# How can I calculate the base current required for my multistage connected PNP transistor to obtain an obtain an output collector current Ic = 1.5A?

I need to design a circuit that provides an output of 11.5V with a current of 1.5A. This output is drawn from a TIP42 PNP transistor. The transistor gets its base input from a NPN2222 transistor.

The NPN2222 transistor is connected to a 3.3V and cannot draw more than 1mA. The output of the circuit needs to be 0V for any values below 2.4V input, therefore both transistors need to operate in switching mode.

I understand the circuit design however I'm not sure how I can obtain the ideal current flowing into the PNP transistor to obtain the required output. I have tried to analyze the graphs etc however the formulas are confusing me a bit since both transistors operate in switching mode. I'm also unsure how to deal with the output collector current of NPN and inwards base current of the PNP transistor.

Here are the datasheet links for transistors:

I have come up with a schematic (I'm trying to find Ia as per schematic) to solve the problem:

• Could you clarify where exactly the 2.4V come into play? Commented Feb 29 at 16:45
• Please tag this as a homework question, if it is, and show your work and where you are stuck. If it is not h/w then you would surely not be using a TIP42 in this role. Commented Feb 29 at 16:52
• @JonathanS. The 2.4 V is a threshold value. It will be input in place of the 3.3V voltage input. Commented Feb 29 at 17:23
• @Coding247 The threshold will neither be accurate, nor will the circuit switch rapidly. It'll instead transition between "on" and "off" with a soft curve (no matter what resistor values you choose). Is this acceptable? Commented Feb 29 at 17:37
• Yes that should be fine Commented Feb 29 at 18:57

To make an NPN transistor conduct at 2V4, you have to place a two resistor divider on the input, like 3k3 and 1k (or 1k2 for sooner conducting.)

The calculated Rb for the PNP is 360ohm, but rather use 220 or 270 ohm since you don’t know the exact Beta, or measure the real Beta of your Tip42 at Ic=1.5A. I calculated with Beta=50.

• This definitely helps understand the formulae better, however the PNP transistor is required to take the 12V input which means it needs to drop Vce (sat) <= 0.5 so that the load can still operate at atleast 11.5V. So the transistor would be in switching, so will the Ib = Ic/Beta formula still be valid? Commented Feb 29 at 17:26
• The formula is valid in switching also, but to decrease Vce drop to minimum you have to drive the base more then beta ratio. Try, and if it is not sufficient use lower Rb Commented Feb 29 at 17:48

From this OnSemi datasheet for the TIP42 you can find in a number of places where they spec its use as a switch using $$\\frac{I_{_\text{C}}}{I_{_\text{B}}}=10\$$. You may be able to get away with a higher value. But I'd play it safer and stick with that value. This means in your case that $$\I_{_\text{B}}=150\:\text{mA}\$$ to the PNP BJT. That's quite a bit.

The best way to cut that down, and it will work in this case, is to use the NPN BJT with it's full active mode $$\\beta\$$ value and to select one that guarantees $$\\beta\ge 150\$$. This means the NPN BJT will not be able to be used as a switch. It must remain in active mode.

This is the way to set that up:

simulate this circuit – Schematic created using CircuitLab

You will have to find an NPN that can guarantee $$\\beta\ge 150\$$ at the collector current of $$\I_{_\text{C}}=150\:\text{mA}\$$. Even the 2N2222A can probably get there. But you also have to handle $$\300\:\text{mW}\$$ dissipation in the NPN BJT. A TO-92 package is about $$\\frac{200^\circ\text{C}}{\text{W}}\$$ so that's already $$\60^\circ\text{C}\$$ rise over ambient. Very hot. So it would pay to find an NPN BJT in a better package than a TO-92 here.

I've added $$\R_2\$$ to pick up some of the dissipation load. It's not strictly necessary. But if you don't include it, then you will need to find a still bigger NPN package and maybe even provide heat-sinking for it. So the addition of $$\R_2\$$ allows you to off-load some of that heating to a resistor that may be cheaper/easier.

Note that $$\R_1\$$ itself needs to be half a watt all by itself. No avoiding that.

$$\R_3\$$ is a stopper resistor. It's optional here. But it may help avoid ringing/oscillation by dissipating some of transition energy when switching.

$$\R_4\$$ is there to just keep the circuit off if the input control is left floating. It's also not strictly necessary depending on circumstances. But I've added it as a footnote, just in case.

You can replace $$\Q_2\$$ with a Darlington arrangement. This helps guarantee enough $$\\beta\$$ and would likely allow cheaper, more readily available NPN BJTs while still guaranteeing high $$\\beta\$$.

simulate this circuit

$$\Q_2\$$ in the above schematic will still get hot. But perhaps that is okay because you are using cheap readily available NPN BJTs.

Bottom line is that you are pushing the limits given your need for output current and the limitation presented by the maximum drive current you want to support. It pushes things. But not beyond the breaking point. So the above shows a few ways you might consider BJT arrangements.

You might also wish to consider the use of a MOSFETs for a circuit like this. The recombination current goes away and this may make an appropriate circuit a little easier to achieve. MOSFETs commonly can tolerate gate-to-source voltages around the $$\12\:\text{V}\$$ range you need. So extra protection efforts won't be needed to complicate the final result. So that may be a very good approach here, if you are willing to consider MOSFETs. (They cost more, but they work well too.)