LIDAR Jammer

[jason]

Hi Sam,

Great work! As a suggestion, you may wish to change out your linear regulator to a switcher, as this will reduce the heat dissipation quite dramatically... With the liner regulator you are dropping 13.8v - 5 v, or 8.8V which is a hefty amount. I would suggest you look at the LM2675 switcher IC (+ inductor, diode, etc) by National (TI), or the self contained modules by TI (i.e. PT5101 series), etc.

Secondly, you may wish to place a larger cap (47uf or 100uf) on your 5V rail in order to provide some localized charge for the LEDs... 0.1 uf are pretty small and really meant to reduce switching noise, etc.

Happy Holidays and thank you for your ideas. Here are my thoughts:

The 7805 is dissipating 0.15(13.8-5) = 1.32 watts. Turns out this is well within the device's dissipation capability without a heat sink, and there is actually a small heat sink on the PC board. I ran it overnight sealed in the plastic case with 15V on the input side and it was fine in the morning. The choice of the 7805 in this project was predicated on simplicity and low cost. Thereanks was no requirement for efficiency (1.32 watts is lost in the noise in a motor vehicle). And even if heat were to become a problem, a $0.05 1-watt resistor of 10-20 ohms in series with the input power leads will safely move some of the power dissipation out of the 7805 leading to an even wider heat margin than currently exists.

The second change you suggest seems like it may any have some potential for bumping up the pulsed output power a bit. I will put my scope on the 5V rail and switch it on to see if there is any drooping occurring. Certainly the bandwidth of the 7805 is not sufficient to correct any IR droop caused when the IR LEDs are active so if this is more than 50mV or so, I'd consider it a worthwhile change. I believe a precise needed value of charge can be calculated, leading to a specific minimum value of capacitor, but the values you suggest seem reasonable at first glance.

Jason

[jason]

wow great work please keep us updated as to the results. I would assume that it would be possible to increase the number of leds just thinking of the distance a lidar can operate might need to be brighter? to be able to blind the reciever at 1000M?

There is nothing stopping you from modifying the circuit to use more LEDs.

But remember, you're not trying to blind the LIDAR unit, you're trying to confuse it. In order to confuse it, you need to return pulses of light of the right wavelength and at the right pulse rate to prevent the LIDAR unit from accurately measuring distance (from which it computes speed). So the LIDAR unit is looking for reflections from the pulses of (laser) light the operator is bouncing off your vehicle. The jammer only needs to generate pulses which are roughly equal in power (more is fine too) to the pulses generated by the LIDAR unit. The LIDAR unit will see these pulses and not be able to tell them apart from its own pulses. Since the timing will be essentially random for the pulses generated by the jammer, they will prevent the LIDAR unit from accurately measuring distance and thus prevent it from computing your speed.

So while you may be able to increase the output power by adding more LEDs (and certainly you can drive the existing LEDs harder than I am driving them right now since I'm only operating them around 60% of their maximum rated power), you would probably be better off doing things like removing your front license plate, or putting a non-reflective cover over it if that is not possible. This will reduce the strength of the pulses returned to the LIDAR unit which also means that the brightness of the LEDs in the jammer will be relatively brighter.

--
Jason

[arakis]

Hi Sam,

Great work! As a suggestion, you may wish to change out your linear regulator to a switcher, as this will reduce the heat dissipation quite dramatically... With the liner regulator you are dropping 13.8v - 5 v, or 8.8V which is a hefty amount. I would suggest you look at the LM2675 switcher IC (+ inductor, diode, etc) by National (TI), or the self contained modules by TI (i.e. PT5101 series), etc.

Secondly, you may wish to place a larger cap (47uf or 100uf) on your 5V rail in order to provide some localized charge for the LEDs... 0.1 uf are pretty small and really meant to reduce switching noise, etc.

Happy Holidays and thank you for your ideas. Here are my thoughts:

The 7805 is dissipating 0.15(13.8-5) = 1.32 watts. Turns out this is well within the device's dissipation capability without a heat sink, and there is actually a small heat sink on the PC board. I ran it overnight sealed in the plastic case with 15V on the input side and it was fine in the morning. The choice of the 7805 in this project was predicated on simplicity and low cost. Thereanks was no requirement for efficiency (1.32 watts is lost in the noise in a motor vehicle). And even if heat were to become a problem, a $0.05 1-watt resistor of 10-20 ohms in series with the input power leads will safely move some of the power dissipation out of the 7805 leading to an even wider heat margin than currently exists.

The second change you suggest seems like it may any have some potential for bumping up the pulsed output power a bit. I will put my scope on the 5V rail and switch it on to see if there is any drooping occurring. Certainly the bandwidth of the 7805 is not sufficient to correct any IR droop caused when the IR LEDs are active so if this is more than 50mV or so, I'd consider it a worthwhile change. I believe a precise needed value of charge can be calculated, leading to a specific minimum value of capacitor, but the values you suggest seem reasonable at first glance.

Jason

Hi the 1.32W is right on the limit of it's capability.. at 65W/C that's 85.8C so you only have ~40C left for ambient temp.... but if the device is enclosed. We are then talking about at 100+C heater in an enclosed space, the air around it will heat up at over 40C even if ambient is 25C.... I use a rule of thumb of 1W for non heat-sunk to220... MHO only obviusly, but Id recomed at least getting one of those clip-on heat sinks...

I'd also place the LED-s in series and power them directly from the 12V battery, and control via a single FET...

[jason]

Hi the 1.32W is right on the limit of it's capability.. at 65W/C that's 85.8C so you only have ~40C left for ambient temp.... but if the device is enclosed. We are then talking about at 100+C heater in an enclosed space, the air around it will heat up at over 40C even if ambient is 25C.... I use a rule of thumb of 1W for non heat-sunk to220... MHO only obviusly, but Id recomed at least getting one of those clip-on heat sinks...

I agree with your figures... However, there is a heat sink on the 7805 in the form of a large pad underneath the TO-220 which connects via thermal reliefs to the nearly continuous ground plane on the back of the board. I should think that this heat sink will be more than enough to handle an extra 0.32W beyond your rule of thumb.

Furthermore, dropping volt or two ahead of the 7805 is trivial with the resistor I mention in my last post and will further reduce the power the 7805 dissipates.

I'd also place the LED-s in series and power them directly from the 12V battery, and control via a single FET...

I'd still need at least two FETs in order to maintain the existing capability of running the diodes in parallel, or alternating fashion (see schematic). This is useful because it allows you to tradeoff pulse frequency for pulse power depending on your situation.

Also, if you connect the LEDs in series, the pulse shapes flowing through them will not be as clean as if they are in parallel due to the additional series inductance of a string of LEDs. Take a look at the images in my next post to see what I mean -- you may not even be able to reach the desired output power in the 83 nS width of the pulses.

A couple more minor points...

It is also nice to have a regulated voltage on the LEDs so that you can precisely regulate the amount of current flowing through them. This should allow you to safely approach the maximum rated power output of the LEDs without fear of burning them out if your car's alternator is running a bit high, for example.

Finally, there is the issue of reliability. If you did somehow manage to burn out one of the LEDs or FETs, the circuit would stop functioning if they were connected in series. It might be something as simple as a rock thrown up by the car in front of you hitting the jammer and smashing a couple of LEDs.

--
Jason

[jason]

The following images were taken from my completed PC Board using a Tektronix TDS-540B oscilloscope and a probe rated for 250 MHz.

I attempted to make two sets of images, one before I replaced a 0.1uF capacitor with a 47uF capacitor between 5V and ground, and one after. Unfortunately, the before pictures were no good because I did not have the image type set properly on the oscilloscope. Nonetheless, I number of interesting details of the circuit's operation are visible.

This first image is the DC voltage on one of the LED anodes, representing the voltage drop across the 4.7 ohm resistor (see schematic):



The peak voltage of the waveform is about 5.1 volts, and the minimum of the dip is about 3.75 volts. A couple of things to note:

• The ramp of the voltage drop clearly shows that some time is needed for the circuit to fully conduct -- probably on the order of 20 nS to reach half of the maximum. That represents a slew rate of about 57.5 MV/S
• The magnitude of the difference in voltage suggests that the peak current flow through the diode approaches 250 mA.
• The high frequency ringing at the end of the pulse may be a measurement artifact.
• The ramp up from 4.7 to 5.1 volts at the end of the pulse is probably caused by the 7805 regulator since it's bandwidth is considerably smaller than the pulse widths in the circuit.

Next, we view the DC voltage on the cathode of the LED, representing the voltage drop accross both the 4.7 ohm resistor and the LED itself:



The peak voltage of the waveform is about 4 volts and the minimum of the dip is about 1.9 volts.

• The peak voltage across the LED is approximately 3.75 - 1.9 = 1.85 volts.

Finally, we get to the AC voltage on the 5V rail:



It appears that the voltage is dipping 300 millivolts when the 6 LEDs are firing (in parallel). This is with a 47uF capacitor present between the 5V rail and ground.

Each pulse is 83 nanoseconds in duration and should flow at least 250 milliamps of current for each diode. That comes out to about 125 nano-joules of energy per pulse. Since the energy stored in a capacitor is (1/2)C*V^2 and we want to limit the voltage drop to, say, 100 millivolts, we can compute the minimum required capacitor value (assuming no circuit losses and full charging between pulses). It comes out to 25 microfarads. Yet there is a 47 microfarad capacitor in the circuit between the 5V rail and ground, so clearly other factors are causing the dip we see above.

Looking at the AC voltage on the 5V rail when the circuit is configured to fire the LEDs in two alternate groups, we get the following:



Now we see that the voltage is only dipping 200 millivolts when each group of 3 LEDs are firing.

In summary, it appears that the replacement of the 0.1uF capacitor on the output of the 7805 with a 47uF unit results in some improvement in the voltage dip. With the 0.1uF capacitor in place, the dip was on the order of 100 millivolts worse than with it present. Based on calculations and experimental results, I doubt the addition of further capacitance would improve things much further.

--
Jason

[ferdinandk]

As you are switching a (relatively) large load very fast, things like parasitic inductance and ESR have be taken into account. I would try to bypass each LED with an 10 or 100nF ceramic cap. And maybe add another 100nF in parallel to the 47uF output cap, just to make sure.

[jason]

As you are switching a (relatively) large load very fast, things like parasitic inductance and ESR have be taken into account. I would try to bypass each LED with an 10 or 100nF ceramic cap. And maybe add another 100nF in parallel to the 47uF output cap, just to make sure.

I'd agree that the ESR of the 47uF capacitor is a factor, as is the parasitic inductance of the LED's leads and PCB traces. Probably on the order of 20-50 nH of inductance per LED based on the length of the LEDs leads and the PCB traces using the old rule-of-thumb of 25 nH per inch. Would be nice to simply neutralize this reactance with a suitable series capacitor, but I'm afraid that's not possible since we're not dealing with sinusoidal signals. Your suggestions make sense, and it's easy enough to add a couple of 100nF capacitors from the 5V rail to the ground plane to see if that results in any further improvement.

[jason]

I went ahead and added a 100nF capacitor to the board, as suggested by Ferdinand. Put it near the middle of the 5v rail to the ground plane (handy via nearby). Then I hooked the board back up to the oscilloscope to see the results.

First, there's the AC close-up of the 5V rail:



Looks like an improvement from the previous 300 millivolt drop to 250 millivolts. There now seem to be more lower-frequency oscillations on the 5V rail, but they should not affect circuit performance.

Next, here's a DC-coupled graph of the collector of one of the FETs (same pickup point I used in my previous post):



Here too we see some changes. Looks like another 0.1 volt or more of peak voltage drop across the resistor, suggesting a higher peak current through the diode (0.1 volt / 4.7 ohms is about another 20 mA of current through them). That translates into improved performance due to a slightly higher peak light output. There is also less ringing at the end of the pulse which I thought may have been due to the 3-inch ground lead on my oscilloscope probe, but looks like it was in the circuit itself since it is nearly gone now.

No doubt further improvements can probably be made by adding still more bypass capacitors along the 5V rail but the circuit already works quite well and I think I'll put it back on my car for the time being.

As mentioned a few posts back, I've added a 1 watt 10-ohm resistor in series with the 12V power cable to reduce the dissipation of the linear regulator. In the Spring, I may add a wireless control circuit to enable me to switch this on and off from inside the car so I don't have to either run a wire through the firewall into the passenger compartment or go outside to switch it on and off like I do now.

--
Jason

[madhawk]

hello friends!

i have seen your post!
nice work!

i work since many years with laser jammer and radar detector (first customer and now i have my own business)

do you have testet your jammer?
i have 3 lidar guns and 2 of them are very very difficult to jamm!!!
(all guns have an jamm indicator.) when you take a "brute force jammer" you get a problem with the police because they can see over the jam indicator a jamming sequence from an laser jammer...
what you guys need is a intelligent jammer who can detect the guns and the true pps from the gan and send a signal with the same pps back to the gun.
my lti trucam has a antijam firmeware and the lidar gun change the pps and your jammer is knocked out...

when you need my help so i can make some tests with your device... (i am from austria!) or when you need other information about the lj´s on market please feel free to contact me.

i am searchin also a other device. i search a device who can read the bus protokol from an radar detector or laserjammer. (to build my own interface)
have anyone a idea?
thank you for your support
best regards from austria!
madhawk aka. RadarWetzi

[traxonja]

Hello jason,

Any progress on this project? If you need an additional tester, I am willing to work with you in parallel.

Have you considered the detection of LIDAR pulses? It would be a good idea to receive the pulses from LIDAR and decide what to send out with your TX unit, as @madhawk suggested...

Best regards,
Trax