# Calculate transistor params as switch to operate within the maximum limits safely

I want to use mmbt2222 and/or 2n3904 as switch with in it's maximum safe limits continuously not pulses

I have tried to go through the datasheet but couldn't figure out what is the important params. to use to calculate the surrounded resistors

I have tried some values with the transistor base and when I add load the transistor burn within seconds
the circuit that I use here

My question as described in the title: How to Calculate transistor params as switch to operate within the maximum limits safely?

EDIT: I mentioned those transistors as an example, they just happened to be in my collection, I am not asking for circuit design for those transistors , I am asking for a generic way to calculate the params based on those trans. as an example only.

thanks

The two most important equations concerning transistor calculations in this circuit are:

• icollector = ibase * hFE where hFE is the DC current gain given in the datasheet.

• vthreshold =~ 0.65 V this is the minimum requirement for the base emitter voltage to make the transistor conduct.

Now, given your circuit, the minimum requirement of 0.65 Vbe is met, we just don't know yet if your transistor is in saturation mode, since we don't know Vce. Therefore we need to calculate the current through the collector, what can be done by using Ohm's Law and the first property above:

ibase = Ub,r/Rb,r = (5 - 0,65) / 1000 = 0.00435 A

icollector = HFE * ibase = 100 * 0.00435 = 0.435 A

The voltage across the collector resistor is thus 27*0.435 =~ 11.7 Volts. In this case this is the exact boundary of the linear/saturation mode. Using the supplied simulation you can play a bit around with the 27 Ohm resistor and you will see that this calculation is a pretty good approximation if the resistor is in saturation mode (which is wanted if you want to make an electronic switch).

In conclusion, he base resistor determines the actual collector current. The collector resistor determines the mode of the transistor, either linear of saturation.

If you're using Fairchild's transistors, anything can happen, they measure the [base] current in volts, so who knows what they put in those packages, chewing gum maybe?

If you're using something less silly, like ON's datasheet

then we can start thinking about this. Your base current is nowhere near 150mA. In fact it's at most 5mA so the much lower first line should apply, so the saturation voltage shouldn't get so high [because of that]. And ON's datasheet also has pretty graphs on this

But in fact, in your circuit Vce does get high, but because you're not giving it enough base drive, i.e. you are barely saturating the 2222 transistor. I get a beta of 90 in LTspice simulation with those resistor values you have, not the forced beta of 10 this BJT should have when saturated (according to its datasheet). Ok, it doesn't get that low for me either, but you're too left of the sweet spot. A base resistor between 500 and 700 ohms should be more reasonable.

You could in fact us a base resistor as low as 300ohms here. If you go below that, the power dissipation in the transistor actually starts to go up, due to the base current.

Pay close attention to order of colors in the folloing graph (as alas LTspice doens't have a more decent way of identifying stepped data). [Forced] beta keeps going down to about 12 at 100 ohms base resistor (magenta line), but the power disspation in the transisotor bottoms on the red line (300 ohms) and slighly goes up after that. The base resistor power dissipation increases consideralbly too after that. Also, whatever your 5V source is (uC?) might not be able to actually drive it with high of base current.

Another interesting thing to note is that the base current is around 15mA when the sweet spot (for power dissipation) is reached at 300 ohms base resistor... pretty much like the ON datasheet suggested on that first line.

Below I've used a behavioral resistor to vary the base resisotr over time instead of using step. You can better see the curve that power has this way... but the value of the resistor is not that obvious. It goes from 10ohm (left) to 1.01Kohm (right).

Also, beware that the thermal resistance (to air) of the MMBT2222 varies quite a bit with the manufacturer (well at least in datasheets). Fairchild gives 357 °C/W, NXP 500 C/W, ON gives 556 C/W. Even with the latter, at 150mW (for a 300-500 ohm resistor) it should survive, but would be very hot at around 85C. Max junction temp is the usual 150C in all these datasheets.

Regarding the 2N3904, that one has a higher thermal envelope of 625mW (in TO-92) but the die of the 2N3904 (or MMBT3904) can only take 200mA collector current. So what happens with that one is that the junction gets destroyed by the excessive current density (you're giving it over 400mA collector current), and then the thermal limit is exceeded so the package goes up in smoke.

The selection process is basically:

1. Check that the transistor model can handle the load = collector current. (Well, you also need to worry about the CE [stand]off voltage, but that's not exceeded here.)
2. Then figure out the base drive (=resistor) needed to get a low Vce(sat)... so that its power dissipation (SOA=safe operating area) isn't exceeded either.
• if I can go down to 300 ohm why do you think the magic smoke escapes !! – Muhammad Nour Dec 6 '15 at 9:44
• @MuhammadNour: Because at 1K with the 2222 the power dissipation on the transistor is higher. Is the 1K that you have if your simulation/diagram what you've actually used in your [physical] circuit? – Fizz Dec 6 '15 at 9:46
• yes I have used 1K and for the load I have used 27 ohms – Muhammad Nour Dec 6 '15 at 9:47
• Try 500 ohms instead for a 2222. Actually a 2N3904 shouldn't have popped at 1K according to the sim (but does at 2K)... but who knows what you've got in your parts bin. – Fizz Dec 6 '15 at 9:49
• I have popped 2 2n3904 & 1 mmbt2222 with 1K :D, I will try to recalculate the whole thing again after finishing my work, thanks anyway – Muhammad Nour Dec 6 '15 at 9:51

Based on your simulation, you're trying to push over 400mA through that transistor.

From the mmbt2222 datasheet, the Vce(sat) at 500mA is 1.6V.

$$Current \approx 0.444mA=\frac{12V}{27\Omega}$$ $$Power_{transistor}=0.444mA \times 1.6V=711mW$$

Also from the mmbt2222 datasheet, Total Device Dissipation is 350mW.

I'd say the extra 360mW is causing some problems...

• what does it means: Vce(sat) at 500mA is 1.6V, do you mean I can't use 12v @ 500 mA! – Muhammad Nour Dec 6 '15 at 3:01
• It means that according to the datasheet, you will drop 1.6V through the transistor at 500mA. This transistor was not really designed to handle currents that high. 12V is not a problem.. the Vce_max (called V_ceo in the datasheet) is 30V. The current is causing the power problem. – Daniel Dec 6 '15 at 3:07
• If you use the above equations, you can see that the power goes down if you increase 27 ohms to something higher. – Daniel Dec 6 '15 at 3:08
• I really don't want that much power I just want to know how to calculate the params to operate in the safe limits and the transistors I have mentioned in my question happened to be the ones that I have right now – Muhammad Nour Dec 6 '15 at 4:18
• It's kinda laid out for you there for your situation... make the power dissipated less than the total device dissipation. I can't walk you through every calculation you need to guarantee that your part will work in every situation. There's too much to cover. – Daniel Dec 6 '15 at 4:27