ou probably have a general under-standing of how current flows in acircuit. But did you know that bybrushing up your grasp of how cur-

rent really works, you could diagnose duty-cycledcircuits such as fuel injectors and solve a variety ofelectrical and solenoid-related puzzles? By using alittle logic, you can uncover the roots of bafflingsymptoms and failures, without resorting to parts-swapping—which is expensive, time-consuming,and boils down to an admission of ignorance.Parts swapping is also of limited utility. Sure, you

may know that replacing the noise-suppressionrelay on LH 2.4 Volvos usually fixes a no-start con-dition; but what happens if the system is a Motronic4.3 on a 1996 BMW? What if you’re not dealing witha defective component, but an electrical connectionwith high resistance instead? It’s better to outsmart,rather than trying to outguess it.In any pulsed current-flow system, conventional

methods of measuring voltage drops leave much tobe desired. In a fuel-injection circuit, voltage dropsare averaged to a lower DC value, because the peri-od when the injector is OFF—and no currentflows—is greater than the “ON” period, when cur-rent is flowing. This makes finding high-resistanceconnections difficult in duty-cycled circuits.A graphical representation of the actual current

flow in an operating circuit is a better method for

diagnosing the circuit’s operational characteristics.In a steady-state current-flow circuit, as in fanmotors, starter circuits, or lamp circuits, measuringvoltage drops across each connection in the circuitcan reveal high resistance.I will show you how this works by taking you

through a current-flow analysis of the problem men-tioned above: a defective noise-suppression relay ona Volvo with an LH 2.4 system. This relay is rou-tinely replaced by experienced Volvo technicians inthe event of a no-start condition when there is sparkand dim current pulses through a “noid” lightattached to a injector connector. I’ll show you howan analytical approach to waveform analysis canhelp you diagnose circuits—without “inside knowl-edge” on the pitfalls of any particular system.I used a Fluke 80i-110s current probe to measure

the injector current waveforms on the LH 2.4 sys-tem. On a late model BMW the probe was used tofind excessive current in a shorted circuit with sev-eral branches (only one of which was problematic).The Fluke current probe lets you monitor currentflow in a non-invasive manner, which saves a greatdeal of time over the old cutting-and-splicingmethod. The “patients” were a 1989 Volvo 740Turbo and a 1985 535i BMW with a short circuit.I would like to thank Alberto at Ital Mechanica in

Huntington Beach California for presenting me witha problem Volvo that I was able to use for analysis.

42 November 1997

Y

EI x R

CurrentEvents

43November 1997

Dropping ResistorsAny system with low ohm injectors needs current

regulation to prevent overheating of the coil of wireinside the injector. An easy way to do this is byadding resistance in series with the injector. Thislimits the voltage which is available to the injectorby “dropping” it across the power resistor; hence theterm “dropping resistors.” This power resistor dropsvoltage and limits current flow at the same time.Resistance is good when it’s part of the circuitdesign, but when it’s not, there is trouble. Volvo uses dropping resistors on LH 2.0-2.2 and

LH 2.4 systems with turbocharged B230FT engines.These turbocharged engines utilize low ohm injec-tors (3 ). All non-turbo Volvo engines feature highimpedance (16 ) injectors, which do not usedropping resistors.

Magnetic Field StrengthThe strength of a magnetic field surrounding a

solenoid depends upon the peak current throughthe coil and the number of turns of wire per unitlength. Since the number of turns remains thesame, only current flow is the variable which wewill deal with in the automotive industry. A loss ofcurrent can result from either high resistance in thecircuit or low battery voltage. Battery voltage iscritical during cranking. In order to create a magnetic flux density large

enough to open the pintle of the injector, there mustbe sufficient battery voltage without any unintendedvoltage drops in the circuit. Voltage drops comefrom additional resistance introduced into the cir-cuit. Any resistance in a circuit will produce a volt-age drop and thus limit current flow. Kirchoff’sSecond Law states the following: “The sum of indi-vidual voltage drops around a complete series cir-cuit is equal to the applied voltage.” With the addi-tional resistance caused by a poor internal connec-tion to the Noise Suppression Relay (Figure 3), wehave an additional voltage drop in the injector cir-cuit. Figure 1 illustrates a typical current path forVolvos with B230FT engines (this number identifiesthe engine as being turbocharged).

Calculating Current FlowWe begin by looking over the electrical schematics

for this Volvo and taking ohm readings from the twomajor components in the system, the dropping-resis-tor and the injector. With these readings we can cal-culate current. (Yes, there’s another device here thatadds resistance to the injector circuit: the transistor.A DMM can’t calculate transistor resistance, howev-er, so we’ll find this number another way.Meanwhile I’ll give the transistor a value of 1 sothat we can continue with Ohm’s Law and calculatethe current flow to the injector circuit.

Figure 1 shows the following resistance: 6+3+1=10 .With a voltage source of 13.5v and Ohm’s Law, we cancalculate current flow: I=E/R, I=13.5/9, I=1.5 amps.Figure 2’s waveform indicates that the actual currentflow is close to the calculated current flow of aworking system.

No Fuel FlowThe no-start condition on the Volvo resulted

from a high resistance contact between the polesof the relay (Figure 3). The engine cranked over; ithad spark, injection pulses, and fuel pressure; butno fuel flow. Why?Because a defectiverelay limits currentflow. And high resis-tance added to theinjector circuit makesa no-start. Notice howthe inductive kick-back voltage is loweron the no-start condi-tion. The lower thecurrent flow throughthe injector, the weak-er the electromagneticfield surrounding thecoil of wire, and thelower the kickback.

B+Injector

Dropping Resistor

Noise Suppression Relay, no resistance

18

6Ω

3Ω

Coil Resistance

1.3 Amp's current flow

1Ω

B+=13.5V (idle V) R t= 6+3+1Ω's R t=10Ω's It= 1.35A

I = E/R

E I x R

Pintle is open Fuel flows

Figure 2

Figure1 LH 2.4 Injector Current And Fuel Flow

60V. Peak

Sufficient current flowthrough oneinjector

1.5

0

1989 Volvo 740 Turbo LH 2.4Strong Injector Current Flow

Here we see voltage and current flowthrough one injector during starting.

This much current is necessary to openthe pintle and allow fuel to flow.

45

Reduced CurrentFlowLet’s look at what’s

happening with cur-rent under a no-startcondition due to alack of fuel flow. Byreading the currentwaveform in Figure 4,we see that the peakcurrent in this circuitis only 0.5 amp andthe inductive kick-back voltage is now40 volt peak. Thestrength of a magnet-ic field surroundinga solenoid dependson the peak currentthrough the coil.Figure 4 shows howpeak inductive-kick-back voltage hasbeen reduced by thehigh-resistance con-tact in the relay.With reduced cur-rent flow from a fee-ble magnetic fieldcomes less inductive-kickbackvoltage. Even though I had aninductive-kickback voltage, theweakness of the field preventedfuel flow—there simply wasn’tenough juice to open the pintleinside the fuel injector.Figure 3 shows how I used

Ohm’s Law to calculate the cir-cuit’s total resistance and theresistance between relay con-tacts. A DC RMS voltmeter couldbe used in pulsed DC circuits tomeasure the voltage drop acrossthe relay. To do this, peak-voltagedetection must be on, since thisis not a steady-state current mea-surement. Without peak detec-tion, the RMS value of themetered-voltage drop across therelay would be much less.On-time characteristics may

differ, depending on the transis-tor, so the voltage drop is not asimportant as the current flowneeded to make the circuit oper-ational. A rule of thumb is thatVolvo LH 2.4 Turbo injector cir-cuits need at least 0.7 to 1.0 amp

current flow through the injectorto make them open. Any less willfail to inject fuel into the cylin-ders. Even if you are workingwith high resistance injectors of14-16 impedance, the same 0.7to 1.0 amp rule will apply.

—By Lester Bravek

B+Injector

Dropping Resistor

Noise Suppression Relay, high resistance

18

8Ω

6Ω

3Ω

Coil Resistance

1/2 Amp current flow

1Ω

B+=10.5V (cranking V) R t=8+6+3+1 R t=18 It=.58A

I = E/R

E I x R

Pintle remains closed No fuel flow

Figure 3 LH 2.4 Injector Current Flow—No Fuel

Figure 4

40V. Peak

Sufficient current flowthrough oneinjector

1989 Volvo 740 Turbo LH 2.4Weak Injector Current Flow

A .5 amp current flow through each injector could not create a magnetic

flux density strong enough to open the injector pintles, so there was no fuel flow.

CurrentEvents

1.5

0

46 November 1997

1The noise suppression relay contacts looked nor-mal, yet the relay was adding high resistance to thefuel injector circuit. These relays may look normal,but an intermittent case of high resistance connec-tion can wreak havoc with the injector circuit. Don’ttake chances with a marginal relay.

To measure current flow through the fuel injectors,I used the dropping resistor located on the left handside wheel well, near the front of the vehicle. If youare unfamiliar with component placement onVolvos, this appears to be the standard location onmodels that are equipped with dropping resistors.

Placing a current probe around each individual wirein a splice point will show you which of the severalwires has the most current flowing through it. Thisspeeds up the fault-finding considerably. This BMWcircuit had two splice points inside the instrumentcluster area similar to the one pictured.

After entering the passenger compartment through abulkhead connector, the circuit split into twointerior splice points behind the instrument clus-ter. I could have saved time by consulting a BMWfactory electrical troubleshooting manual, whichgives splice-point locations inside the vehicle.

Current flow waveforms give a visual representation,as well as a numerical value, to the current in thecircuit. With this information, I was able to jumpfrom wire to wire looking for the faulty circuit. Evenwith this additional help, the whole process stilltook two full hours.

Finding the faulty wire with the current probe waseasy. The hard part was tracing it back through thebundled wiring harness. Following the wire backand forth behind the instrument cluster from onesplice to another was time-consuming. I eventuallyfound the fault in the shifter area.

2

3 4

5 6

CurrentEvents