quiet/stops working when the relay switches.
What is this c/I control unit?
So on the first pulse the relay activate and times out correctly, then after that no more pulses from the c/I unit?
Also, you can most likely eliminate the most of the circuitry that switches the 555.
 

AnalogKid

Joined Aug 1, 2013
9,248
In the post #1 schematic the output signal from the remote control unit goes to the non-inverting input of an unknown device. What is the logic polarity of this signal? That is, is the 555 supposed to be triggered when this signals goes high or low?

There are other issues with the schematic, but this one sets the rest.

Also, what does c/l mean?

Separate from that, does the remote control make a short pulse output, long pulse, stay active as long as the remote button is pressed, other - ?

Is the output of the 555 supposed to be the same pulse width no matter how long or short the signal from the remote is? If so, then a 555 has issues.

ak
 

sghioto

Joined Dec 31, 2017
2,360
The full story.
From the link the TS posted:
"To expand more I will start at the very beginning
So my car has c/l but only on the key, so decided to upgrade to remote c/l and got one of the readily available aftermarket kits online.
I have had no problems integrating this with the cars current c/l when on a test fitting.
It was after this that I noticed the optional remote boot release option like on most new cars, and I thought it would be as simple as a few more wires and a solenoid (late model felicia's already have a manual boot release mechanism in place to work with)
Now this is where the problems start; the springs in the current manual release mechanism are too strong for solenoid to actuate, and with limited space available my only option was a motor with a small arm to gain some leverage.
So still should be quite simple, a relay driven by the pulse emitted from the remote c/l unit. No not that simple the pulse form the unit is pitifully small somewhere around 25mV and 2mA for 1.8 seconds
Again this can be worked around using an op-amp as an amplifier to trigger a transistor witch triggers a relay (it took up a lot of time working this circuit out) I will include this in my sketch
Now this is the problem solved you would think, but again No. Operating the motor for 1.8 seconds can leave the boot release mechanism in the unlocked position and the boot being unlatched. Now this could be solved by pressing the key fob's boot release button until the boot mechanism is latched again, but roughly speaking the mechanism would land in unlatched position 25% of the time.
So I figured by using a 555 chip I could have the motor always end in a place that leaves the boot latched (2.05seconds leaves the motor in the same place every time, a park position if you will)"

Same circuit modified with less components.
If the RCU is loosing power something is probably wired incorrectly.
1613696594271.png
 

Thread Starter

electrichelmut

Joined Feb 17, 2021
5
Power loss issues were being caused by a faulty battery (It is silly I did not check it in the first place)
I am now trying to work on creating a stable voltage for the circuit, and to protect the componants from voltage/power spikes( i.e that I have no filter capacitor in my original circuit)
The central locking unit, did not come wit a internal schmatic, and trying to find one would be impossable, it is just a cheap unit mass produced in china for the car aftermarket industry.
The reason the circuit is so convoluted is because as I came across a new issue, I tended to just add componants and not remove the redundent ones
 

sghioto

Joined Dec 31, 2017
2,360
It appears extremely unlikely that the output is only 25mv. For the original circuit to work the output must be at least over 1 volt
as the op amp circuit as configured has a gain of only 1.3
If that's the case then most likely you can remove the op amp and you don't need a zener.
1613751664531.png
 
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