Saturation means that the transistor could support more collector current than is available. You can get a lot done with a fairly basic model of a transistor, and we're going to stick to that here in this answer. I'm glossing over some semiconductor physics and being a little loose with the terms, but getting into the details would just cause confusion for anyone just trying to learn the basics.
#7 SEGMENT COMMON ANODE FULL#
So what would happen if you gave it the same 1 mA base current but didn't allow it to have the full 50 mA collector current? That's basically what saturation is. For example, if you put 1 mA into the base and out the emitter, and you find that you can then put 50 mA into the collector and out the emitter before the C-E voltage starts going up appreciably, then the transistor has a gain of 50. That ratio of "a lot" to "a little" is the gain. In the case of a NPN, such as those shown here, a little bit of base to emitter current allows a lot of collector to emitter current. To a first order approximation, you can think of a BJT as having some particular current gain. The latter is referred to as "saturation" in bipolar junction transistors (BJTs). Generally when you're switching things on/off with transistors, you want the transistor to be either full off or full on. This is actually a pretty good naive question.įirst, some background. It was asked in a comment "Why would you want to run the switches out of saturation?". Added about transistors not in saturation Again, consult the datasheet for the real values for your LEDs. Most 7-segment LEDs are designed for this and you can usually go up to 5x or so peak current relative to the average. In that case, you can increase the current when on so that the average is still bright enough. For example, with three multiplexed digits the LEDs will be on at most 1/3 of the time. In that case, the average on-time will be the reciprocal of the number of digits.
If you are multiplexing multiple of them, then you also want to switch the anodes so that only one digit is enabled at a time. If you are just driving a single 7-segment digit, then the above applies directly. Again, that leaves a lot of slop from 12 V. That means the supply can go as high as 4.3 V + 10 V + 2.1 V = 16.4 V. (200 mW)/(20 mA) = 10 V, which is the maximum C-E voltage drop at 20 mA. The maximum supply voltage is limited by the power dissipation capability of the transistor. 12 V is therefore obviously plenty, and you can tolerate significant sag in that 12 V without the LED brightnesses changing in a noticeable way. That means this needs a minimum supply voltage of 4.3 V + 1 V + 2.1 V = 7.4 V. Let's say we want at least 1 V across the transistor so that it is past the low voltage regime. We already know the emitter of a transistor that is on will be at 4.3 V. This method works over a wide supply voltage range. 20 mA is usually the maximum continuous current such LEDs are rated for, but check the datasheet. Due to the gain of the transistor, most (around 98% or more) of this current will come from the collector, which means thru the LED.
The emitter current is therefore (4.3 V)/(220 Ω) = 20 mA in this example. When on, the base of a transistor will be driven to 5 V. The LED current is controlled by the emitter resistor value, and is largely independent of the supply voltage. All you need is a single NPN transistor and emitter resistor per segment. The SEGx lines are 0-5 V digital outputs from your microcontroller.
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