Electronic – Why won’t this PNP Transistor trigger


The circuit below should be taking a 3.3V signal from an MCU on MCU_LS12 and outputting a 12V high-side signal.

The output is always 12V. On scoping the base to the output transistor is not being pulled to ground "enough" – only going 12V then to 11.5V.

What am I missing? Input signal on LS12 is a 3.3V from an MCU, sending in a 50% square wave for testing. Why is Q6 not dropping Q8s base to ground? What can I change? Is it the divider?

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Best Answer

Let's draw out the schematic using the EESE editor (as you should have done):


simulate this circuit – Schematic created using CircuitLab

I gather you wired up \$Q_8\$ wrong. As Andy points out, a normal PNP can still act as a PNP transistor if you reversed it. But usually with much worse \$\beta\$ (due to the way things are doped and physically built in a BJT.)

However, what Andy may have missed [assuming I can take you seriously that you are using an MJD127G (datasheet)], then this is a Darlington!! You don't reverse those and expect much. You need to get those arranged correctly!

Since you've mentioned that you've used \$R_{LOAD}=200\:\Omega\$, I will go with that. This means a mere \$I_{C_8}=60\:\textrm{mA}\$. Here's an important chart from the datasheet:

enter image description here

The \$V_{CE_{SAT}}\approx 800\:\textrm{mV}\$ at this current. So you cannot seriously expect better than about \$11\:\textrm{V}\$ across \$R_{LOAD}\$. Ever. You need to plan on that. And less, if your collector current significantly increases.

Note that they use a \$\beta=250\$ for saturation! Pretty significant. But this is a Darlington. So that's to be expected. If your load current is truly only \$60\:\textrm{mA}\$ then your base current only needs to be \$250\:\mu\textrm{A}\$.

Now, it's pretty clear you are also using a Darlington for \$Q_6\$! What?? Oh, well. That thing has a minimum \$\beta=5000\$ at an \$I_C=10\:\textrm{mA}\$! Are you sane? The base current required for \$Q_6\$ here, in this emitter follower configuration is \$50\:\textrm{nA}\$ (assuming that at these low currents that the \$\beta\$ holds (probably not.) In any case, you don't have any base current to speak of on \$Q_6\$.

So what's the value for \$R_{22}\$? It's \$R_{22}=\frac{3.3\:\textrm{V}-1\:\textrm{V}}{250\:\mu\textrm{A}}=9200\:\Omega\$. However, accounting for, say, \$50\:\mu\textrm{A}\$ for \$R_{25}\$, I'd use a \$7.2\:\textrm{k}\Omega\$ there. The value of \$R_{25}\$ should source at most \$50\:\mu\textrm{A}\$, so I'd stick something a \$22\:\textrm{k}\Omega\$ there. (I'd very tempted to make it much larger. But what the heck. Stick with this.) So, again, \$R_{22}=\frac{3.3\:\textrm{V}-1\:\textrm{V}}{250\:\mu\textrm{A}+50\:\mu\textrm{A}}\approx 7.2\:\textrm{k}\:\Omega\$.


simulate this circuit

If you increase the load, just follow through with the calculations.

Why are you using Darlingtons?? Ah. Now you mention you might have a load getting upwards of \$3\:\textrm{A}\$. So it makes sense.

Let's redo things for that kind of load:


simulate this circuit

That Darlington will drop more voltage and will now dissipate a fair amount of power. In fact, it will dissipate more than you dare to apply!! Take a look at the thermal resistance and also the maximum operating temperatures! Assuming you don't do something very special on the board itself to dissipate better, you can't dissipate more than about \$1.5\:\textrm{W}\$ on that device.

So while all the numbers work out "semi-okay," you have several problems.

  1. Dissipation on your Darlington is simply several times too high.
  2. You will lose about \$1.5\:\textrm{V}\$ from your high side supply rail, reaching to your load. If you can live with about \$10.5\:\textrm{V}\$, then that may not be such a problem. But there it is, assuming the Darlington doesn't just burn itself up first.

Other than that, seems okay.

You need to deal with dissipation. This is one of those cases where a MOSFET starts to look pretty good.