A complete mystery with a fuel injector pulse generator

jamie_s said:

Well i posted a schematic that will work.

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Are you certain? Where are the component values? Will a PIC be able to drive the 3055 adequately? How will the IRFZ34 work with a couple of amps and less than 5 volts Vgs? Maybe an IRLZ34 (or related) would be a better choice as it is logic level.

My point remains that the original poster has not presented an accurate schematic of what he has been trying. There have been several good suggestions, but we don't know which one(s) have been incorporated. As best I can tell, the electrical characteristic of the injector are still based on supposition, which makes giving any assurances that particular design will work a bit problematic.

John

jpanhalt said:

Are you certain? Where are the component values? Will a PIC be able to drive the 3055 adequately? How will the IRFZ34 work with a couple of amps and less than 5 volts Vgs? Maybe an IRLZ34 (or related) would be a better choice as it is logic level.

My point remains that the original poster has not presented an accurate schematic of what he has been trying. There have been several good suggestions, but we don't know which one(s) have been incorporated. As best I can tell, the electrical characteristic of the injector are still based on supposition, which makes giving any assurances that particular design will work a bit problematic.

John

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I agree that TIP120 is the only one of the three devices mentioned that has a chance of working in that circuit without possibly having to select a working device from a batch.
Also, the flyback voltage may damage the transistor unless some sort of flyback clamp is added to protect it.

Hi Ian,

yes I have researched the info from the microchip website.

MAX input 75 mA
MAX output 25 mA

This is in regard to the 10/12/16F pic family

I did away with the reed relay and 9014 transistor and replaced it with a single component - IRL510A - a power MOSFET. It's probably overkill as it can handle up to 5.6 A. It finally works the way I want it. Thanks so much everyone for your assistance.

Lastly, I am still unsure on how to select the proper transistor using the Hfe Dc current gain.

I computed the current running through the power transistor at .21 A using standard methods Ohms law and such, and then looked at the specs to see if that is within the components range (Max collector current 100 mA), so it is out of range and will not work. This correct? However, KeepIt has suggested another method using Hfe, but I am still unclear about it. Exactly, how is Hfe used to see if it is within component range?

If I am supposing that if the output is 5V and has a max output of 25 mA, then using the bottom chart on the 9014 datasheet, I have type A, so the Hfe is 60-150. What does this mean exactly, that the component will have a gain up to 150 times the input base current? Then compare that to the max collector current and see it is handles? Why is the range so large?

Thanks!

insert the injector between +12v and the mosfet, not between the mosfet and ground

1. Agree with Jamie
2. FET's don't have an Hfe. So lets keep the Apples with the Apples and the Oranges with the oranges.
3. A reverse biased diode needs to be acrross the injector to avoid damaging the FET when the FET turns off.

I'll try to get back to this later. Both transistors and FETs.

The datasheet for the IRL510A is here; http://www.datasheetcatalog.org/datasheet/fairchild/IRL510A.pdf

Note, that it says: LOGIC LEVEL GATE DRIVE
that fixes Vgs between 1 and 2 Volts. Guess what those magic numbers are?
It also means the FET will be on when Vgs > 2V, since Vgs Max is what we are concerned about.

Now look at figure 2: At about 2.5 V Vgs, the device can sink about 1 Amp; 10 ^0 = 1.

Note, Vgs leakage current of 100 nA and and input capacitance of 180 pf. What this means is that UNLESS you provide a path to ground for at least this amount of current, the device could turn on by charging the input capacitance.

If some bipolar device is driving the FET, leakage from the Bi-polar device could turn things on and you need to find a place for that to go. That was the purpose of the gate to ground resistor.

As with any port, you have to pay attention to source voltages and currents available, sink voltages and currents available and leakage.

Does this make sense when using N-channel Fets to switch to ground now?

Oh yes, you are correct jamie, that is what i did in the implementation. you bring up another question though. why wont the injector between the ground and transistor work? just curious. also, do you know anything about Hfe referring to the questions in the previous post?

Iawia said:

Oh yes, you are correct jamie, that is what i did in the implementation. you bring up another question though. why wont the injector between the ground and transistor work? just curious. also, do you know anything about Hfe referring to the questions in the previous post?

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The mosfet is a low side switch therefore will switch the load to ground. having the injector between ground and the fet is not good practice and will also cause the fet not to switch correctly depending on voltage drop across the injector.
Hfe makes no difference with a FET, basically with your "L" fet (logic levev) and with an 0-5v signal from the PIC it is either on or off.

Hfe only applies to the transistor (different to a FET / MOSFET) and basically is the gain of the transistor,
easiest way to understand the basic fundamental is
if a transistor has a hfe of 100, 2 ma of current on the base pin will result in approx 200ma of current between collector and emitter.
a transistor with hfe of 200, 2ma of base current will result in approx 400ma of current between collector and emitter and so on.

Note what I said in post #26: FET's don't have an Hfe

I also asked did you understand how to do the FET switch design?

The BIGGER problem with your transistor design is that Ic (max) was way too small for the injector. When you use FETs or Bipolar transistors you also have to pay attention to SOA or Safe Operating Area. That may include different parameter for a pulse response.

Iawia said:

Oh yes, you are correct jamie, that is what i did in the implementation. you bring up another question though. why wont the injector between the ground and transistor work? just curious. also, do you know anything about Hfe referring to the questions in the previous post?

Click to expand...

When you put the load between the N-channel MOSFET source and ground, the transistor is acting as a source follower. Since the transistor requires a couple of volts Vgs, more or less, to turn it on, that voltage will appear between gate and source. So, if you drive the gate with 5V, you will only get 2 or 3 volts across the load. If the source were to try to rise higher, the transistor would turn off due to insufficient Vgs.
Regarding Hfe (current gain) in a bipolar transistor: Keep in mind that Hfe is low when the transistor is saturated. Most transistors specify the saturation voltage (Vce(sat)) with a "forced beta" of 10 (20 for a few types), i.e., Ic/Ib=10. So, just as an example, if you want a bipolar transistor to saturate with Ic=100mA, you will need 10mA of base drive.

A switch using an NPN transistor with the load connected to the positive supply and the transistor's C connected to the load and E connected to ground.

β or beta is another term for Hfe.
β has a range, so your circuit has to work for all β's in range.
Ic = β * Ib or the Collector current is equal to hfe * the base current
Ie = (β+1) * Ib; Not that it matters.

When a transistor is a used as a switch, you want it to saturate and therefore Vce(sat) comes into play. This would be the voltage drop when the transistor is fully on.

You have a load and therefore you know the current you need to turn on the load. (Ic)

The transistor you select must have an Ic greater than your load and a SOA (Safe operating Area) associated with the transistor. The latter is usually seen for power transistors.

You could, for instance, select a transistor with a Ic of 100 mA for a LED that draws 10 mA. Suppose this transistor had a β (minimum) of 100. So this means that we need 10 mA/100 of base current at least, but their are variations there too.

So, let's just make this a genneral purpose switch up to IC of the transistor. In that case an Ib of 100 mA/100 is required. 100e-3/100 Amps.

Now, you have to make sure that you have enough current from your logic element multiplied by β (minimum) to equal 100 mA.

If that test passes, you can then select the resistor.

You would need a base resistor of R <= (Vhi - Vbe)/Ib; Vbe is the base emitter drop. 0.6 V to 0.7 V is customarily used.

You would also check that the Max Ib would not be exceeded.

You were warned earlier that β can be slightly lower at saturation.

Darlington transistors can have Hfe's in the thousands.

Your earlier problem was that the transistor you selected had an insufficient Ic.

Thermal issues is yet another problem. One thing to remember is that when a transistor heats up, it conducts less and as it conducts less, it heats up more etc. This is called thermal run-away.

I want to emphasize the need to use Ib=Ic/10 when driving a transistor into saturation. The attachment is extracted from the 2N3904 datasheet.

Roff said:

I want to emphasize the need to use Ib=Ic/10 when driving a transistor into saturation. The attachment is extracted from the 2N3904 datasheet.

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Thank you Roff, Jamie, and Keep. I am very grateful for your help.