Converting automotive indicator signal to v DC

Thread Starter

The Mad Hat Man

Joined Apr 21, 2020
13
Having read the following thread, I thought it better to start a new thread, rather than start old arguments ;). https://forum.allaboutcircuits.com/threads/converting-square-wave-to-dc-voltage.122935/

I have a requirement to power a <300mA camera from each vehicle indicator signal. To make it more awkward, the flasher unit is before the indicator switch, therefore the signal (left or right) is, I believe, ostensibly a square wave. Without loading the circuit too much (it will effect the flashing rate), can I connect a bridge rectifier across the bulb circuit to provide 12v DC to power the camera? Because this is for a road/race car, size, weight and complexity is at a premium. I would like as simple a solution as possible, as switching the camera on and off at the same rate as the indicators, is not very effective :D.
Can you assist and give any guidance?
I have looked at Arduino options, but it gets bulky very quickly.
 

SamR

Joined Mar 19, 2019
2,096
Going by the seat of my pants... the signal goes from the lever to the flasher so grab it before the flasher and simplify the circuit. Why grab it after the flasher? A diagram of what you are doing would help.
 

dendad

Joined Feb 20, 2016
3,514
Just wire the camera across the indicator lamp. The car is already DC, so no bridge rectifier is needed.
But I cannot get my head around why you need to do this.
 

Hymie

Joined Mar 30, 2018
819
You could try such a technique, but you would need to add a capacitor across the output of the bridge rectifier to maintain the 12V between indictor flashes. But the problem is that the capacitor value would need to be very large to prevent the load volt drop between flashes turning off the camera.

Say that the indicators are flashing at a rate of 1 per second, with a 50% duty cycle and that the camera will switch off once the voltage falls below 9V drawing 300mA. Using the volt drop formula for a capacitor constant current discharge of V = (I x t)/C, where I =amps, t= discharge time and C = Capacitance in farads.

So you have an allowable 3V drop over ½ second with a current draw of 0.3A; this results in a required capacitor value of 0.05F (50,000µF). Bear in mind that the flasher unit would need to withstand the peak capacitor charge current which may alter the flasher unit behaviour if it is a bi-metalic switching type.

Since the vehicle electrics are dc, you only need a diode and not a bridge rectifier to provide the camera voltage. The diode is required otherwise the indicator lamp would remain illuminated between flashes.

You would probably be better triggering a 555 timer using the indicator signal, to give better control of the camera on/off timing and a more cost effective solution.
 

dendad

Joined Feb 20, 2016
3,514
Can you please explain in a bit more detail what you want your application to do.
Are you using the flasher signal to take a photo or power each camera?
What sort of camera are you using?
It still is a bit odd to me.
 

Reloadron

Joined Jan 15, 2015
5,424
The turn signals are pulsed DC and yes, since many turn signal indicators are thermal changing the load will change (or stop) the flash rate as when a lamp fails. More elaborate LED signals just use a slightly different system for timing. This would go much better with a schematic for tha make and model you have?

Anyway you could take the signal going to a bulb and just insert a low pass filter which will give you a DC level. What needs known is the normal flasher rate? Using a simple LP filter of a resistor and cap the cap will charge to the average value so if the lamp is 12 volts (about) and the duty cycle is 50% (on / off equal time) the cap will charge to about 6 volts. This will take about 5 time constants but I don't expect you to grasp all of that. Just knowing the timing is important. This is also just a single approach. There are several ways to go about doing what I "think" you may want to do.

Ron
 

Thread Starter

The Mad Hat Man

Joined Apr 21, 2020
13
Sorry folks. Maybe I did not explain very well.
The flasher relay is BEFORE the indicator switch, so the output, either Left or Right, is the "flashing" signal for the lights. I want to switch on a video camera to assist in side visibility; hence the need to switch each video camera on when the indicator stalk is actioned. this gives the problem that the only signal available is the "square" wave available to the light bulbs. this is what I need to convert to a steady state 12V DC (+/-10%) to switch the camera on.
A seperate switch, rewiring the indicator circuit or a bulky electronic system is not an option.

Thanks for your responses.
 
Last edited:

Thread Starter

The Mad Hat Man

Joined Apr 21, 2020
13
Yes it does! Your posts are a bit confusing.
I still cannot figure out what you are doing. What do you need to power cameras from the flashers????
Well. That is an interesting reply, as you cannot have seen my vehicle.
I have checked the circuit in my vehicle and it goes 12V to the flasher unit, up to the indicator switch, which switched the signal to either the left hand or right hand indicator lights.
I am sorry for your confusion, but my facts are correct :p.
I need to turn on an external video camera (think reversing camera system) when either the LH or RH indicators are energised. There is a camera on either side of the vehicle.
 

Reloadron

Joined Jan 15, 2015
5,424
OK, so two camm correct? Right and Left? Should that be the case maybe build or buy a module known as a "Retriggerable One Shot" timer. The idea being once triggered by an active turn signal it will turn on the video for a preset time interval and once the turn signal is off it will time out and shut off the camera.

"Upon application of input voltage, the time delay relay is ready to accept a trigger. When the trigger is applied, the output remains de-energized. Upon removal of the trigger, the output is energized and the time delay (t) begins. At the end of the time delay (t), the output is de-energized unless the trigger is removed and re-applied prior to time out (before time delay (t) elapses). Continuous cycling of the trigger at a rate faster than the time delay (t) will cause the output to remain energized indefinitely.

So your first turn signal pulse starts the timing interval on its falling edge. The timing interval continues with additional pulses. On the last pulse, turn signal off, following a short time duration the cam shuts down. They can be bought as off the shelf modules made for use on a 12 VDC system. Macromatic makes them as well as other companies. There are also inexpensive off the boat versions from China.

Ron
 

Thread Starter

The Mad Hat Man

Joined Apr 21, 2020
13
thanks guys. I see what you are suggesting, however, unless I am stoopid (No need for a response ;)), surely a couple of these would do the same, and off the shelf, easily fitted and effective - https://www.carbuilder.com/uk/0-to-60-second-time-delay-relay-for-demist - Using Ignition power for load voltage supply and indicator trigger to set the initialisation pulse. It appears this is identical to what @Reloadron suggested??
 

Reloadron

Joined Jan 15, 2015
5,424
Yes, looks to be what I suggested. The only thing to be careful with when looking at this stuff, with a focus on timers is to make sure the timer in question does exactly what you want it to do. Your link looks to be a one shot and yes, it should work.

Ron
 
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