Yes, isn't the rule of thumb that the diode should be rated to at least the solenoid's current rating, and at least double the operating voltage?

The peak reverse voltage that a snubber diode sees is exactly the power supply voltage, so a rating of twoX or threeX the supply voltage is fine.

Post #19 in this thread deals with the diode's current rating.

 

I

When I tested it on an LED it worked real good, I could change the frequency and the duty cycle and the LED would do as the module would direct it to.

However, once I tried the PWM on the solenoid the LED would turn-off, but the NE555's built in LED was still blinking telling me it was working, the LED though that I connected after the transistor would turn off.

Quite possibly the reverse energy of the solenoid was reverse biasing the LED, if it was across the solenoid, especially the energy retained in a 30a device.
The effect of the diode will retain the solenoid at PWM transition to zero.
Max.

 

Would the heat sink still be required even though it will only be on for about 1 second every hour when in use?

What heat sinks work well with the MOSFETs?

I'm confused. You said that you are going to PWM the solenoid during the time it is switched on. The power dissipation in a FET used as a switch in a PWM circuit is all about the power dissipated as the FET turns on and then turns off each PWM cycle; less about the power wasted by RdsOn.

To make it easy on the FET don't PWM it. Then the heatsinking requirement stems from I^2*Rds. With Joey's FET, @ Id=100A, Vgs=10V, 10V, RdsOn(Max) = 1.3 mOhm, so it would be dissipating 30*30*0.0013 = 1.2W, which would still benefit from being on a heat sink. I think that the FQP30N06L is too underrated.

 

We really need a schematic to get much farther. A high-side switch using your P-channel might be just fine. I would use at least a small heat sink, one that screws to the tab.

Let me see if I can draw a schematic of it, but i have it very identical to the first schematic i posted.

 

A high-side switch using your P-channel might be just fine. I would use at least a small heat sink, one that screws to the tab.

How would the P-channel be wired up on the schematic?

 

the 1n4007 diode is way under rated, a solenoid with 30 amps current through it when switched off will have a much greater reverse curent spike than the 1 amp it will handle.

So the current diode is too small for the 30 amp solenoid? I do apologize if some of my questions seem rather dumb but I am still in the learning process of this.

 

So the current diode is too small for the 30 amp solenoid?

And, the 30 amp solenoid current is too big for the diode...

 

And, the 30 amp solenoid current is too big for the diode...

Now that's the sense of humor that people can have a good laugh with and appreciate

What diode would be a good fit for the 30 amp solenoid?

 

Ok here is my best attempt at the schematic.

What diode is recommended that will work with the 30A solenoid?

Do I need a bigger resistor as well?

As for the N MOSFET (RFP30N60LE) I should be getting them on Friday, I can hook them up parallel or order another one that can handle the solenoids 30A. I went to check our local Radio Shack and they did not have any high Amp N-MOSFET's.

Thanks to all for the the great info thus far. I do apologize if I do not understand all the terminology that is being given, but I am picking up most of this stuff as I read it and research it.

 

What diode would be a good fit for the 30 amp solenoid?

@MikeML seems to have an idea of how much energy your solenoid will store, and therefore how much will have to be absorbed by diode. I don't, so I cannot size it for you.

But, you can minimize the power dissipated by the diode, and shorten the "spike" time, by adding a small resistance in series with the diode.

Back of the napkin calculation:

Assume max Vds of 40V on the FET (the part I recommended). V+ is 12V, assume Vf of the diode is ~1.2V (at 30A). This leaves a margin of 40V - 12V - 1.2V = 26.8V.

26.8V / 30A = 0.893 ohms. I'd choose 0.75 ohms (or closest available).

Upon FET turn off, instantaneous V across the resistor will be 30A * 0.75 Ohms, or 22.5V. Instantaneous R power will be 30A^2 * 0.75 ohms or 675W (no, you don't need to pick a huge resistor, since this will be a *very* short pulse). Instantaneous D power will be 30A * 1.2V, or 36W. Notice the resistor will absorb most of the energy, and very quickly.

 

Ok here is my best attempt at the schematic.

Oh, so now you went and added an LED. I hate mission creep!

In the calculations above, you must insure the peak voltage across the R and diode (in series and at 30A) does not exceed the reverse breakdown voltage of the LED.

And, don't you think you need a R in series with that LED?

 

Oh, so now you went and added an LED. I hate mission creep!

In the calculations above, you must insure the peak voltage across the R and diode (in series and at 30A) does not exceed the reverse breakdown voltage of the LED.

And, don't you think you need a R in series with that LED?

I can take the LED off, I had just added it so I could see the pulse and duty cycle of the PWM.

I bought the LEDs pre-wired in a package and they already have a small resistor on the positive side of the wiring. Sorry I did not think to include that on the schematic.

 

R1 is way to big. A high-current, low Rds,on FET has significant gate capacitance. The large R1 will cause slow turn-on and high power dissipation in your FET.

Unless there are other objections, I see no reason why you cannot eliminate it entirely.

 

Unless there are other objections, I see no reason why you cannot eliminate it entirely.

I agree, however I'd move it to between the gate and ground. This is a failsafe to turn the MOSFET off in the event of a failure of the PWM signal.

 

How would the P-channel be wired up on the schematic?

Like this:
from here.

More details here.

 

I agree, however I'd move it to between the gate and ground. This is a failsafe to turn the MOSFET off in the event of a failure of the PWM signal.

Like this?

 

Like this?

Yup.

 

@MikeML seems to have an idea of how much energy your solenoid will store, and therefore how much will have to be absorbed by diode. I don't, so I cannot size it for you.

But, you can minimize the power dissipated by the diode, and shorten the "spike" time, by adding a small resistance in series with the diode.

Back of the napkin calculation:

Assume max Vds of 40V on the FET (the part I recommended). V+ is 12V, assume Vf of the diode is ~1.2V (at 30A). This leaves a margin of 40V - 12V - 1.2V = 26.8V.

26.8V / 30A = 0.893 ohms. I'd choose 0.75 ohms (or closest available).

Upon FET turn off, instantaneous V across the resistor will be 30A * 0.75 Ohms, or 22.5V. Instantaneous R power will be 30A^2 * 0.75 ohms or 675W (no, you don't need to pick a huge resistor, since this will be a *very* short pulse). Instantaneous D power will be 30A * 1.2V, or 36W. Notice the resistor will absorb most of the energy, and very quickly.

When I wire the resistor in series with the diode, would the resistor go in front of the diode or behind it?

Sorry if this is an annoying question but I have never wired a resistor and diode in series before.

 

I agree, however I'd move it to between the gate and ground. This is a failsafe to turn the MOSFET off in the event of a failure of the PWM signal.

Absolutely correct

When I wire the resistor in series with the diode, would the resistor go in front of the diode or behind it?

Sorry if this is an annoying question but I have never wired a resistor and diode in series before.

Makes no matter.

 

We need some more information about the solenoid.

What is its DC resistance? (12V/30A is 0.4Ω)?

What is its inductance? Closed? Open?

Part number or data sheet would be helpful.