So I am working on a project that involves magnets and I need to control the high and low side of each coil. What I got was a S8550 PNP transistor and they keep smoking out on me every time I try to pass power through it. Same thing happened when I started with a 2n3906 PNP.
But I digress, on my meter reads a Hef of about 400 but compared to what? Is it a 400 gain at 1A or 1mA? This is my first work with transistors on a multi-meter and I am new to datasheets. Once I figure out the what the gain is compared to how should I use that to choose the correct resistor for a circuit I am building?
For reference this the circuit I am working on and can't seem to not kill. Note: I have changed the 2N3906 to the S8550 and I keep changing the 4.1K ohm. Also with an LED it works fine, its not until the coil goes in does it fail. If it helps the coil reads 5 ohm resistance.

 

It's likely a gain of 400 at 1 ma, and you can't blow 3.6 amps through a 2N3906.
18V/5 ohms is 3.6 amps.

The usual way is to assume the transistor has a gain of 10. That pours enough current through the base to make the collector act like almost nothing getting in the way of the current. Now, stick to low current loads or buy some bigger transistors.

 

my meter reads a Hef of about 400 but compared to what? Is it a 400 gain at 1A or 1mA?

That's a good question. My meter has an hfe option, but it's generally not useful because you don't know what the collector current is. If you want a meaningful beta measurement, you need a curve tracer.

For reference this the circuit I am working on and can't seem to not kill. Note: I have changed the 2N3906 to the S8550 and I keep changing the 4.1K ohm. Also with an LED it works fine, its not until the coil goes in does it fail. If it helps the coil reads 5 ohm resistance.

As connected, the snubber diode won't protect the transistor. Put it across the CE of the transistor. Cathode on the emitter, anode on the collector.

 

The usual way is to assume the transistor has a gain of 10. That pours enough current through the base to make the collector act like almost nothing getting in the way of the current.

Ok I wasn't sure the best way to handle that but that makes sense.

18V/5 ohms is 3.6 amps.

I didn't even think of putting those two numbers together. That is a lot more amps than I expected and will now account for. I sure you can tell I am new to the site and my god you guys work fast, barely 2 minutes and you solved the problem that plagued me for weeks.

 

As connected, the snubber diode won't protect the transistor. Put it across the CE of the transistor. Cathode on the emitter, anode on the collector.

It should protect it. The coil starts producing voltage to keep the current flowing and as soon as it rises above 0.7V the current has a path through the diode. To kill the current more quickly, either more diodes or some resistance should be added. You can let it build up to some reasonable fraction of the transistor breakdown voltage. The higher the voltage the faster the field energy will be dumped.

 

But I digress, on my meter reads a Hef of about 400 but compared to what? Is it a 400 gain at 1A or 1mA?

In this case it doesn't matter. The Hfe (not Hef) is relevant when operating the transistor in the linear region. Here you are wanting to use it like a switch which means either in cutoff or in saturation.

As others said, in this app you want to drive it into hard saturation which generally means you assume a beta of about ten (even less if it's a high power transistor).

 

It should protect it. The coil starts producing voltage to keep the current flowing and as soon as it rises above 0.7V the current has a path through the diode.

The way I see it, when the switch opens, the inductor will try to continue conducting and the voltage on the collector will rise until the reverse breakdown voltage of the snubber diode or transistor is exceeded. Am I missing something obvious??

 

Am I missing something obvious??

I think so.
Any inductor as a DC load can be isolated into a little circle consisting of the inductor and a diode when the switch opens, whether it is a transistor or a mechanical switch. No reason to complicate it more than that.
Perhaps if you re-draw the circuit with an NPN opening up the ground on a relay coil, it will jump into focus for you that protecting the transistor at the transistor requires better planning that just shunting the inductor with a diode.

 

I think so.
Any inductor as a DC load can be isolated into a little circle consisting of the inductor and a diode when the switch opens, whether it is a transistor or a mechanical switch. No reason to complicate it more than that.

Even when one end of the inductor is connected to ground? If that's the case, I need to do some refresh...

 

two rules:
The current in an inductor cannot change instantaneously
The voltage across a capacitor cannot change instantaneously,

What do those statements mean?

 

two rules:
The current in an inductor cannot change instantaneously
The voltage across a capacitor cannot change instantaneously,

What do those statements mean?

non sequitur

 

When the switch opens, the current tends to continue flowing in the same direction. With one end of the coil grounded, remove the ground symbol and see that the inductor will suck up (positive) current through the diode from what you think of as common (and circle it around).

 

When the switch opens, the current tends to continue flowing in the same direction. With one end of the coil grounded, remove the ground symbol and see that the inductor will suck up (positive) current through the diode from what you think of as common (and circle it around).

What if common is connected to earth ground?

When the inductor is switched low side, I saw the voltage spike caused by the field collapsing as being absorbed into the supply. I never thought of it as a loop...

 

Well, in the inductor example, the circuit will see R of the inductor for a brief moment. (transistor smokes)

 

What if common is connected to earth ground?

What if common is connected to my left elbow?

Irrelevant. The ground symbol is a human construct meant only for convenience when making drawings. Learn to divorce your mind from conventions and think only of how the energy flows, you must.

 

Well, in the inductor example, the circuit will see R of the inductor for a brief moment. (transistor smokes)

That is not a foregone conclusion if the snubber diode works as WBahn and #12 are indicating.

We know the current continues while the field collapses. I'm trying to figure out how it goes into ground (which could be earth ground) and back through the anode of the diode to complete the loop...

 

What if common is connected to my left elbow?

Irrelevant. The ground symbol is a human construct meant only for convenience when making drawings. Learn to divorce your mind from conventions and think only of how the energy flows, you must.

I don't follow the analogy. Earth ground is at zero potential and I don't see how the inductor could force current to flow out of zero potential to a higher potential.

 

Replace the ground symbols with a piece of wire.

Earth ground doesn't have a blinkety-blank thing to do with where the electrons go. They are inside the coil when the switch turns off and they want a way to flow until their energy is dissipated. They will not flow into or out of the planet because the planet is not the other end of the battery.

 

Replace the ground symbols with a piece of wire.

Earth ground isn't an abstract concept. It's a 0V potential. Current that goes there can't come out unless it flows to a more negative potential. The other end of the inductor is at a positive potential while the field is collapsing.

 

Current that goes there can't come out unless it flows to a more negative potential.

Do you really think current can only flow one way at a ground symbol?

Earth ground is at zero potential and I don't see how the inductor could force current to flow out of zero potential to a higher potential.

You were a lot closer with this idea. The inductor CAN force flow out of a common point because energy was stored in the magnetic field of the inductor as the current ramped up when the switch was first closed. That makes the inductor act like a tiny generator or battery at the moment the switch opens. The energy is already in the magnetic package. That's what is trying to resolve to its quiescent state.