Technical electrical

[CECAA850]

This is not a speaker or amp question, but I thought I'd pose it here. The collective knowledge of this forum never ceases to amaze me.

Automobile manufacturers are using Pulse Width Modulation to regulate everything from fan speed, pump pressure, and light output (to name a few). Is there a passive device (diode, cap, etc) that can be wired in line to block completely a PWM signal while allowing normal voltage (if on the wire at other times) to pass?

Thanks, Carl.

[CECAA850]

I found THIS

[DrWho]

What are you trying to accomplish?

[mustang guy]

This is an interesting question. From what I can tell, use of a low pass filter is the passive way, and a PWM to DC converter is the other. I dabble in RC, and the servos all get PWM to DC conversion via my Futaba controllers on the planes. Using the low pass approach introduces delay as I understand it.

I am actually reading up on this as you are trying to figure it out. I have a specific problem with a car which was blowing out the JL converters, which were kinda expensive. Now I have disabled the day/night function of the GPS, and it is allways bright. A big pain!

[CECAA850]

What are you trying to accomplish?

I'm trying to disable a daytime running light. Why is a whole different story for another thread. Here's how the system works. There's 1 power wire that goes to the low beam bulb. A module puts battery voltage on that wire when low beams are called for. During the daylight hours, that same wire has a PWM signal to dim the bulbs. I know a way to make that wire have battery voltage at all times if I want, but I haven't figured out a way to eliminate the PWM voltage while still letting normal battery voltage through.

[CECAA850]

This is an interesting question. From what I can tell, use of a low pass filter is the passive way, and a PWM to DC converter is the other. I dabble in RC, and the servos all get PWM to DC conversion via my Futaba controllers on the planes. Using the low pass approach introduces delay as I understand it.

I believe the low pass filter will allow normal voltage through but not eliminate the PWM voltage. Bear in mind (you and DrWho) that I have a VERY LIMITED knowledge of circuitry when you respond, thanks.

[Guest " "]

Interesting article. Based on the tec details, the passive filter that comes to mind is the standard auto noise filter that comes in the clear tube. Inside the tube is an inductor, capacitor, a diode wired to the base of a power transistor . Noisy dc goes into the collector and clean dc exits the emitter. Basic t filter. Cost between 10 and 25 bucks depending on the current requirements. Radio shack sells them, so does parts express. eBay had them as well. But once you kill the pwm signal by filtering it out, then you need to install relays and switches to turn the device on. The t filter would be great to protect a non pwm device from pwm signals with out the need for relays and switches if that is what you are trying to do.

[WMcD]

If I understand correctly - - which is not always the case.

The article is okay as far as it goes but the filter description is a bit misleading.

Look at it from the transmitting end in terms of power. We can have the square pulses (a square wave) on all the time which is to say the first pulse doesn't turn off until the next one starts. That is full power sent over time.

If we want half power sent, we reduce the length of the pulse so that the power is turned on only half the time. I'd call the the duty cycle or 50%.

Why is this a good way to do things? The answer is that we could set up full or constant on power and then put in a dropping resistor in series. Because of the loss or voltage drop across the resistor, the load will get half the power and and the dropping resistor gets the other half. If we have an incandescent bulb, it glows less brightly. The problem is that we are wasting the power in the dropping resistor, and it gets hot. There is a voltage drop across the resistor and current through it, at the same time.

Let's suppose you have an ordinary light bulb circuit at home and a wall switch. The wall switch never gets hot. That is because when it is open, there is a voltage across it, but no current through it. Therefore no power consumed. When the the switch is closed, there is no voltage drop across it because it is like a good wire. There is current through it, though.

These three situations are described by our familiar concept of Power consumed = Volts (droped across a device) x Current (through the device). If you have zero current or zero voltage, there is no power consumed. Ergo the wall switch never wastes power. This is an important concept in that it shows a pure switch doesn't consume power and doesn't get hot.

The next concept is that if you wanted to dim the incondescent lamp, and you had very very fast fingers, you could operate the wall switch.on and off very quickly. But nobody is that fast. Because we're slow, the lamp just blinks.

But a transistor or SCR switch can act very quickly.

One example is an SCR or Triac dimmer used as a replacement for light switches in AC service. It is fed the sine alternating voltage but does not turn on until sometime after the sine wave has started, or at least started from the sine wave's zero point. How long the delay is established by dimmer knob. So, in a way, this is similar to the square wave "width" theory. Please note that to the extent the sine wave is modified in time, the max voltage stays the same.

Now we have to look at the the receiving end device. There is always some sort of integration. This means accumlating and, often, draining off.

In an incandescent lamp, the filament takes some time to heat up and to cool. Lets stick with the square wave. The lamp gets hit with full voltage and draws full current (the sending switch is closed). But then the voltage and resulting current drop to zero when the switch causing the pulse opens. The filament cools (light output goes down and the filament goes from white hot (maybe it never got there) to red hot. Then it gets hit with another pulse and warms up a bit.

Overall, this is integration. In a way, the statistical analysis of stock market prices called a moving average is the same. By integrating over limited chunks of time, we are removing short term spikes and dips by averaging them. We're removing high frequencies.

But what to do when we really want a d.c. level at the receiving end to drive some device with half power -- which is to say, reduced voltage? Less than the peaks of the square wave.

Maybe I'm getting too simplistic here . . but:.

Suppose we had a 12 volt charger and a rechargable battery with some load attached which we only want to feed 6 volts.. The rechargable battery starts off being totally discharged. We could charge up the battery unit it comes up to 6 volts and remove the charger. Then let ithe load draw the battery voltage down to 5.9 volts. Then connect the charger and let battery voltage go back up to 6.0 volts. Over and over.

This is just a concept and in fact batteries don't like being charged and discharged to half voltage ratings, because of the electro-chemistry involved.

So we have to find an electrical device which accomplishes the same thing. That is a capacitor.

The circuit shown uses a capacitor in place of the rechargable battery described above. The ancients decribed capacitors as an accumlator going back to the days of Leyden jars were used to accumulate static charges - Leyden jars are capacitors. How much charge they could store refered to their "capacity") The reason for the resistor is more difficult to explain.

But, again, the capacitor is charged up when voltage and current are applied (the resistor limits the current), and then discharges though the load a little bit. Then gets hit with the next pulse. It integrates out the changes (max and zero) of the applied square wave. If you have a Leyden jar with a lot of capacity, the load does not discharge it very much or the load is not relatively large, and it is going from 6.0 volts down to 5.99999 volts. But it is about half of the max voltage on the pulses.

In a way, what we have is a form of digital sending encoding and digital receiving - at least on the transmission line we only have 1s and 0s of varying duration.. On the encoding end, we modulate the width of the pulses according to the power to be sent and delivered. On the receiving end, the voltage of the capacitor, is proportional to the width of the pulses.

But I've digressed quite a bit. Smile. I'm really trying to show that many concepts and words add up to something understandable.

I think in your situation, the sending circuit controlled by the car's computer, is sending pulses, let us say 100 percent "on" for half the time,for "dim" and very long duration pulses for full on. The latter is to say, as above, pulses so long that they never turn off.

Therefore, no, there is no circuit to do what you're asking. This is because you can't manipulate on the receiving end, what the transmitter is sending.

Let me add that I understand that automotive systems are moving to almost a situation of ethernet contol of accessories. So maybe in some situations a cable bundle is sending power and also digital information of what to do with it. That does not seem to be the case here, though.

WMcD

[JJkizak]

My first assumption on the headlights being on all the time was that the oil companies won and we didn't. I still have that assumption.

JJK

[CECAA850]

William, if I read you post correctly, the answer is "no" there is no such device. I can leave it "as us" or use a capacitor to, in essence, have the device full on, all the time.

I believe I'll look into tricking the computer by way of manipulating the two inputs (park brake and sunlight sensor) that tell the computer to turn on the DRL's. The problem that will be that my low beams will be on and I don't necessarily want that. I can think of a very easy way to accomplish everything I want using manual switching but I'm trying to avoid that scenario. Thanks for all the input and explanations.

Carl

[Quiet_Hollow]

What's the year and make?

[CECAA850]

What's the year and make?

'12 Chevy Cruze. I think I found a work around. The DRL's can be turned off by toggling the headlamp switch to the off position. I'm ASSUMING it sends a ground to the BCM, the BCM sees the ground pulse on that particular wire and shuts them off for that key cycle. I was thinking of installing a relay that uses the starter input to energize the relay and have the relay output a ground to tie into the BCM wire that receives the DRL "off" signal. Still in the thought process though.

[T.H.E. Droid]

Kind of crude, but why not simply run the low beam wire(s) back through a switch? (or a switched relay if you want to avoid loss in the extra wires)

[russ69]

My GTO had a fuse you could pull, it just killed the daytime lights and was labeled as such if I remember correctly.

[CECAA850]

Kind of crude, but why not simply run the low beam wire(s) back through a switch? (or a switched relay if you want to avoid loss in the extra wires)

The thought had crossed my mind but I was looking for something that involved no user input.

[CECAA850]

My GTO had a fuse you could pull, it just killed the daytime lights and was labeled as such if I remember correctly.

Yes, many were fused that way. The good old days.......................

[DrWho]

Kind of crude, but why not simply run the low beam wire(s) back through a switch? (or a switched relay if you want to avoid loss in the extra wires)

The thought had crossed my mind but I was looking for something that involved no user input.

Going with the extra relay concept, it sounds like all you really need to do is be able to identify whether the headlamp signal is AC or DC...

If you DC block the signal (with a series cap), then only the AC component will pass through. There are various ways one could detect the presence of an AC signal - the way shown in the attached schematic rectifies this AC signal into a DC voltage, and then a comparator is used to detect this level and drive a relay.

When the headlamp is turned fully on, or fully off, the voltage at Vout1 will be high. Since this is hooked up to the negative side of the relay control, no voltage is present across the relay control pins (since the positive side is connected to the battery).

When the headlamp signal is doing it's PWM thing, the AC component sneaks through C1 and gets rectified by the two diodes. C2 holds up this DC voltage and R2 sets the time constant at which the cap discharges (this will be the time it takes to go between running lights and turning the lights off completely or full on). R1 and R6 are a voltage divder used to compare against the rectified voltage. C5 is a filter cap and should be placed close to the input of the comparator. (C2 should be close to the comparator too). R9 is a current limiting resistor for driving the relay. You should also add a 100uF cap across the power pins for the comparator (just to make sure power transients don't hurt the part).

The only tricky part is finding a comparator that can run on "12V" and drive enough current to turn on the relay. Car power is generally quite nasty, so you should use a part rated for >40V. Linear Tech should have something.

Have you measured the voltage range and frequency of the PWM signal? That will be very important for making sure this circuit will work. I'm sure there are more elegant solutions, but this is intentionally a bit overkill since the rest of your car is a bit of an unknown. The cost to you also isn't as big of a deal as building several million of these circuits...

Btw, insert a million caveats about not guaranteeing anything on this. It was an interesting problem so I thought I'd sketch things up and simulate it.

Also, should you desire to disable the function of the circuit, you could add a switch across R2 to short out the input to the comparator.

[superedge88]

A simple relay is all that is needed then, don't over complicate with trying to"filter"out signal.

What are you trying to accomplish?

I'm trying to disable a daytime running light. Why is a whole different story for another thread. Here's how the system works. There's 1 power wire that goes to the low beam bulb. A module puts battery voltage on that wire when low beams are called for. During the daylight hours, that same wire has a PWM signal to dim the bulbs. I know a way to make that wire have battery voltage at all times if I want, but I haven't figured out a way to eliminate the PWM voltage while still letting normal battery voltage through.

[JJkizak]

Set up an alternator with a handcrank on a workbench with a headlamp circuit and a switch. Rotate the crank as fast as you can so the meter shows 14.6 volts then turn on the headlight switch. You will break your wrist as soon as the switch is turned on. That's the answer "why."

JJK

[CECAA850]

A simple relay is all that is needed then, don't over complicate with trying to"filter"out signal.

What are you trying to accomplish?

I'm trying to disable a daytime running light. Why is a whole different story for another thread. Here's how the system works. There's 1 power wire that goes to the low beam bulb. A module puts battery voltage on that wire when low beams are called for. During the daylight hours, that same wire has a PWM signal to dim the bulbs. I know a way to make that wire have battery voltage at all times if I want, but I haven't figured out a way to eliminate the PWM voltage while still letting normal battery voltage through.

I'm not following.