Simple fuel injector ciruit.

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; Vce_­1; Vce­₂; Vbe_­1; Vbe₂

 + Current through: collector-emitter ( Ic_­1 & Ic₂); Base-emitter(Ib 1 & 2).

Laws applied: Ohm’s Law in Series and Parallel circuit.

R13 and R16:

R13=R16=1kΩ; V_signal=5V; V­_be= 0.66V(average typical voltage)=> V_{R13 or R16}= V_signal - V_be

                                                                                                             V_{R13 or R16}= 5 – 0.66=4.34V

I_base= V_R13/R13= 4.34/1000= 0.00434 A= 4.34mA

The result is supposedly the same for I_base2 and V_R16 because we R13=R16 and the two transistors are the same.

R14 and R15:

R14=R15=470Ω; V_supply=12V; Vce is negligible.

V_LED1= 1.8V => V­_R14= V_S-V_LED1= 12-1.8=10.2V => I_{R14 or c1}=V_R14/R14=10.2/470=0.0217A*

 V_LED2=2.1V => V_R15= V_s-V_LED2=12-2.1=9.9V => I_{R15 or c2}=V_R15/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:

V_R14=12-1.785=10.125V   I_c=10.125/469=0.0218A relevant to*

But V_{R14 actual}= 9.08 => Ic_actual= 9.08/469=0.019A (Justified)

V_R15=12-1.846=10.154V   I_c=10.154/469=0.0219A*

But V_{R15 actual}=5.93 => Ic_actual=5.93/469=0.0126 ~ 0.013A (justified)

Similarly:

V_R13=4.94    I_b=4.94/977=0.005 (justified)

V_R16=4.93    I_b=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.

Experiment No.6, 7, 8 and Transistor. (part3)

Experiment No.8
Vary the base resistor and measure changes in voltage and current for Vce, Vbe, Ic and Ib. Then Plot a load line.

 The following circuit is set up on a bread board: 

With LED and transistor stay the same, the experiment is to obtain a variety of results(Vbe; Vce; Ib; Ic) from a variety of Rb( ranged from 2k to 1M). Hence, plot a load line so we can determine the switching and current amplifying characteristic of this transistor.

Vbe stabilizes from 0.6-0.8 indicating a behavior similar to a forwarded rectifying diode.
 +At low Rb values like 2k or 47K or even 1k (tested); Vce are all very low and lower than Vbe; indicating that C-E junction is switched on and connected nearly as good as a conductor. This is because as Rb is low( in this case the following load line graph will show that from 2k to 47k the transistor is in the saturated region) hence the base current is large enough to eliminate the depleted margin between base n emitter, making it easier for the current from collector to emitter. 
Notice that even though the 47k is proportionally bigger than 2k but the amount of Amplified Ic is about the same (7-8mA), this tells me that 8mA might be the practical amplifying limit for this transistor. But the thing is, 47k and 2k are both near the margin of base current where it kinda breaks through the breakdown point of Vbe on a B-E diode(refer to diode forward Voltage drop break down point), allowing the best current amplified and only results in minimum voltage drops( Vce).
+At high value of Rb (220k; 330k; 1M) we don't see Vce really low anymore, they are now significantly big, and bigger than any of the diodes forward Vd( LED n B-e diode). This is because Rb is too large, its impact on eliminating the depleted region B-E is small that Ib only make the base a little bit more positive than the emitter. As a result, the amplified current is rationally smaller and it takes more Voltage drops to amplify.
+ From Rb= 47k to 1M, the Ib value seems unreadable on the multimeter, hence specific result can be calculated for need of determining the Beta value.

The Load line:

 The graph tells me that when operating @ Rb = 220k is the lowest that the transistor doesn't seem to be in the Saturated REgion anymore. So i guess where it starts is about 100k and above the transistor will not be so efficient as it starts to draw more than 0.3 to wayy more than 2.0 V on Vce. And while being at it, the Ib is quite low and it doesn't make positivity much of the base to the emitter, hence it doesn't help Ic much.
 Lets believe that in the saturated region the beta is maximum. Then: Beta= Ic/Ib.
Rb=2k; Rc=470; V=5V; Vbe=0.8V. Vrb=V-Vbe= 5- 0.8= 4.2V=> Ib=Vrb/Rb= 4.8/2000=2.4mA.
Beta is Ic/Ib= 8/2.4=3.33
 On the other hand:
 Rb=330k: Vbe=0.69V. Vrb=5-Vbe=5-0.69=4.31V=> Ib=4.31/330000= 1.3e^-5 approx~ 13 uA.
So the above statement about "in the saturated region the beta is maximum" is not entirely true. I reckon its more complicated than just Ic/Ib.

Experiment No.6, 7, 8 and Transistor. (part2)

Experiment 7 

  0.797 is approximately identical to the forwarded diode Voltage Drop of 0.66, where from base to emitter is actually a p-n conduct which behave like a normal rectifying diode( only partially).
   Supply voltage of 15V goes through R1 of 10K Ohms that leaves only a small amount of available voltage enough to push the small current through b-e, in order to make the connection between c-e. This is the part where a transistor operates as a switch.
  Because as we vary the supply voltage or the resistor before base, voltage drop @ base will slightly change, plus we can't really determine the value of the resistance between a likeliness of diode, then a voltage drop will come in handy:
As 15V goes through R1 and goes through b-e, the total voltage drops between these 2 will add up to near 15V. We know the VD of the b-e is 0.797V, then we can get the Voltage Drop across the resistor: 15 - 0.797 = 14.203V. Hence the Current through base: I= Vr1/R1= 14.203/10000= 1.4mA approx~. 
Knowing the base current is one of components which help determine how big the amplified current will be. Since VD across c-e is impractical to determine using the same method, we will have to advance deeper into the operation of the transistor.
However, on the practical field,  we can always measure using Multimeter- connected across collector and emitter, and a reading of 46.3mV obtained. When a small current goes through base, a low voltage drops between c-e indicates that very small effort it takes to conduct as this act almost as effective as a conductor, hence this really low resistance allows a bigger, amplified current to go through.

 In this graph, the straight line is the resulted load line of where you choose your configuration of resistor sizes in relation of base current and amplified current. 
A- Saturated Region. This is when the transistor works at its best. Only a small increase in Ib(which is the Base current) can result in a steeply effective Ic( Amplified Current) switched on and amplified. (Take a point on the load line in A region aims straight down the Ib line (which is the base current)). Also, we can see that only a VERY small amount of Vce (voltage drop) across what supposes to be nearly as good as a conductor, then we have ourselves a transistor working @ in other word:" turned on at maximum level.

B- I mention "near Cut-off" because it is supposed to be below the forward base current level. This "cut-off" means similar to "the whole city of New York's electricity line has been cut off for several hours for maintenance due to excessive usage". It's a "cut-off", where no current flows through the base as the transistor is simply switched off.

Active Region: In between Saturation and Cut-off. This is the range of Ib and resultant Vce where the transistor is not operating at its best, usually when varying between wide range of resistor configurations which effectively alter the current outcome. As we can see a small rise in base current leads to a very significant rise in voltage drop between what we need as a conductor(Vce).

Power dissipation of a transistor: 
Our transistor is physically a component on a circuit board but NOT necessarily in term of "loads" or "consumers" etc... In fact, it's a JUNCTION which when operates will eventually have 2 loaded consumers: They are: Base-emitter and Collector-emitter. Each component has its own Voltage drop( Vbe; Vce) and Operating Current( Ib; Ic), therefore we actually have 2 components which have power dissipation. P= Pbe + Pce= Vbe x Ib + Vce x Ic. 

Beta: These are the amplified current gain from the base current, calculated by: B= Ic/Ib. 
Where in saturated region the transistor will experience maximum current gain.

Experiment No.6, 7, 8 and Transistor. (part1)

1, What is a transistor?
Remember the basic p-n junction we learn from normal rectifying diode? This concept is used again in a revolutionary electronic device called Transistor, the base of TODAYS compact, advance electronic devices which aren't meant to be without the transistor. Now, the transistor uses the p-n basis in a more advance way: instead of just p-n, its either p-n-p or n-p-n, which they called the bipolar semi-conductive junction. 

We can start by consider a transistor is like 2 diode back to back. The goal is to use a small amount of amperage, to active a sleeping high amperage just for the sake of turning thing ON and OFF things... / the language of our digital world that has never changed.

Best explaining about the transistor is to go and actually do:
Experiment no.6: 
  First thing we were told: that there is a base, a collector and an emitter.
In NPN, small current flows from Base to Emitter ( as p to n), which turns on the connection between Collector and Emitter, which allows a bigger current to mount from collector to emitter.
In PNP, small current flows from Emitter to base ( as p to n) which allows an amplified current from Collector to Base ( p to n). 

So when having a transistor in your hand, it is important to know the layouts if you have no clue what type of transistor it is, by using a multimeter set to diode test mode, and check n label each 2 terminals. Put them together as a form of 2 diodes back to back you will be able to figure out where is where or NPN or PNP type. 

While doing this, if there are 2 forward diode you have spotted, then the one with a slightly higher Voltage drop is from the emitter whether it is anode or cathode. Why? Because higher voltage drop means harder to push through, as the current through this spot is slightly restricted than the amplified current through from the collector. Plus, remember this is a semiconductor and a amperage amplifier, so its obvious that when small current trigger the big current, you would expect the collector- base junction or collector- emitter junction to act pretty much like a conductor wouldn't you?


So with that being said, this is what we would expected: 

Experiment No.5 Capacitor

1. What is a capacitor?
An electrical component which can stores and release electrical charge. It consists of 2 metal plates and an insulating layer in the middle. The 2 metal plates are spaces for charges from the battery to accumulate. When a capacitor is connected to a power supply or a battery, electrons start to flow into the negative plates and charging up the capacitor. And when disconnected with the power source, the charges releases as there is nothing to hold the positive charges and the negative charges together, hence they tend to find an alternate way to get together. Note that the plates attract the charges from the battery as if accumulating on it is the closest the positive charges and the negative charges can get together. The truth is, the metals plates are nothing but 2D conductor which is area that stores charge instead of a wire. It is the insulator that attracts the charges, by having opposite charges facing the metal plate, though insulated, which will attract the opposite charges from the battery to near it.

2. Types:
Non-electrolytic capacitor : Regardless of polarity
Variable capacitor: Adjustable
Electrolytic capacitor: Large capacitance. Polarity correct. High leakage. Big. Dangerous.
Tantalum capacitor: Large capacitance. Polarity correct. Small & Expensive.

3. Identifying Capacitor size:
   Unit: Farad. 1 Farad equals 1 Amp-second at 1 Volt.
 Follow the EIA code and look it up with the capacitance table to determine the potential capacitance.

4. RC Time Delay or "Charging time".
Use the formula: T= R x C x 5.
Time constant: Why 5 is in the equation? Because whatever the total charging time, the time it takes to charge a capacitor up to 2/3 of its potential capacitance is 1/5 the total charging time.
Capacitance is in Farad. Farad is in the favor of the amount of charges(Coulomb) and Voltage, hence Amperage. Because it takes up Voltage to push the flow, Resistance will definitely affects Amps( Coulomp/sec) hence affects time.
Capacitance obviously affect charging time since more space means more charge can be stored, more time to reach full capacity.

5. Practice circuit:

The task revolving the charging time of the capacitor, this is the circuit where we can test with different variation, to see the relationship between reality and theory.
As we can see the opened switch is nothing of a circuit so when the capacitor is hooked up with the resistor and 12V it will start charging up.
When the switch is closed, remember that flow of charges always finds the easiest, fastest way to get together oppositely. So we can prevent the charging by closing the switch( bridge) for the sake of convenience in our timing experiment.
With different sizes of resistors and capacitors, we need to calculate their charging time first then compare them with the actual result. Checking charging time like this is a good way to determine whether the capacitor is working properly or not.

The 

The charging graph resemblances a type of exponential graph.

Experiment No.3 and 4 - Zener diode

A Zener diode is a normal diode which lets current flow from positive to negative(conventional) in forward direction. In reverse, pn layer is created so that current is restricted but when there is sufficient voltage pushing through the boundary then the zener diode limits the current by trying to keep the voltage at a certain level.
Experiment No.3

Take a look at what we did to justify the operation of the zener diode:
  The Zener is a 5V1 400mW, so basically the advertise tells us that this zener will limit the voltage drop through it by 5V. To test this, we wired it in the diagram above, then use the Voltmeter to check the zener. As we expected the Vd turned to be 5V even if we vary the supply between 10V and 15V.
  Then we try to put the zener in forward, it flows just like a normal diode, Vz is recorded as 0.848, a fairly low Vd for a forward diode, so it mean there is good flow with little difficulty.
   Normally, when a zener is reversed, a resistor also needs to be wired in series, there should be no flow across it with just low supply voltage. In our case our "zener" voltage is 5V, so when the supply reaches 5V it will start to flow in little through the zener. So it means that a zener with a "avalanche" voltage can be in range of various voltage value, for different voltage regulating purpose.

Experiment No.4

Both V1 are 5V indicating that the zener diode breaks down @ 5V in reverse and will continue to keep the voltage across it to 5V. V2 are almost the same 0.758 to 0.78 tells us that normal rectifying diode regulates a much smaller voltage drop and will vary slightly due to Voltage supply changes( 10 to 15V). As Vs increase from 10 to 15, the two diodes keep to their own voltage drops, hence the resistor obtain all the voltage drop to its own power.


"In the reverse direction, no current will flow until the voltage impressed across it is equal to the zener voltage. At this point, a current will flow and an extremely small increase in voltage will cause a large increase in current."

(Read more: http://wiki.answers.com/Q/Why_does_a_zener_diode_work_in_reverse_bias#ixzz1VHCzbmv6

 This statement relating to the voltage-current break down graph of a diode: We an see that @ the point of Vz, a modest increase in Voltage(reversed) can cause a huge climb in current, as nearly as similar to the exponential behavior of a forward bias!!! This means that after break-down point a diode if not to be broken down in a destructive way like low voltage normal diode, then break-down voltage is a substantial way to regulate a much larger voltage without having to add many small Vd diodes in forward - So this is the way they came up with a zener didode, operational when available voltage towards it reaches it's own BREAKDOWN voltage, by modifying the layers in p-n junction that would enable the diode not to self-destruct @ breakdown voltage like normal rectifying diode.
If voltage available for the zener is lower than its breakdown voltage, it won't breakdown and conduct any current. Vice versa, if avalanche current is too high that according to P=IV that exceeds the power dissipation limit of the zener, it will self-destruct from the reversed voltage breakdown regulating behavior.

Experiment No.2 Diodes n LED part 2

  Identifying a diode without using the multimeter: the only way is looking @ the sign on the cover. It often reads: 1N4007 and major black color, and there is a bit of a different color, so that's where the cathode is.
  Now a diode is to be wired in a circuit which has 1 resistor, 1 diode or 1 LED, and supply of 12V

                   
Knowing the specification for diodes is very important because when hooking up with Supply, it might destroy the component. In the data sheet for 1N4007 it shows: Average Rectified Current @ Absolute Maximum Rating of 75'C IS 1.0 Amps: this means that the operational Amperage which can safely flow through the diode is 1.0 Amps without overheating n eventually breaking the diode. This also explains why we have to put a resistor in series with the diode in order to regulate current flow in the whole circuit to lower than 1 Amp, for example: a 1K resistor. Using Ohms law, its easy to figure out that we shouldn't put a diode straight with the supply voltage because as R is minimum, I will be REALLY big and will certainly greater than 1Amp.
Next, our job is to figure out the Amperage actually flowing through the circuit:
      Since the actual resistance through a diode is inconsistent to determine, we have to resort on Ohms Law n series circuit law. In a series, current flow equally through every component. So, we need to find the voltage drop across the resistor. We already know 0.66 V of the diode in forward, otherwise we can measure it too. Obtain that, and subtract with the Supply we have the Vd of the resistor. And take that value divide by the resistance it self then we got the total Amperage. Both calculated(theoretical) and measured value are pretty much 0.005A, just as we expected. This is wayy below our limit of 1.0 A, so its safe.

By replacing an LED into the circuit, it makes no different than some light emitted and more Voltage Drop is drawn for the LED. The LED uses more Voltage to emit lights that is actually electromagnetic radiation, so to me it's reasonable that it's power dissipation is higher, hence requires more Vd, not necessarily because of resistance.

Again, the operation of a diode, can be comprehend by basic electron flow and conventional current. Conventional current says electricity flow from positive and negative, but what actually happens is electrons flow from negative to positive. If we apply the case of diode where there are P-N layers; holes that attracts electrons and vice versa, then the statement of conventional current is not entirely incorrect! Holes are actually atoms those are missing electrons, thereby positively charged. So the conventional flow is saying positive charges flow from positive to negative.