Maker Pro
Maker Pro

Injector Driver Circuit

voltagedrop

Apr 22, 2021
12
Joined
Apr 22, 2021
Messages
12
Good day folks,

After weeks of prototyping, burning up components and tools, I am looking for help with a circuit I am trying to build. My greatest obstacle I think is trying to combine 2 different voltages in a circuit.

Long story short, I am attempting to build a circuit to drive and pulse electronic fuel injectors from an outboard marine engine. These injectors fire at 55 volts, which is supplied by the ECM. I am attempting to "drive" the circuit from a standard injector tester, which is designed for typical 12 volt injectors. The reason I am attempting to drive from this tool is it already has options for various tests (various pulse width's, and # of pulses, which is useful on certain tests, as well as a continuous, which is what I am interested in now to attempt cleaning these very expensive injectors).

The injector test tool outputs a constant +12V, and pulses the - side of its outputs. Therefore I was using it to drive the base of a PNP BJT (2N4920), with that BJT then connected to the base of a NPN (2N3055). The 2N3055 is connected to a boost converter set to 55 Volts (i believe I can turn it down to around 30 volts after things are working, and the injectors will still function properly, as the engine does this under normal operation (30volts at cranking, 55 volts at running). My lastest thought was that my bench top power supply could not handle feeding the boost converter as I know the injectors will draw a fair bit of initial current, so I replaced the power supply with a good ole 12volt "car" battery. No change.

My circuit seems to pulse a LED just fine as intended. But when I install the injector in place of the LED, I get 1 hard activation (as it should), then tiny little pulses. On the scope, I see just under 50 volts on the first pulse, then down to 5 or so for everything after the first pulse. The 2N3055 gets smoking hot (will boil a drop of water), so I tried 2 in parallel attempting to spread out the current, and they both got just as hot.

Attached is a diagram of my latest attempt. The 2 capacitors were not "spec'ed" in any way, but were just added in an attempt to get the circuit working, as were 2 I had laying around with a high enough voltage rating. Circuit works the same with or without them.

Further, I did some testing on the injector testing tool, and found it was capable of outputing approx. 15 amps, before the pulsed voltage output dropped off below 10 volts.

Thanks for any help
 

Attachments

  • Injector Driver.pdf
    39 KB · Views: 11

Harald Kapp

Moderator
Moderator
Nov 17, 2011
12,774
Joined
Nov 17, 2011
Messages
12,774
There is no way Q1 can drive Q2. Any base current driven into Q2 by Q1 has no return path.
Your circuit should look somewhat like this:
1663913042040.png
R2 is the current limiting resistor for teh base of Q1.
R3 is approx. 10 × the value of R2. It is meant to ensure that Q1 is really off when V1 doesn't supply the pulsed voltage. R1 is a placeholder for the fule injector.
Add capacitances and EMC protection components ass required.
In this circuit the base current going into Q1 can return to V1 via the emitter connection. As long as V1 and V2 do not share a common ground it is completely irrelevant whether V1 switches the "+" side or the "-" side. The only thing that matters is that current is drivenm into the base of Q1 - or not.
 

voltagedrop

Apr 22, 2021
12
Joined
Apr 22, 2021
Messages
12
Ah. Ok. I tried similar circuit to this, however the injector tool (V1) in your diagram, and the boost converter, were both being supplied by the same bench top power supply, and therefore a common ground as I'm sure the input and output sides of the boost converter are not isolated. This resulted in the injector tool burning up (2 L7805 regulators inside the tool released the magic smoke). I will try it with separate power supplies. Thanks.
 

voltagedrop

Apr 22, 2021
12
Joined
Apr 22, 2021
Messages
12
There is no way Q1 can drive Q2. Any base current driven into Q2 by Q1 has no return path.
Your circuit should look somewhat like this:
View attachment 56307
R2 is the current limiting resistor for teh base of Q1.
R3 is approx. 10 × the value of R2. It is meant to ensure that Q1 is really off when V1 doesn't supply the pulsed voltage. R1 is a placeholder for the fule injector.
Add capacitances and EMC protection components ass required.
In this circuit the base current going into Q1 can return to V1 via the emitter connection. As long as V1 and V2 do not share a common ground it is completely irrelevant whether V1 switches the "+" side or the "-" side. The only thing that matters is that current is drivenm into the base of Q1 - or not.
Harald Kapp, thanks for your help. Worked as intended after following your lead. Need to do a little tweaking with R2 as I didn't have a high power resistor of the correct value so she gets real hot, but besides that works great. Thank you muchly.

If your willing, would you be able to explain why the 2 power sources have to have isolated grounds. I have tried finding an answer, but every result I come up with is to do with AC/"household" wiring.

Thanks again.
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
12,774
Joined
Nov 17, 2011
Messages
12,774
would you be able to explain why the 2 power sources have to have isolated grounds.
... because you stated that
The injector test tool outputs a constant +12V, and pulses the - side of its outputs.
When you connect the power sources with common ground, this common ground would short-circuit the pulsed output of the injector, thus making the output seemingly active all the time.
 
Top