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enter image description here

These two waveforms are the power dissipation vs time graphs on two bjts (npn and pnp). I got these using LTspice ( alt + right click on the bjts ).

The thing is, while considering will my transistors be able to dissipate those powers without burning themselves, which value of the powers should I use? For example they peak around 3 W, however their average is (ctrl + letf click on the label) shown as around 1W. So, can I consider this 3 W peak as too short to damage the transistors?

transistors are : 2n2222 and 2n2907

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    \$\begingroup\$ What is the timescale for these peaks? 3 W on for a minute then off for a minute is far different than 3 W on for 0.01 sec then off for 0.1 sec. \$\endgroup\$
    – nanofarad
    Nov 30 '19 at 16:50
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    \$\begingroup\$ @ζ-- silly me, time axis is not shown, I will edit in a minute. \$\endgroup\$
    – muyustan
    Nov 30 '19 at 16:56
  • \$\begingroup\$ For class B power amplifier the power dissipation in the output transistors is equal to around \$P = 0.1* \frac{V_{cc}^2}{R_L}\$ and the peak of instantaneous power is around \$P_{peak} =\frac {V_{cc}^2}{4*R_L}\$ for a purely resistive load. \$\endgroup\$
    – G36
    Nov 30 '19 at 17:21
  • \$\begingroup\$ -g36 I use +9v and -9v as supply voltages. So vcc^2 is simply 81 right? \$\endgroup\$
    – muyustan
    Nov 30 '19 at 17:22
  • \$\begingroup\$ Hence you must choose the output stage transistor is such a way that they can handle such a power dissipation (\$P = 0.1*\frac{Vcc^2}{R_L}\$) plus the quiescent current power dissipation. Also transistors the Vce voltage Vcc<<Vce_max and current rating IL_max << IC_max. \$\endgroup\$
    – G36
    Nov 30 '19 at 17:24
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The thing is, while considering will my transistors be able to dissipate those powers without burning themselves, which value of the powers should I use?

Before considering Safe Operating Area and transient thermal impedance, first look at average power, here it is 1W. If average power is enough to make the smoke come out, there's no need to check the other conditions.

Check transistor datasheet, it specifies maximum disipated power of 1.2W at Tc=25°C.

"Maximum dissipated power" as specified in a datasheet is not exactly what it says, because it implies that the case is cooled by a perfect heatsink of infinite size which magically holds the case at 25°C. This ain't gonna happen in the real world. Unless maybe you dip it in liquid HCFC coolant or something.

"Maximum dissipated power" is useful for comparing transistors, but it doesn't tell you how much your transistor will be able to ... dissipate ... in real world conditions, with a real heat sink, inside an enclosure that may be hot, maybe with low airflow, dust, cat hair clogging the vents, etc.

In this case, your problem is that 2N2222 is a low-power transistor, which means its silicon chip is quite small, so it will have a small contact area to the metal case, and the case is also small and not designed to transfer heat efficiently. This explains the enormous ThetaJC, or Thermal Resistance Junction To Case, of 97°C/W. This means with one watt average the chip will be 97°C hotter than the case. And you won't be able to keep the case at 25°C, so your transistor will burn.

This huge ThetaJC comes from the way the TO18 package is constructed. Consider the long and complicated path heat has to travel to get from the chip to the case... This is a very old package...

enter image description here

So you need a transistor with better heat transfer between the chip and the case. For example, a SOT89.

enter image description here

In this package, the chip is mounted on the large copper slug in the middle, which is directly exposed, so you get a "Thermal Resistance, Junction to Leads" of about 5-10 °C/W, which is way better than 97°C/W. You still have to cool the center lead by using PCB copper area as a heat sink, but at least if you keep it cool, the chip will not be 100°C hotter than the leads as in the 2N2222 case. It will only be 5-10°C hotter than the center lead if it dissipates 1W. This is how, with proper heat sinking, a tiny SMD package can safely dissipate a lot more power than a TO18.

If you make a push-pull emitter follower with your transistors, the center pad will be the collector, that is the supply voltages, which is nice because you can add copper area for cooling.

Likewise TO220 has good thermal resistance junction to case because the chip is soldered on a copper slug, and heat travels through in the correct direction (depth-wise through a large cross-section area of copper) and not in the bad direction (lengthwise through a thin metal plate).

enter image description here

In your case, for your average power, I'd go with SOT89 transistors in SMD if space is tight, but that would be pushing it a bit. Most likely DPAK. If you want to prototype it, use TO220 or TO126 transistors and a small heatsink. Even a very small clip-on heatsink will do. Or a TO220 with no heatsink at all, but you will burn your fingers. When prototyping, it is always nice to not burn fingers.

The important thing to remember is that temperature difference between case (or heatsink) and air helps cooling because it moves air by convection. However, temperature difference between chip and case does not help at all, it is a part of your thermal budget that you spend but it does not help cooling. This is why manufacturers optimize packages for low RthJC.

Once you have picked a transistor and package that can safely handle the average power, then you will have to check safe operating area and transient thermal impedance.

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  • \$\begingroup\$ Thanks for such effort, I will examine through your answer carefully. In mean time, what are your new thoughts about requirement of a heatsink with these two things changed; 1: new transistors are TIP41C and TIP42C complementary pairs (datasheet : st.com/resource/en/datasheet/tip41c.pdf ) and 2: new average power is 350mW. I think with these two changes, it may not require a heatsink. \$\endgroup\$
    – muyustan
    Nov 30 '19 at 23:35
  • \$\begingroup\$ by the way, push-pull amplifier assumption is correct, I actually am dealing with it. But, i don't plan to implement this in a pcb, it will be builded on a breadboard. \$\endgroup\$
    – muyustan
    Nov 30 '19 at 23:39
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    \$\begingroup\$ Yes no need for heat sink at 350mW. In fact 2N2222 would work at 350mW too. \$\endgroup\$
    – bobflux
    Dec 1 '19 at 21:28
  • \$\begingroup\$ but it will be hotter than tip41c because the difference in the package type, to-220 is better with cooling. Am I right? I've changed transistors for other reasons btw. \$\endgroup\$
    – muyustan
    Dec 1 '19 at 21:31
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    \$\begingroup\$ Correct. Without heat sink, what matters is "thermal resistance junction to ambient" ie how much surface area is in contact with air, so a bigger package wins. With heat sink, what matters is thermal resistance from junction to case and then to heat sink, that depends how the package is built to efficiently transfer heat, and what material you put between it and the heat sink. \$\endgroup\$
    – bobflux
    Dec 2 '19 at 0:22
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It doesn't look feasible without some sort of improved cooling beyond just a bare transistor, looking at the average power alone. I don't have hard numbers on how short is short enough for peak power dissipation, but half-millisecond pulses seem sufficiently short relative to the thermal time constant of the device.

Let's take a look at the datasheets for the 2n2907:

enter image description here

and the 2n2222 (I wasn't able to find a 2n222 so assumed it was this particular typo): enter image description here

Let's assume that the average power is 1 W for each BJT, ignoring peak power at the current moment. Assuming the transistors are in room-temperature air, the TA (ambient temperature) value controls the maximum rating, meaning that you can only dissipate 400 mW/625 mW on the two transistors before failure may occur.

While the TC values look promising (as they both exceed 1 W), they assume that you can somehow keep the transistor case cooled to 25 deg. C. This is going to be pretty hard, and you're going to run into your derating curves soon enough.

Let's continue assuming that ambient conditions are 25 deg. C still air. Following the T_C dissipation rule, we find that the total device dissipation derates to 1 W at \$25\,\text{C}\ + \frac{500\,\text{mW}}{12\,\text{mW/C}}\$, or about 65 degrees centrigrade (this is 40 degrees above ambient).

We thus need to dissipate a watt of thermal energy from the transistor body, while operating at a case temperature of no more than 40 degrees above ambient, for a thermal resistance of 40 deg. C per watt. According to some of the data in this answer, a good heatsink will barely pull that off, considering every assumption we've already made so far. If the device is enclosed in a case, or ambient temperatures are higher, bets are already off.

See if you can use a larger transistor with a suitable heatsink, such as a TO220 or even a TO3 (which might be overkill), to get some more margin over your operating conditions. Alternatively, if you have a PCB with a good ground plane or copper pour, see if you can use an SMT device with a large thermally-dissipating tab (such as a DPAK device) to improve conductivity down to your PCB.

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  • \$\begingroup\$ however, I cannot understand where 500mW came from. edit: I got it, it is 1.5 - 1 = 0.5, right? \$\endgroup\$
    – muyustan
    Nov 30 '19 at 23:47
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    \$\begingroup\$ @muyustan The device dissipation at Tc=25 deg is 1.5 W, with a derating of 12 mW/deg C. Given that we need to dissipate 1 W, we can derate by up to 500 mW, which allows us to establish the Tc that we will target with our heatsink. \$\endgroup\$
    – nanofarad
    Nov 30 '19 at 23:49
  • \$\begingroup\$ yes, i actually edited my comment, thanks. \$\endgroup\$
    – muyustan
    Nov 30 '19 at 23:49
  • \$\begingroup\$ What would be your new thoughts about requirement of a heatsink with these two things changed; 1: new transistors are TIP41C and TIP42C complementary pairs (datasheet : st.com/resource/en/datasheet/tip41c.pdf ) and 2: new average power is 350mW. I think with these two changes, it may not require a heatsink. I would go and calculate myself but in the datasheet, the way of providing the relevant data is a little bit different from the examples in your answer. \$\endgroup\$
    – muyustan
    Nov 30 '19 at 23:55
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    \$\begingroup\$ @muyustan The datasheet doesn't have a derating curve, so I took some "typical" values--Rja is on the order of 60 or 70 C/W. Taking the worst case, your TO220s should end up at roughly 25 + (70*0.375) = 50ish degrees centrigrade--warm but not unacceptable. If you want better cooling you can always throw on one of those little TO220 heatsinks if your lab has them handy. \$\endgroup\$
    – nanofarad
    Dec 1 '19 at 2:03
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You can't average current for heat dissipation

Heat is usually a function of the square of currrent. It is not proportional to current, so you can't average.

What you can do is compute heat based on peak current being continuous, then multiply by duty cycle.

If it were a sine wave, you could use RMS.

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    \$\begingroup\$ I don't relate these to my question. Where is the current? \$\endgroup\$
    – muyustan
    Dec 1 '19 at 20:18

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