The low amperge sound just like when in a car the igniton capacitor(condensor in the old days) is bad. Even with a BJT or Mosfet as a switch wouldn't a cap help for this?

 

The low amperge sound just like when in a car the igniton capacitor(condensor in the old days) is bad. Even with a BJT or Mosfet as a switch wouldn't a cap help for this?

The cap is there to provide a path for the primary inductive current and prevent arcing when the mechanical contacts slowly (relatively) open, which would reduce the output high voltage.
It has no effect on the primary current.

BJTs and MOSFETs don't need the capacitor since they generate a clean, rapid, open circuit with no arcing when used as a switch.

 

The cap is there to provide a path for the primary inductive current and prevent arcing when the mechanical contacts slowly (relatively) open, which would reduce the output high voltage.
It has no effect on the primary current.

The automotive books disagree with you on that. While it does save the points from wear and arcing, the biggest thing is it stores ~400V for the next opening of the points, to make the spark hotter more amperage in the spark. Even electronic ignitions use a cap to store the BEMF from the primary for the next spark. Since the whole coil system is like a flyback.

 

more amperage in the spark.

True.
I thought you were referring to the primary current.

the biggest thing is it stores ~400V for the next opening of the points,

No, it doesn't store it for the next opening.
When the points open, the primary/capacitor voltage jumps to about 400V due to the stored primary inductive energy, which then generates the spark at the secondary, discharging the capacitor. That's how a flyback circuit works.

Even electronic ignitions use a cap to store the BEMF from the primary for the next spark.

Well, the transistor ignition I installed on my old GTO had no primary capacitor and it worked well, so I'm not sure why they are needed on present day ignition circuits.
Do you have a schematic for such a circuit?

 

Do you have a schematic for such a circuit?

No it's a solid state cap built into the modules, like the HEI from GM.

No, it doesn't store it for the next opening.
When the points open, the primary/capacitor voltage jumps to about 400V due to the stored primary inductive energy

Since the running voltage is ~8V, due to the ballast resistor or resistor wire used after the ballast was eliminated, just where does this 400V come from? According to the books it is from stored voltage in the cap. I must be missing something in your reply, your saying the 400V is stored in the coil windings?

 

For the past week I've been trying to get rid of the back EMF. I tried many things, such as: zener diodes across coil and FET, shottky diodes, rectifier diodes, supression zener diodes, capacitor, with different ratings. I even had to put multiple in series to increase voltage back. Despite all that I've only managed to lower the back EMF to about 160V (400V previous). Also I had to put components in other direction than normal, and don't even know why, Example in picture.

I've also lowered dwell resistance so low that the EMF is at the same time as pulse, but I guess that can cause troubles.

Anyone got any idea what can still be done? Maybe I've got a mistake in circuit.

 

Where are you measuring this back emf?

Update:
I posted one solution to reducing the voltage spike on the MOSFET Drain, in an earlier post #20

 

Where are you measuring this back emf?

Update:
I posted one solution to reducing the voltage spike on the MOSFET Drain, in an earlier post #20

Ok I did that.
I'm measuring the ouput/secondary voltage.

Now I'm confused, is the back EMF the spike on the left side or the right side (in picture)? I always thought it was the one on left.

Also why is my output current not connected, like on the source or MOSFET. Example in picture.

 

just where does this 400V come from? According to the books it is from stored voltage in the cap. I must be missing something in your reply, your saying the 400V is stored in the coil windings?

We may be talking apples and oranges.

If it's a capacitor discharge system, than the several hundred volt primary voltage is generated by a DC-DC converter and stored on a capacitor
This is then discharged across the coil primary (typically with an SCR) to generate the high voltage on the secondary.

If it's a Kettering type flyback ignition, then the energy is stored in the inductance of the primary due to the primary current.
When the primary current is suddenly stopped by the points or transistor switch, the stored inductive energy tries to keep the current going, which causes a large voltage spike at the primary This is then transferred to the secondary by transformer action to generate the HV secondary spark.
For this system the capacitor is only needed if it uses mechanical points (to eliminate sparking across the points when they open).

 

Anyone got any idea what can still be done? Maybe I've got a mistake in circuit.

You circuit has no path on the output for the discharge energy so naturally the voltage will go very high.
You need to simulate the capacitive load that the fence will typically provide.
You can estimate that by the capacitance formula for a wire above a ground plane.

 

Now I'm confused, is the back EMF the spike on the left side or the right side (in picture)? I always thought it was the one on left.

hi luki,
I did explain earlier.

When the MOSFET turns ON, [left side] it connects the low end primary coil of the transformer to 0V, a magnetising current flows thru coil.
This magnetising currents rate of increase is limited by the inductance of the transformer,
ie: the primary ON current does not instantly go from zero to maximum. So Emf= -L * di/dt .... 'a low emf = smallish spike'

When the MOSFET turns OFF, the primary current drops to zero, almost instantaneously , So Emf= - di/dt ... ' a high emf = large spike'

Choose a MOSFET with say a 120Vdc Drain/Source specification, use a MOV or TVS, from the Drain to 0V, to limit the Drain spike to around 80V


E

 

You circuit has no path on the output for the discharge energy so naturally the voltage will go very high.
You need to simulate the capacitive load that the fence will typically provide.
You can estimate that by the capacitance formula for a wire above a ground plane.

I calculated the capacitance, so I just put it at the ouput right. Is that considered as fence voltage drop?

The MOSFET can take 100V, and TVS drops it to 80V, so thats alright I guess.
One thing that still bugs me is ,why is the ouput current not constant, as it shows the picture in my previous post. Its just 2 spikes on ON and OFF.

 

hi liki,
Can I suggest that you change the plotted area background colour from black to white.?
I find that viewing blue and green traces very difficult to read on a PC screen.
E

 

I calculated the capacitance, so I just put it at the ouput right. Is that considered as fence voltage drop?

Yes, it's a load on the output.
What length of fence did you use and what value of capacitance did you get?

It certainly will cause a drop of the output voltage.

 

Yes, it's a load on the output.
What length of fence did you use and what value of capacitance did you get?

It certainly will cause a drop of the output voltage.

I've got 6.85 nF for 1000m. The drop is around 560V on 3KV.
I'll be using fence charger at home, we'll rarely use more than 1000m of wire (gallagher wire), so I think 3K is enough. Also theres very little undergrowth.

 

hi luki,
If you are fencing for sheep or other woolly animals, you may need a higher voltage.

E

 

hi luki,
If you are fencing for sheep or other woolly animals, you may need a higher voltage.

E

It's only for cattle.

The output current I(C5) is not constant, like the Id(M1). Is this a problem.

 

hi,
I don't see that will being a problem.
E

 

It looks like there are just spikes for ON and OFF. And the max current is there only for a fraction of a ms. I want to get 1J out of it so I would need current to be at max (100ma) for around 4ms.

 

I want to get 1J out of it

Why 1J?