“If we need to switch a larger current, then we will need to use a larger transistor, which means a physically larger package that can dissipate heat more easily.
Author: Elizabeth Simon
If we need to switch a larger current, then we will need to use a larger transistor, which means a physically larger package that can dissipate heat more easily.
When discussing bipolar junction transistors (BJTs), the focus is on low-power devices used to amplify small signals or switch smaller currents, but what if you want to switch larger currents? After all, one of the things we might want to do with a microprocessor is to use the output to run a DC motor or activate a solenoid. The 2N3904 we discussed last time is limited to 200mA collector current. It does not need too many motors or solenoids.
The transistor circuit turns on the solenoid (the model is drawn in LTspice, source: Elizabeth Simon)
The answer should be obvious. We need to use larger transistors. Generally, a device in a TO-92 package cannot dissipate a lot of power, so when I say “bigger,” I mean physically larger packages that can dissipate heat more easily (ie, lower thermal resistance) ).
For example, let’s take a look at the STD1802 transistor data sheet from ST Microelectronics. As usual, it is recommended that you print this data sheet or open it on the screen to make it easier to follow the discussion below.
This part is packaged in a DPAK box. This is a surface mount device (SMD) with a large lug on one side and two leads (and a third lead) on the other side. The large lugs extend below the parts and provide an effective way to dissipate the parts. The lugs of this package type are designed not to rely on connecting an external heat sink, but to be soldered to the PCB. Then, use the copper area on the PCB to spread the heat to the device.
On page 2 of the data sheet, we see that these transistors have a maximum collector-emitter voltage of 60V and a maximum collector current of 3A (peak 6A). These levels should be able to handle most tasks. Of course, we must keep in mind the maximum power consumption of 15W. But please wait a moment, under the condition of 15°C, TC=25ºC or lower. As a reminder, TC usually refers to the case temperature, which is usually higher than the ambient temperature (especially if you want to dissipate any power).
Just for fun, let’s take a look at the thermal resistance and see how it matches the maximum power dissipation. The only thermal resistance we get is junction-to-case. This is what we need for this particular calculation (although, as we will see, it is generally not very useful). As a reminder, the formula for calculating junction temperature is as follows:
After substituting, we get TJ = 25 + 15 * 8.33 =149.95ºC, which is just below the maximum junction temperature of 150ºC.
Of course, we still don’t know how much power is actually dissipated at room temperature, because it depends on the housing temperature, which may be higher than the ambient air temperature. There is a graph (Figure 2 on page 4) that tells us the percentage of maximum power that can be dissipated, but this is also based on the case temperature. To really know how much power we can dissipate, we need to find an application note or something similar so that we can understand the possible differences between the situation and the environment.
Earlier, I mentioned that one of the features of DPAK is that it can use the copper heat sink on the board. This is a complicated subject, but I was able to find this special subject book from Infineon. I also found this “International Rectifier” application note and “ST Microelectronics” application note.
The ST Microelectronics application note is for MOSFETs, but it has specific information about the DPAK package used by this BJT part (we replaced the “drain pad” with the “collector pad”, we should be very close).
If you want to continue learning, the DPAK section will start on page 5 of the ST application notes. First, they show the recommended PCB footprint of DPAK, and then show the relationship between RθJ-PCB and the area of the drain pad. Under the recommended minimum footprint (about 50mm2), it appears that the thermal resistance of the PCB is about 62ºC/W. The power consumption calculation on the next page shows that the maximum power consumption is 2.4W, but this assumes that TJMAX is 175ºC. Because our TJMAX is 150ºC. The maximum power consumption is Pd = 125/62 = 2.02W.
This is much smaller than the 15W value implied by the data sheet. The PCB temperature should be close to the ambient temperature, so this may be a good estimate that we can really dissipate. What if we need to dissipate more power? According to this application note, if we increase the size of the drain pad, we can increase the power consumption. Fortunately, this application note includes a beautiful graph (Figure 4) that shows the allowable power dissipation for different junction-to-ambient temperature differences and drain pad area.
(Source: ST Microelectronics application note)
One thing these figures show is that although an increase in pad area increases power consumption, you will soon reach the point of diminishing returns. Looking at the graph, I see that the maximum power consumption shown is 5W. It seems that our maximum power consumption in this part will be about 3.8W. All of this means that we are still far from the 15W shown in the data. sheet.
This way of specifying the maximum power dissipation makes me feel a bit deceptive. Careless engineers believe that they can use this part in a design that consumes about 15 W of power. The moral of this story is to read the specification carefully and conduct a “reasonability check” if possible (because we just did).
But do we really need to consume so much power to turn on the solenoid? Suppose we will switch a 2A load and the transistor will be in a “off” or fully saturated state. It can be seen from the data sheet that the maximum collector-emitter saturation voltage is 300mV at 2A. Therefore, we can multiply the voltage drop and current to get the power (2A * 300mV = 600mW). is that OK?
Well, if we take a closer look at the table, we will find that IB is 100mA, and in the next row, VBE is 1.2V. We need to add the power consumed by the base current (100mA * 1.2V = 120mW) to our total power. This means that our total power consumption is 600 + 120 = 720mW, which is much smaller than the 15W we discussed earlier, and also much smaller than the 2W we calculated using the application notes.
Therefore, now that we have determined that this may be a suitable transistor, let us simulate the schematic shown earlier as follows:
Results of running SPICE simulation (Source: Elizabeth Simon)
From the simulation results, it is obvious that there are other problems to be solved in this circuit. First, we see that when the transistor is turned off, the peak output voltage exceeds 70V, which exceeds the maximum collector-emitter maximum rated voltage of 60V. This problem can be easily solved by adding a buffer circuit. Another problem is that we need about 100mA of current to drive the base to turn on the transistor. I don’t know any microprocessor that can drive such a large current.