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Flip-Flop switching between two injectors

malc9141

Member
Hi Guys

Would someone kindly look at this diagram and tell me if its right? The idea is, for each injector (think v. low inductance coil), a fast pulse is overlapped by a slower pulse, then next time, the other injector is treated the same, and back again. The fast pulse is a bigger current to get a fast response (FIRE), but the slower dwell (HOLD) with a lower current allows a quick mechanical reversal. Thank you! MalcScreen Shot 2021-05-10 at 10.32.37 am.png
 

augustinetez

Active Member
You sure you've got the right IC number in the drawing?

A 4093 in my books is a Quad 2-input NAND Schmitt trigger and the pins you show as outputs are inputs and vice versa.

Might also want to look up LM1949, it is a dedicated injector driver chip that will take care of the peak and hold currents.
 

malc9141

Member
Thanks. Yes, I know the chip OK. But if I have the wrong IC Title, what is the correct name of the Flip-Flop I should have? And if we ignore my (wrong) title, is that Flip-Flop OK? Should it be a 7400 ?
 

AnalogKid

Well-Known Member
Most Helpful Member
The pinout matches a CD4013 dual D flipflop. Wiring diagrams are difficult to decipher. It would be better if you drew a schematic with the functional symbols for the flipflops.

ak
 

crutschow

Well-Known Member
Most Helpful Member
If you actually want a flip-flop then the CD4013 dual D-FF would likely be the appropriate one.

And please do not draw your schematics with gaps in the wires (where have you ever see that done?).
Just draw the wires crossing and, if there is a connection, use a dot that covers the junction of the two wires.
 

AnalogKid

Well-Known Member
Most Helpful Member
1. What are the purposes of R4 and R7?

2. Why do you think you need an attenuator formed by R4 and R5?

3. If you think C5 and R23 form a delay network, they don't.

4. What does the diagonal line between clk1 and clk2 indicate?

ak
 

malc9141

Member
To Analog Kid: I am not an electronics expert otherwise I would not be asking. I am using a feed-in of a circuit I was advised about on this Board.
And I did think the C5 R23 formed a delay, so thanks for the criticism.

I wonder if anyone wants to help? I have put in the Dots. I should have changed the title of the IC to CD4013 but have overlooked this.
Screen Shot 2021-05-10 at 2.35.59 pm.png
 
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augustinetez

Active Member
A bit more info about what you want to use this for would be a start.

Reason for asking - depending on final use, there could be other factors to consider like dead time between injector activation, length of activation and other things that the project could throw up that may have an impact on the design of the circuit.
 

malc9141

Member
I can explain. There are two (pairs of) high pressure (ie very fast acting) injectors. Those in each pair fire at the same time, at a certain interval (t), but the other pair fire half the interval later. The interval (t) depends on engine speed and is a minimum of 0.02 sec. It would be longer at slower engine speeds. The injectors are open for around 0.002 sec.
We can take a signal off the flywheel and use it to fire both Pairs, or (easier) have two signals per rev, equally spaced.
We used the simple single signal with a single cylinder experimental engine. But with a four cyl boxer arrangement, we fire two opposite cyls then the other two, half a rev later. ( I hear you say "but.." Well, its a two-stroke).

The total width of the pulse as you show it, is altered by a pot on the 555. But the short pulse is fixed.
This is therefore the "accelerator".

So it seems easiest to use the same circuit as before but add a Flip-Flop so it fires alt injector-pairs. (The diagram shows as if there are two inj but there are two pairs). So at this stage, I seek advice on how to connect the Fire & Hold lines, so as to get alternating Injector-pair action.

The system we had developed long ago before the Texas Instrument circuit appeared, seemed OK. The TM 1949 is more or less what we had constructed. It still needs quite a lot of work to fit it in and I don't quite understand it.
 
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malc9141

Member
Yes. I think it is exactly that. Between A & B, it is t. But we have created two equally spaced input signals per rev. So t is half the time of a full rev. But that's OK.
After leaving (3) it is split into two (overlapping in time), one very short (about 0.005 ms or less) and the variable "accelerator" pulse from 0.0 to about 0.04 ms which starts at the same time (zero) but continues.
When seen on an oscilloscope, the final shape is a sharp, brief, rise and then a fall to a plateau, before falling again back to baseline.
The pulse length is determine by a Pot at Pin 7 on a NE555. This is split into Fire & Hold but the Fire is made to be short. The total duration of pulse is Hold. The interval (t) is determined by the flywheel speed x2 since there are two signals per rev. This means the shortest t is .01 sec (not 0.02 as I might have said).
 
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AnalogKid

Well-Known Member
Most Helpful Member
1. I almost understand your timing description. To remove all doubt, please post a timing diagram like the one in post #10 with five lines, one for each of the MOSFETs and one for the 555 signal. Indicate which timings are adjustable, and which ones are based on the R3-C5 delay. With that information, analysis should be rapid.

2. I can re-arrange R3-C5 to produce a short delay. What delay time do you need? And, is this delay to be adjustable?.

3.
thanks for the criticism.

I wonder if anyone wants to help?
Really - ?

In a Q & A forum, neither your text nor your image are communicating your concepts and questions effectively. There is no more helpful feedback than that.

ak
 
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AnalogKid

Well-Known Member
Most Helpful Member
First pass. This is a re-draw of your original schematic. As it is, it will not do what you want. All four MOSFETs turn on and off together, within 1 us. R3-C5 has no effect; it is overpowered by the 555 output stage.

Corrections later.

ak
!Injector-Control-1-c.gif
 
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Pommie

Well-Known Member
Most Helpful Member
Don't follow the circuit at all but wonder why MOSFETs connected to Q and /Q can both be on at the same time and why C5 and R3 don't cause a delay. Just curious.

Mike.
Edit, see the C5 R3 problem now.
 

AnalogKid

Well-Known Member
Most Helpful Member
To clarify, all four MOSFETs change state at the same time; two turn on and two turn off. This is not what the OP wants.

R3-C5 are driven by a (relatively) zero-ohm voltage source, the 555 output. To get a delay, you would reverse C5 and R3, and take the clock signal from the R3-C5 node. This is a traditional R-C delay network.

ak
 

crutschow

Well-Known Member
Most Helpful Member
To get a delay, you would reverse C5 and R3, and take the clock signal from the R3-C5 node.
That's problematic with the CD4013 as it requires a minimum clock rise time for proper operation.
 

AnalogKid

Well-Known Member
Most Helpful Member
TS: My guess is that you want an injector to turn on fast with high current, then shift to a low current holding mode, then turn off at the same instant that the other injector turns on:

1. M2 turns on
2. M1 turns on and M2 turns off
3. M1 turns off and M4 turns on
4. M3 turns on and M4 turns off
5. M3 turns off and M2 turns on
etc.

If this is correct, there is a potential problem with steps 2 and 5. There could be a dip in the injector current if M2 begins to turn off a microsecond before M1 turns on. Fortunately, a problem called cross-conduction does not apply to your application because M1 and M2 are in parallel.

There is no electrical difference between M2 and M1 coming on sequentially as you describe, and at the same time in parallel, because M2 carries the maximum injector current. So a better control method is to turn on M2 and M1 together. Then, after a short delay, turn off M2 and leave M1 on with the holding current. A possible issue is that M2 and M4 will have individual turn-off timers, so there could be a small mismatch. If this is not critical as long as both are on for a minimum required time, then things just got easier.

Another issue is using an R-C delay circuit with the 4013 clock input, because that input does not have hysteresis to interface with a relatively slow edge rate. The above changes fixes that. Also, we need to know the injector voltage and current, to add transient protection to the MOSFETs.

Need that timing diagram.

ak
 

AnalogKid

Well-Known Member
Most Helpful Member
Here is a re-work of the design, based on assumptions. Note that the operating voltage of the circuit and the exact part number of the FETs change the FIRE-x on time. With the components shown and VCC = 12 V, the turn-off delay is approx. 4 ms.

Another way to do this uses a hex inverter instead of the dual flipflop. In that version the delay time is independent of the type of MOSFET, and much less dependent on the operating voltage.

ak
!Injector-Control-2-c.gif
 

shortbus=

Well-Known Member
Most Helpful Member
I guess as a car guy I don't understand this whole idea. IIf this is for fuel injection in an engine.

The usual thing is to chose an injector with a large enough rate in pounds per hour to run the engine, then you would have the programing of the ECM change the time open for each desired value of performance in the engine. Not by using more than one injector.

Places selling injectors have online calculators to help, just a couple of many.
 

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