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

  • \$\begingroup\$ @Tut I'll fix the R2/R1 mistake, good call. I'm actually looking at using a 2n2222 maybe? Since an IGBT has an insulated gate, should I take any prudence in selecting a heave duty transistor? I think the TIP31 was overkill. The whole design is pretty bad, but I needed to start somewhere. I'm already learning, and, well, that's really the point. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 14:42
  • \$\begingroup\$ @Tut why do I keep getting this wrong!!??!! This is ridiculous! With 1k, it does go out of saturation by the numbers. And good call on 4.7k, I don't know how I missed that either. I'm really striking out on this one. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 15:22
  • \$\begingroup\$ @Tut: could you just write up that the 47 is supposed to be 4.7 and that I missed the 10k, it was supposed to be 1k? This is really bad work on my part, and I apologize for the repeated simple mistakes. I'd delete the whole thing but I think you deserve the "correct answer". \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 15:29

4 Answers 4


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.


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.
  • \$\begingroup\$ I put the 4.7k on, and it let the smoke out of the BJT when I disconnected the "points" cable from ground. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 17:26
  • \$\begingroup\$ Just checked the circuit: the 1k 1W resistor turned into an .87 ohm resistor somehow. I double checked my order, and it says it's 1k, so at least it wasn't another blatant oversight. I'll rebuild it tonight and let you know how it works. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 18:08
  • \$\begingroup\$ Not enough information here, but without the coil resistor your IGBT may be fried. Check your current. \$\endgroup\$
    – Tut
    Commented Oct 23, 2013 at 18:11
  • 1
    \$\begingroup\$ I found the problem: I didn't look closely at the bands on the resistor, and trusted my shipping information from Digikey. They are 1 ohm resistors. The bill says 1k ohm, 1W, but they measure at 1 ohm. What a strange set of coincidences: the 47k instead of the 4.7k saved my BJT for the short term, then when I changed it out, the incorrectly sent resistor problem kicked in and nailed me. Again, once I get this sorted out, I'll try to follow up. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 18:53
  • \$\begingroup\$ Hooked it up with the correct components (1k and 4.7k resistors), replaced the smoked out BJT, and everything worked well down to 5ohms. I'll see if it starts the car tonight. Thanks again for the help! I know 1/4W would have gotten the job done (it's what I'm using now), but I wanted to go overkill. It probably only added to the death of the first BJT. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 19:30

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.

  • \$\begingroup\$ This is all really good. Turns out the 1k was a 1ohm, causing lots of problems, and the 47k was supposed to be a 4.7k (listed a 4k7 on the diagram, and I'm just not used to it, I messed up). Thanks for putting the time in on this answer! \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 19:36
  • \$\begingroup\$ That explains pretty much everything! I am glad that you could fix it quickly. Now 4.7K for Rb while keeping the 1K for Rc will work perfectly. If you find it useful, please, accept my answer. \$\endgroup\$ Commented Oct 24, 2013 at 9:02

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.

  • \$\begingroup\$ Even after fixing my calculation error and using 25 as the gain, I still get (12.38-(6.3m*10k) = -50V, so the device should still go into saturation mode. You found several (really junior grade) mistakes that I made, but it doesn't explain why it's not in saturation, and that's really what's got me perplexed. I agree lowering the resistor value will probably make it work, but I want to be able to calculate this stuff out next time and have it work as expected, like a real engineer. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 14:51
  • \$\begingroup\$ Oh, I thought you were Tut. So you found a couple of my errors, Tut found some other errors. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 15:09

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.

  • \$\begingroup\$ @OlinLanthrop I'm not looking for the IGBT to go into saturation. The TIP31 isn't going into saturation when I hook it up, but when I run the numbers it seems like it should. I don't know why my real circuit isn't following theory. Tut and subsequently pjc50 pointed out that I had the wrong gain, but even at the low side of 25 for the current gain I calculate that the TIP31 should be into saturation. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 15:12
  • \$\begingroup\$ Also, the spec for the IGBT does list a a Vce(saturation) spec. Why do they have this spec if IGBTs don't saturate? And the reason I'm using an IGBT is that it's a good stepping stone into the world of IGBTs, which I hope to use for higher power motor control within the next year. So it's a lower power application that I hope to learn from, though there are indeed a lot of designs that use BJTs for transistorized ignition systems. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 15:17
  • \$\begingroup\$ Tut just found the problem: I used the wrong number in the calculation, which explains everything. \$\endgroup\$
    – Bob
    Commented Oct 23, 2013 at 15:24
  • 1
    \$\begingroup\$ @Bob: Your question was too long, so I didn't read most of it, so misunderstood which transistor you were asking about. The IGBT datasheet may show a "saturation" spec, which is really the maximum voltage it will have accross it at some current. The actual BJT in the IGBT never saturates. From a quick look at R1 and R2 you can see you are asking for a gain of around 47, which sounds like a stretch for a TIP31. TIP31 is a strange choice for that circuit. \$\endgroup\$ Commented Oct 23, 2013 at 16:08

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