# Turn signal detector

I'm trying to figure out the simplest design possible. I want a video camera to activate on my truck when I activate either left or right turn signal. Each side mirror has its own camera. So basically I want detector circuit capable of detecting 12 volt pulses and send the output high to activate a relay. Once the turn signal is deactivated, output will go low and stay low until pulses are detected again. I was planning on using 555 timer detector circuit but I just want a single pulse on the output and not multiple pulses. Any information greatly appreciated, thank you

• 555 re-triggerable one-shot Commented Jun 5, 2022 at 2:24
• Retriggerable astable circuit Commented Jun 5, 2022 at 2:24
• unkle, the basic idea is just as already written. This is the central idea -- a re-triggerable one-shot. Concept has been around forever. Way before electronics. However, you also need a way to detect and generate triggers from the turn signals. You need a way to translate those triggers into an input that the re-triggerable one-shot accepts. You then need to convert the one-shot output into some kind of camera power or enable to control. You have provided nothing much so the comments provide in-like-kind value -- nothing much in short. Write more. Get more. Or are you just buying a product?
– jonk
Commented Jun 5, 2022 at 3:37
• To be clear, do you want the left turn signal to activate only the left camera, and right for right, or do you want either turn signal to activate both cameras? Commented Jun 5, 2022 at 14:49

Here's a simple schematic. The idea is to use a capacitor across the relay coil to prevent it from dropping out in the turn signal off-period.

Considering a turn signal flash frequency of 1 Hz and a duty cycle of 50 %, the duration of 'on' time and 'off' time would be 0.5 s.

The time constant with the specified RC values (185 Ω and 2200 μF) would be 185 x 0.0022 = 0.4 s.

The capacitance value of 2200 μF would suffice since the drop-out voltage of the relay would be fairly low.

'NO' contacts of the relays K1 & K2 are to be used to activate the cameras.

• You need a current limit resistor in series with the capacitors to protect the contacts and diodes from high charge current (e.g. 22 ohm / 2 W). D2 and D4 are not needed.
– Jens
Commented Jun 5, 2022 at 15:14
• @Jens - Hi, Thank you very much for the feedback. Commented Jun 5, 2022 at 17:31
• Hello vu2nan, thank you for the idea. That is basically what I was thinking was a simple rc circuit. Charge the capacitor on 12 volt on and discharge capacitor on 12 volt off and keep the relay energized. Now I'm just wondering the reasoning behind the 2 diodes? D2 seems like a protection diode but not sure about D1 Commented Jun 5, 2022 at 18:58
• Anytime, unklemanny87! D1 prevents the capacitor C1 from being discharged by the turn signal lamp when it is off. Commented Jun 6, 2022 at 2:59

I'm trying to figure out the simplest design possible. ….
So basically I want detector circuit capable of detecting 12 volt pulses and send the output high to activate a relay.

Although an 555 would work, for me it is not the simplest way possible.
Using a relay directly as vu2nan answered would need larger capacitors and may not give you independent timing control.

Proposed Circuit:
I propose you use a transistor and a RC circuit polarizing its base to maintain it conducting for longer time, easily achieving 1-10s of turn-off delay, after the turning signal is switched-Off.

Timing-related components are circled in blue, below:

You can simulate it in FALSTAD here adjusting parameter values as C=100uF and R1=10K (for time constant) and even coil resistance (click/“edit” the component).

Making the circuit a reality:
Most low-power bipolar NPN Transistor as 2N2222 or BC337 can be used to drive relays with 80R~200R coils. Diodes can be any as 1N4001 …4007. R.diode = 1K is here to protect an overcurrent in the transistor base, if “R1” (ahead) is replaced by a trimpot.
About R1: Maximum recommended value is R1 </= 10K. It can be smaller, so it could be replaced by a trimpot. R1 directly affects the base current of Transistor, so for higher-current relays and lower HFE transistors, this could be a problem, as when R1 > 10K or HFE < 100 or R.coil < 100R.
Somewhat larger timings = simply increase C; an increased R2>22K has lower influence.
Warning notice: This simple circuit does not include a schmitt trigger feature, to switch-off the transistor (and relay) faster. Instead, then current decays proportional to capacitor voltage decay. Because of that, larger Turn-off timings should be limited to T ~= 10s.
Don’t use this circuit for T ~= 100s, as transistor will be partially conducting for longer time (UPDATED: BJT dissipation will be ok up to 300mW = 50% max in small TO92 transistors) and a slow coil voltage swing is not recommended for relay contacts. If that is your case, then a 555-based circuit is preferred.

RC circuit is basically what I was thinking but wasn't too sure about how to implement it.
Now I'm wondering if there is an advantage towards using an NPN vs just driving the relay directly from the 12 volt turn signal circuit like vu2nan showed.

Yes, there are in fact two advantages:

1. Capacitance is multiplied by HFE of the transistor.
In this case, the 100uF would behave as 10.000uF, where 100uF+BC337 is much smaller and cheaper than 2200uF.
2. Time Constant factor is amplified. The release voltage of the relay is ~= 1/2 nominal (for you, 12V/2 =50%) or about 1x TimeConstant (nominally 1TC = 63% drop). On the proposed circuit, transistor stops conducting at ~ 2V, which is about 80% of voltage drop, increasing the conduction time to switch the delay: 1TC < delay < 2TC.
Feature (1) is the most relevant.

This updated circuit here changed some values for R.diode, R1, R2 and C; it enables you to simulate, graphically experiment, and “see” these effects happening.
It now enables experimenting with cursors different Capacitances (50uF < C < 1000uF) for different cycling Frequencies (10mHz < F < 1Hz, 20% duty cycle), and HFE (up to Darlington values).

For example, if someone aims to obtain as above a Delay = 10s using the same coil resistance of R.coil = 100Ω, you would need about 100,000 uF (0.1F) of “direct capacitance”! That is 170 times bigger than the simulated 600uF, confirming the “advantages” 1 and 2 explained above, with important savings in size, cost and parts availability with the proposed circuit.

Finally, a huge file from Chemi-con, provides a size reference for different capacitances on the same capacitor model. For pricing and availability, use search tools as this Mouser one.

• Hello EJE, thanks for the response. Like I mentioned to vu2nan, an RC circuit is basically what I was thinking but wasn't too sure about how to implement it. Now I'm wondering if there is an advantage towards using an NPN vs just driving the relay directly from the 12 volt turn signal circuit like vu2nan showed? Commented Jun 5, 2022 at 19:04
• Yes there is: you can see the transistor as a capacitance multiplier - (1) first feature: Capacitance is multiplied by HFE. In this case, the 100uF would behave as 10.000uF, where 100uF+BC337 is much smaller and cheaper than 2200uF. There is (2): The release relay voltage is ~= 1/2 nominal (for you, 12V/2) or about 1x TimeConstant. On the proposed circuit, transistor stops conducting at ~ 2V, which increases conducting time 1TC < delay < 2TC. Feature (1) is the most relevant.
– EJE
Commented Jun 5, 2022 at 19:53
• Another point: if C=2200uF and R.relay=200Ω, TC= 0.4s. For more than 1s, you may need 4700~10000uF, not counting the large inrush current whenever capacitor is charged (poor diode). For me this is a No-Go. Then, my suggestion to use a “capacitance multiplier” as suggested, while much simpler than a 555-based timer.
– EJE
Commented Jun 5, 2022 at 19:58
• @unklemanny87, I updated the post - see additional info there. Obviously I would choose my circuit as the “NPN capacitance multiplier” is a mid-point solution of “Diode+Capacitor” and “Full-555-based” circuits., but it is your choice. And I forgot to mention a design “detail”: that if your truck is 24V based, the LM555 needs overvoltage protection (V<16V), where NPN continues to work fine.
– EJE
Commented Jun 5, 2022 at 22:26