Tech Note 011:

Acquiring vacuum signals with your DSO
Submitted by Bill Lakow

Find the largest and the most central vacuum port in the intake manifold and connect your transducer with a short hose.

The transducer converts the positive and negative pressures into voltage values that we can displayed on the lab scope.

Figure 1 shows a random display of vacuum signals. This is not very helpful in diagnostics because we do not know which cylinder created each signal. If you use the PDA50A lab scope with Quick Ignition, you can identify the pulse of each cylinder by using the sync probe to select a cylinder to reference from.

In the second two examples, the firing order is 1-6-5-4-3-2. The vacuum signal occurs before the ignition signal.

In Figure 2 cylinder #4 was used as the trigger so that the vac signals are displayed in the firing order. The PDA ignition scope provides the scope channel. This function provides secondary ignition and lab scope display at the same time and the ignition and lab scopes are both triggered from the ignition sync signal. The firing order 1-6-5-4-3-2 and the vacuum signals are 4-3-2-1-6-5.

The intent of these examples is to help you acquire vacuum signals for analysis. Once the signal is recorded and identified, the diagnostic process can begin.

When analyzing ignition parade patterns, we look for consistent and repeatable shapes that are within acceptable ranges.

Figure 4 The ignition patterns appear to be good, but the vacuum signal does not have a consistent and repeatable pattern.

A cylinder balance test indicated that two cylinders were not working. A physical compression test ( cranking ) indicated that all cylinders were equal and had acceptable compression. At cranking RPM everything seemed ok, but at almost 1900 RPM the vacuum signals are showing a problem. Two cylinders are not getting enough air and fuel to run with the other six.

Figure 5 illustrates another quick way to view vacuum signals on cars that have analog MAP sensors. Figure 5 shows vacuum pulses captured with an AC Pass Filter probe and the ignition sync probe. The firing order is 1-3-4-2 and the sync probe is connected to cylinder #4 so the vacuum signals are displayed in the firing order. Notice the third signal from the left, cylinder #4. This cylinder had low contribution on a cylinder balance test and had low reading on a relative compression test and the vacuum signal is lower than the other cylinders.

Related AES Links:

Fluke PV-350

Mitsubishi Coil Failure Analysis with a Low Amp Current Probe
Submitted by Bill Lakow

Figure 1 coil current tested while cranking during the no start condition. The rapid rise in current indicates that the ignition coil is shorted internally. The failed coil was replaced less than two months ago.

Figure 2 displays the coil current with the new replacement coil. The current signal appears to be good except that the current level is abnormally high and the ignition coil is extremely hot to the touch. The ignition wiring was inspected and the factory ignition coil resistor had been by-passed. The wiring was repositioned to its proper place and the resistor is now limiting the current flow to the ignition coil and ignition module. The high current flow and increased heat in the ignition coil may have contributed to its early failure.

Figure 3 shows a good ignition  coil current trace at an  acceptable current level. The ignition coil is not over heating now that the current flow is reduced.

Using Current probe for the initial diagnostic. See current go straight up = low resistance.

Current probe watching the work load.

Typical Saturation Injector signal
Submitted by Bill Lakow

Pay close attention to every detail of the injector trace. The voltage level and inductive kick can be measured with the cursors available in the utilities menu in the Interro PDA.

Watch for a slight increase in voltage during the ON time, when the circuit is grounded. The inductive kick should be two to three times greater than the supply voltage.

Some systems limit the inductive kick through the PCM when the injector is turned off.

The injector signal may have all the critical dimensions that indicate that it is opening, yet there could be a reduction in or no fuel flowing through the component.

DUAL PLUG IGNITION: 1991 Ford Ranger 2.3L EFI 

Comparison primary current waveforms of the left and right side coil packs.

The DIS module in this system uses four power transistors one for each primary winding. Figure #1 testing primary current with low amp prob. The right side coil pack in this system is responsible for firing the air/fuel mixture that makes the power. Red trace, #1 primary winding. Blue trace, #2 primary winding. Both winding and power transistors are good in this coil pack. The primary windings are pulling 5.2 amps probe conversion is 100mv=1amp.

Figure #2 left side coil pack. This side is responsible for emissions reduction. Blue trace, #1 primary winding shows high resistance problem leading edge. Trailing edge looks normal. Red trace, #2 primary winding leading edge looks normal trailing edge is the one with the problem. I suspect that the transistor is partially shorted. The NPN transistor completes the circuit by grounding the primary winding and charging the coil. As the coil charges the magnetic field builds, the transistor then opens the circuit, the field collapses rapidly, and induces the high voltage in the secondary winding plug fires. Looking at the trailing edge you can see a glitch and a sloping down to zero amps. This glitch is setting trouble code #18 in memory erratic input to ECA. This code in memory on dual plug systems puts the ECA into FMEM or limp in mode.

An Electrical Question
Submitted by Jorge Menchu, AES

When connecting a light bulb to positive and negative terminals of vehicle battery, the light lights up, and stays lit up. Yet when connecting a 10 ohm resistor to the positive and negative terminals of battery the resistor wants to smoke and burn up.

The resistance of the bulb is approx 10 ohms.

Why is the the 10 ohm resistor burning up instead of carrying the load like the light does?

Jorge's Reply:
The problem is the physical size of the resistor. Get a larger 10 ohm resistor. The small one can not handle the heat created by the current.

The light for example gets very hot. The filiment gets so hot that it glows.

A small resistor might get so hot that it burns. So, in the case of the resistor - get larger. Here is how you figure the size.
First calculate wattage w = v * a
w = 12 * 1.2
w = 14.4
Next, go to the electronics store and get a 10 ohm resistor that has a rating of a minimum 15 watts. It will be bigger than one that is rated at 1/4 watt!
(Read below for more info about this missing decimal.)

Jorge 

Tony asks: YOU CANNOT MEASURE THE RESISTANCE OF A LIGHT BULB, BECAUSE WHEN POWERED UP IT HEATS UP AND THE RESISTANCE CHANGES. YOU MUST FIGURE IT OUT WITH OHMS LAW. IT WILL NOT HAVE THE SAME RESISTANCE, IT WILL BE MUCH MORE.

Jorge Responds: This is correct!

Tony continues: THIS IS WHY THE 10 AMP RESISTOR SMOKES WHEN HOOKED TO A BATTERY.

Jorge Responds: This statement is not correct. First, the question being asked is "why does the 10 ohm resistor (that they selected) burn up when put across the battery?"

The question is not "what is the best resistor value to simulate the resistance of the light?"

The answer to the question being asked is found in the watt value of the resistor that is burning up. One of the basic rules of electricity is - anytime current flows through resistance energy is released as heat. If the power source is 12 volts and the resistance is 10 ohms then there is 1.2 amps. 12 volts at 1.2 amps offers a watt (power) rating 14.4. 14.4 represent the amount of energy that the resistor must dissipate. If the resistor cannot dissipate this energy it will burn up. So, to prevent a 10-ohm resistor from burning up you must get one that has a watt rating of greater than 14 watts.

Tony continues: YOU MUST KNOW THE VOLTAGE AND THE AMPERAGE TO FIGURE OUT THE RESISTANCE.

Jorge Responds: If your goal were to calculate the resistance of the light bulb when it is lit up then you would want to use a current probe. Take the current and plug that into an ohms law equation  with the voltage as you state. The answer will be higher than the resistance measured when the bulb is cold.

Now, if you notice my calculations above are different then those listed on the site. On the site page I forgot to include the "." in the amp value so that needs to be fixed!

Tony continues: OK, what if you would figure the resistance of the light bulb with it on, and get that ohm resistor and hook it to the battery. Would it smoke

Jorge Responds:  If you figured the resistance of the light when it was warm and got an equal value resistor and then connected the resistor directly across a 12 volt battery would it smoke?

That depends on the physical size of the resistor. For example, if you go to Radio Shack and bought a 1/4 watt 15ohm resistor, you would find it to be physically small and might burn up. A resistor of 1/4 watt used on a 12 volt circuit can only handle 20mA before it exceedes the 1/4 watt rating. At 15 ohms and 12 volts we need a (12volts/15ohms=0.8amps....0.8a * 12v=9.6 watt) 9.6watt or greater.

To confirm this go to your local electronics store and get a 10 (or so) ohm resistor in 1/8 watt, 1/4 watt, and 10 watt versions.

Jorge