Inverting PWM help

Thread Starter

EmpirePerf

Joined Oct 29, 2010
4
Hey guys,
Im new to this site and am definetly not an electronics genious by any means. We are a perfomance shop that builds mostly all diesel race trucks.
Now for the problem, we have swapped injecion pumps on a truck to a different model. They are both pwm controlled on the ground side. the factory pressure regulator(solenoid) is normally open. On the new one its normally closed. So basically I need to invert the signal and am having a hell of a time figuring this out. Its a 12v system, both regulators are operating at 330hz.
I tried irf510 without any luck there. I right now have 2 tip3055's in parallel. I have a diode accross the solenoid. The signal appears to be inverted however its not a good signal. Seems when it shuts off the signal falls to 0v then imediatly ramps up to about 9v then up to 12v for the on time. I dont know how to clean the signal up so the off time is actually off.
I could be going about this totally wrong.... thats what im not sure, i just dont have any help at all from anybody on this one. Any help would be much appreciated. Thank you very much.
 

jpanhalt

Joined Jan 18, 2008
11,087
The answer may be much simpler than you think. I have a multimeter that reads "duty cycle." At a certain point, it inverts. Can you diagram exactly what you are trying to do (e.g., schematic). Ask yourself, how does the chip know whether its "off" is the injectors "on" or "off."

John
 

Thread Starter

EmpirePerf

Joined Oct 29, 2010
4
Its not the injectors tho. its the pressure regulator. If i run the new pump with the duty cycle as it is for the factory pump, i get 29000psi of pressure. which is not good. so if the duty cycle is inverted then we get 5000psi like we need.
Im taking the factory signal wire to the base of the transistor. emmiter to ground, collector to regualtor. 12v supply to regulator. I have a diode from the supply voltage to signal wire to stop the voltage spikes. and a 10k resistor from the signal wire to 12v.
 

jpanhalt

Joined Jan 18, 2008
11,087
As you obviously know, a common emitter (emitter grounded) transistor is inverting. That is, as base goes high, transistor turns on, collector goes close to ground. I am sure a schematic will still help, though. Do you have any resistance in the collector circuit?

Maybe Sgt. Wookie will chime in. He really knows automotive and truck systems and has helped several posters here with injector problems.

John
 

thatoneguy

Joined Feb 19, 2009
6,359
You may want to add a 100Ω or so resistor to the base of the transistor to limit base current so the transistor doesn't burn out.

Otherwise, it should work if I am envisioning this correctly. Output to old fuel pump is +PWM, this is applied to base of transistor (be sure to add current limiting). Battery connects to fuel pump, with the ground being switched off and on via the Transistor at -PWM (inverted). Is this correct?

If not, please describe/draw schematic.
 

SgtWookie

Joined Jul 17, 2007
22,230
I'm not certain, but I believe the PWM output from your module is likely open-drain or open-collector. This would allow the module to sink current, but not source current - or if it IS capable of sourcing current, not very much. Basically, it's acting like a switch to ground that opens and closes, and the injector regulator provides the current path from +V (~14v from the electrical system).

Have a look at the attached.

PWMsig is a "mystery box" inside your module that's generating your 330Hz PWM, the PWM signal inside the box is shown below.

For the moment, I've assumed that the output from your box is M1's drain, an open-drain power MOSFET - but it certainly could be a transistor. The signal at point "Original" is shown in the plot, below. The original injector regulator is represented by the inductor OrigRegul.

To the right of that circuit, the drain of M2 (which is the same as M1 in the left schematic) is now connected to a 1.4k pull-up resistor. The signal at this point will be the same as point "Original".

I've added a voltage follower consisting of two transistors; one NPN and one PNP, to amplify the relatively weak current available from R3.

R4 keeps the gate of the new MOSFET, M3, from "ringing" when the MOSFET gate is switched on and off quickly. The signal at point "New" is shown in a plot on the bottom. You can see that "New" is inverted from "Original".

I haven't added any transient suppression to the circuit; it's pretty basic - although the transistors are rated for >=60v, but the IRFZ34 is only rated for 55v. Keep 60v in mind, as transients up to 60v can occur during "load dumps", such as turning off headlamps.

For a quick test, you could use a 2N2222/PN2222 for the 2N5055, and a 2N/PN2907 for the 2N2905A. Your IRF510 might actually work, depending on how much current the regulator requires; you haven't provided that information.
 

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Thread Starter

EmpirePerf

Joined Oct 29, 2010
4
Thank you guys very much for the help, i will try this out!

i do have a 10k resistor from battery voltage to the collector circuit.

Thatoneguy- yes that is correct

sgtwookie- thank you! i dont honestly know what the current draw is yet... that would probably help a lot. Ive been trying to get a better understanding of this the last several days and exactly what info i need to make this work. I will let you know how comes out or if im having trouble.

Thanks again
 

SgtWookie

Joined Jul 17, 2007
22,230
In the circuit I posted, you could replace M3 with a 2N3055 (R4 connected to base, emitter grounded) but R4 would need to be adjusted to pass a maximum of around 200mA; otherwise you could overheat Q1, the 2N5505. This would allow for roughly 2A sink current through the 2N3055's collector. If you need more current than that, better tell us what your injector regulator requires.

R4 = (14v - (0.8v+0.8v)) / 200mA = 12.6/0.2 = 630 Ohms. You'd need a resistor rated for 5 Watts. You will also need a ~3k resistor from the base to the emitter.
 

Thread Starter

EmpirePerf

Joined Oct 29, 2010
4
In the circuit I posted, you could replace M3 with a 2N3055 (R4 connected to base, emitter grounded) but R4 would need to be adjusted to pass a maximum of around 200mA; otherwise you could overheat Q1, the 2N5505. This would allow for roughly 2A sink current through the 2N3055's collector. If you need more current than that, better tell us what your injector regulator requires.

R4 = (14v - (0.8v+0.8v)) / 200mA = 12.6/0.2 = 630 Ohms. You'd need a resistor rated for 5 Watts. You will also need a ~3k resistor from the base to the emitter.

I ordered the components on the sheet. For now i tried with irf510s. and M3 gets hot. so i need to figure out the current.
 
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