Fuel Injector Armature Circuit
1. Components list:
| Parts | General Specs. | Quantity | Symbol |
| Resistor R13 | 1K Ω | 1 |  |
| Resistor R14 | 470 Ω | 1 | |
| Resistor R15 | 470 Ω | 1 | |
| Resistor R16 | 1K Ω | 1 | |
| Transistor | NPN BC 547, 0.5W @ 25oC | 2 |  |
| LED 1 | Yellow, Vf=2.1V @ 20mA | 1 |  |
| LED 2 | Ultra Red, Vf= 1.8V @ 20mA | 1 | |
| Function Generator | 5V, 3Mhz, Signals Generator | 1 | PWM0-1 PWM1-1 |
| Voltage Supply | 12V | 1 | 12V |
All specs are based on manufacturer’s data sheets.
BC 547 NPN Detailed Specification
Note: Paying attentions to the transistors is most important for inspecting the circuit’s technical behavior, since transistor is (in this case) the most complicated, advanced technological component in terms of structure and operation.
a. Absolute Maximum Ratings:
b. Electrical Characteristics:
2. Setup and Calculations:
a. Setup and Technical Explanation
Wiring diagram:
HOW THE CIRCUIT WORKS:
This is the simulated wiring circuit for Fuel Injector’s LED Armature Circuit.
The 12V supply represents the voltage supply from the battery to 2 LEDs and supposedly to the injector’s armature gates in amplified current. Unfortunately, this is only a diagnostic simulation circuit which armatures the operation of the LEDs that detects the Injectors Operations. Before the LEDs, there are resistors in size of 470Ω in order to restrict the amount of current going through the LEDs. This is because LEDs are diodes and they often have no defined amount of resistance in it, hence the 12V with no flow restriction will eventually exceed the LED’s forwards voltage limit, maximum current and power dissipation limit. And this leads to destruction of the 2 LEDs, though we might see them lit up really BRIGHT…for only a short period of time and then they will be gone forever.
From the cathodes of both LEDs their branches of circuit are basically identical, they connect to the collector terminal of the transistor.
From with the 5V frequent flashing signal to terminal 3 parallel with 4, 5v will be frequently flashed to each base resistor and the base of the transistor itself. And regardless of all the erratic readings that we might eventually have from all the locations created by the flashing device, 5V through a big resistor of 1kΩ will eventually bring through the base a really small amount of current(Ib). That current makes the base more positive than the emitter and plus the designated forward direction is base-emitter, hence the base-emitter will eventually behave like a regular rectifying diode.
In diode’s theory of operation, when there is a small current flow of holes from base to all the electrons from the emitter, it eliminates a small portion of the depleted region between the base and the emitter. Also, from the base to collector there is also a forward bias, and when the small current through base-emitter is on it opens “another gate” between the collector and the base.
But since there is a big potential difference accumulated at collector already, this will allow big potential for a big amount of amperage. And when activated, connection is made from the collector along the base to the emitter, because the base is not very positive to the collector now, the collector has a huge 12V standing behind PUSHING it!!! But that 12V isn’t going to break the reversed collector-base diode so it’s going to push towards the emitter because collector is WAY more positive than the emitter. As N-N, they are the same type of layer hence will act mostly as a conductor when in active or saturated region (check data sheet above).
So in summary, the small current from the generator will turn on the base-emitter of the transistor, allowing the amplified current from the REAL supply voltage to go from collector to emitter, hence turning on all the working components like LEDs on its way. I say “amplified” because that’s what the transistor does. The more base current is put in, the less the depleted region which rationally acts like resistance, hence less voltage’s effort and more current from the supply is flown through. And in the end when all the amplifications are done and the LEDs are all lit up, the 2 emitters go to earth making a complete circuit.
b. Theoretical Calculations:
Starts @ terminal 1 supply Voltage 12V maintained, equally through R14 + LED1(YELLOW) and R15 + LED2(RED) ( “+” is in series). LED1 Cathode + Collector(C1) and LED2 + Collector(C2)
Starts @ Signal Generator of 5V through both terminals 3 & 4 (PWM0-1 & PWM1-1) parallel. 5V through R13 + Base(B1); and R16 + Base(B2). Both Bases go to ground which is terminal 2.
Readings to be collected:
+ Voltage Drops across: R13; R14; R15; R16; Vce1; Vce2; Vbe1; Vbe2
+ Current through: collector-emitter ( Ic1 & Ic2); Base-emitter(Ib 1 & 2).
Laws applied: Ohm’s Law in Series and Parallel circuit.
R13 and R16:
R13=R16=1kΩ; Vsignal=5V; Vbe= 0.66V(average typical voltage)=> VR13 or R16= Vsignal - Vbe
VR13 or R16= 5 – 0.66=4.34V
Ibase= VR13/R13= 4.34/1000= 0.00434 A= 4.34mA
The result is supposedly the same for Ibase2 and VR16 because we R13=R16 and the two transistors are the same.
R14 and R15:
R14=R15=470Ω; Vsupply=12V; Vce is negligible.
VLED1= 1.8V => VR14= VS-VLED1= 12-1.8=10.2V => IR14 or c1=VR14/R14=10.2/470=0.0217A*
VLED2=2.1V => VR15= Vs-VLED2=12-2.1=9.9V => IR15 or c2=VR15/R15= 9.9/470=0.021A*
| Component Elements | R13 | R14 | R15 | R16 | CE1 | CE2 | BE1 | BE2 |
| Voltage Drop(V) | 4.34 | 10.2 | 9.9 | 4.34 | NEG | NEG | 0.66 | 0.66 |
| Current(mA) | 4.34 | 21.7* | 21.0* | 4.34 | n/a | n/a | 4.34 | 4.34 |
As when the base current switch on the collector’s current the collector-emitter junction will virtually act as a conductor. Vce is normally negligible when the transistor is fully open(saturation). So base on the datasheet, the maximum saturation Vce and the knee Vce exceed 0.6V and typically around 0.22-0.25V and minimum is 0.07V, so it’s negligible.
*The values are for comparison purpose only I would not expect these with the actual readings. Because what’s actually coming out of the collector is mainly determined by the saturation of the gate, or current gain i.e. electrical characteristic between collector and emitter.
Adding resistors is the most important step of altering the amount of amperage both signaling and amplified. So the numbers I have added are estimated based on experience in order to bring the transistor’s operation to near saturation region or better yet in the saturation region. That why the Base resistor is not really large in term of the current capacity of this particular NPN BE 547 transistor( about 100mA). And with the current gain of 90-100, a 1kΩ which brings to the base about 4mA will make a HUGE amplification as I would expect the transistor is in the saturated region.
Test procedure:
1. Wiring it up!
This is the schematic created by computer software:
ACTUAL!!!
2. Testing!!!
Base on the recorded reading which is more accurate and practical than hypothesis figures, some recalculations must be done:
VR14=12-1.785=10.125V Ic=10.125/469=0.0218A relevant to*
But VR14 actual= 9.08 => Icactual= 9.08/469=0.019A (Justified)
VR15=12-1.846=10.154V Ic=10.154/469=0.0219A*
But VR15 actual=5.93 => Icactual=5.93/469=0.0126 ~ 0.013A (justified)
Similarly:
VR13=4.94 Ib=4.94/977=0.005 (justified)
VR16=4.93 Ib=4.93/977=0.005 (actual is 0.002)
So there is some unlikely reading in the current that flows through R16 and Base-emitter 2. This could be the use of different LEDs on the same resistors and transistors, this might alter the amperage for the base. I.e. The yellow has a maximum voltage drop of 2.2V, but it is recorded as 5.3V at 0.013A and 12V supply, and the peak current is 30mA with typical of 20mA. Hence this partially explains for some unlikely high voltage drop.
(CHINA YOUND SUN LED TECHNOLOGY CO., LTD)
On the other hand, a supply of 12V is also a reason for “higher than specs” Voltage drop across both the Red and Yellow LED, but still doesn’t explain why the Vd across LED2 is significantly higher than its 2.2V, while the Red one drop 2.2 just slightly higher than its 1.8V. Possibly some faulty inside the yellow LED as well, or more resistance in the red LED (note that the Yellow LED lights significantly than the Red one). Also, in a normal rectifying diode, when the Supply voltage e.g. 5V is being varied to 10V or 15V, the Vd across the diode also increases slightly, because the total current increases and hence needs a bigger PUSH!
IMPROVEMENT
It’s easier to identify and distinguish if using different LEDs color but this results in unlikely voltage drops reading and Current flow, because LEDs with same sizes and different colors have different electrical specs.
Depend on what the objective is, the circuit should be set to same LEDs so it’s more reliable overall and any fault is easier to detect and rectify. Also, different sizes of base and collector resistors can be applied, to achieve maximum current gain and do not need to be in saturation.
Good explainations
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