Why is this BJT not going into saturation?

Question

I built a project based off of this description, but substituted an STGP7NC60HD IGBT. See the schematic below (Note: IGBT symbol doesn't exist as of this post, so JFET was substituted). The coil and points breaker are external to the circuit.

I soldered everything together and tested it on an old car that has its coil and points configured like this circuit (points connect primary to ground). The car cranked, but no ignition.

I took the circuit inside, hooked it up a 12V power supply, and substituted a 12V DC motor for the coil. Not a perfect substitute, but what I had at hand and my power supply wouldn't be able to handle power required to run the coil. I found that when I disconnected the lead that hooks up to the points from ground (the equivalent of the points breaker opening), the motor didn't stop as expected. I took the following readings with the equivalent of the points open:

Base voltage of the BJT: .638V (note that the data sheet lists base emitter saturation voltage at 1.8V).

Collector of the BJT/gate of the IGBT: 12.26V

IGBT collector: 1V

12V DC rail: 12.38V

This means \$i_b=(12.38-.638)/47k = .25mA\$. With a current gain of 100 (EDIT: this is not the correct value! Thanks Tut!), I calculated \$i_c=25mA\$, and \$ v_c = (12.38-25m*10k)=-237V\$, and this being much more than the rail could handle would put the device well into saturation mode.

But these readings indicate the BJT transistor didn't go into saturation. I'm a bit lost, and don't understand why the base emitter voltage didn't go to 1.8V, and why the BJT didn't go into saturation. My solid state knowledge/skills/experience never was very good, and it's been rusting for 5+ years now. Can someone help me understand what I'm missing?

EDIT: I just bench tested the device with a 100, then 40, then 20 ohm resistor as load with pretty much the same result: \$V_{be}=.63V\$, so the BJT isn't going into saturation and shutting the IGBT off.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

There are a few mistakes in the question:

1. The original reference design called for 4K7 (4.7K) ohm resistor for R1. This should fix your problem. Note that the specs for hfe (DC current gain) are typical and not guaranteed, so you need to design in some margin (as was done in the reference design).
2. There was a mistake in your equation: vc=(12.38−25m∗47k) and later vc=(12.38−25m∗10k). You need to use the value of R2 (1K in your schematic). Also note that the result of the calculation is negative but the actual value of Vc will not go below Vce-sat (all other things being correct, a negative value would indicate saturation and the actual Ic would be lower.

The specified 4.7K resistor for R1 should correct the problem.

Edit:

One way of working out an appropriate value for R1 would be to work backwards from the collector. Assuming you want R2 to be 1K (for adequate edge rates for the IGBT in this case):

1. Calculate the saturation current (we'll ignore Vce and call it 0) ... Ic = 12V / 1000 ohm = 0.012A
2. Calculate the necessary base current with some margin (we will use hfe=10 to be safe) ... Ib = Ic / hfe = 0.0012A
3. Calculate the value for R1 needed to achieve the above base current (0.7V assumed for Vbe) ... R1(max) = (12V -0.7V) / 0.0012A = 9417 ohm. With some more margin (allows for temperature, capacitor leakage or ?), 4.7K is a good choice for R1.

Answer 2

I will try to summarize and give some hints into the issues you describe:

1. Actually, there is no such a thing as "base-emitter saturation". The so called Vbe(sat) is an indication of the instantaneous maximum voltage you may apply directly to a base-emitter junction without blowing the base-emitter diode. It is even more mileading, if you were to put the indicated 1.8V for more than a few microseconds, the base-emitter current will blow the device.

   You can safely ignore the Vbe(sat) parameter, as it provides no useful data to you or your circuit under normal/designed operation. I suspect you couldn't get the BJT into saturation in first place, so ended up looking at every parameter in the datasheet and then this one caught your attention. Please, ignore it for now.

   Remember the BE junction is almost in every sense as a simple DIODE. In fact, this single diode will govern the behaviour of your entire BJT. If you go to the datasheet and check the V-I curve of the BE junction, you will see that for low current values (<1 mA), the VBE drop will be around 0.6V. Moreover, your first rule of thumb when analyzing a BJT circuit: the VBE will be 0.6V - 0.8V for most common (practical) base-emitter currents (0.1mA - 100mA).

   So, when first looking at this circuit you will mentally do: 12V (power rail) - 0.7V (vbe drop) which is about 11V, divided by 47K which is about 0.2 mA. Don't worry about precision here. The most importante thing now is grasping the behaviour of the circuit.

2. Now, you will try to apply the (aprox.) linear relationship among Ic and Ibe, assuming the hfe or beta parameter can and will vary a lot from device to device. According to the datasheet, the hfe will be higher than 25, so your Ic will try to reach a minimum of Ic = 0,2mA * 25 = 5 mA.

   However, as you noticed, while the BJT is "ramping up" the collector current up to a minimum of 5 mA, the drop in the 1K resistor will make the Voltage at the collector start moving down. For these values and the hfe(min) of 25, you will get a Vc of 12V - 5mA * 1K = 12V - 5V = 7V (in the worst case, corresponding to the minimum value of hfe). In the best case, your hfe will be higher and Vc could go lower, putting the BJT into hard saturation (Vc = 0.7V).

   I suggest you change the 1K resistor value to a higher value, so as to make sure you will saturate the BJT with the base-emitter current you currently have. If you want to do the math, you will need a Rc of (12V - 0.7V) / 5mA = 2.2K or higher in order to make sure that the BJT is entering saturation mode.

In any case, there is no way that the Vc is stuck at such a high voltage (almost the rail voltage), unless:

• The BJT or any other of the elements in your circuit is not properly wired. I kindly suggest that you check again all the connections and the pin naming of the BJT and IGBT.
• The BJT is damaged, which may have happened due to an ESD shock or overloading.
• One or more of the components you are using have different values than the indicated by the schematic. [This is exactly what happened, according to the author of the question].

I have tried to give a detailed answer as I suspect other people starting with electronics and/or debugging their first circuits may find this information useful.

Answer 3

hFE is listed as min of 25, not 100, at an IC of 1A, which implies an Ib of 40ma.

I can't quite justify this numerically, but that transistor looks too large for that Ib. Values of 10k or 20k seem more reasonable for R1.

Answer 4

IGBTs don't saturate. They are basically a BJT with a FET connected between collector and base. This means the collector voltage can't get less than the base voltage.

A better answer is to use a BJT directly. You don't need a FET to drive it for you in this case.