![]() So it is better to choose Rb > 1mA for Vbe of 4V. This is the bare minimum to turn the transistor on and can be unpredictable because of variations in beta and the input characteristic curves. and the voltage across the base resistor will be 5V-0.7V so Rb = (5V - 0.7V) / 1mA = 4.3 kohms. This is actually the main reason why beta drops as the transistor saturates. As the transistor saturates, there will not be enough supply to allow beta to remain high. We should be safe in assuming a beta of 100 for most general purpose npn transistors. so Rc = (12V-2V)/100 mA = 100 ohm ( you can subtract another 0.1V for Vce if you want but as you can see it will not matter much ).įor the base resistor we start by assuming 0.7V drop across BE when it is on. The Ired drop will be about 2V (varies slightly with actual Ired and current ) and the supply is 12V. In this case, the collector resistor is set to produce 100 mA when the transistor is saturated. So you can begin your design by assuming approximately 0.7V across the BE and then calculate the required values of the resistors to ensure the transistor is saturated. The BE voltage will change only slightly for large changes in the base and collector currents. Essentially this occurs because the resistors are linear devices and BE junction of the transistor is not. In this case, the ratio of the base resistor to the collector resistor ensures that the transistor will be on. In most simple circuits the exact nature of the input curve is not necessary because the biasing resistors will self regulate the DC base and emitter voltages. There is a sharp knee which is generally around 0.6 to 0.7 V. ![]() Having said that, there is no precise turn on voltage. But for silicon transistors, it is always around 0.6 to 0.7V unless you have multiple BE junctions like a Darlington transistor. Technically, you need the input characteristic curve of the BE junction which is not supplied on the given datasheet. But IMO the lower drive requirement of the BC337 is preferable as it places less stress on the uC and also reduces overall power consumption. Of course, the OP stated that his uC can provide up to 40mA of drive current, so a 2N2222 can certainly be used. It is slower than the 2N2222 as a switching device, but I've used it many times at 38kHz and it's fast enough. The BC337 I suggested has higher beta at all levels. Which is why rule of thumb dictates Ic/10 as Ib for transistors of that class of gain. It's the minimum that has to be considered to ensure proper switching. The gain (beta) will drop even further at full on condition with a Vce of ~0.5V, and still further at lower temperatures. At 1V, it drops to 50 and 1V is not yet full saturation. The minimum under those conditions is 100. But the gain of 300 is only a typical value at 150mA and Vce of 10V. The 2N2222 and its plastic version PN2222 is a classic product and one of the most popular switching transistors, and I've used it numerous times. For an 'on' Ic of 100mA, a base current of 3-5mA is appropriate. A BC337 has the same current ratings as a 2N2222 but has a much higher beta so that a lower base drive can be used. ![]() As already mentioned, the 2N2222 has a max Ic of 500mA (800mA by some manufacturers) and is optimised for switching, but its gain is fairly low and you'll have to drive it with a base current of at least 10mA. The BC547B used in that article is a high-gain transistor so that for the required collector current of 20mA in that example, the base current should be 0.5-1 mA.įor your application, the BC547B is not a good choice because its maximum Ic rating is 100mA. ![]() The general rule of thumb is to use a base current equal to Ic/10 for medium beta transistors and Ic/20 to Ic/40 for high-gain transistors. And then there's the matter of tolerance and variations with temperature to consider. But this was measured in a non-saturated condition and that's not good enough for calculating the base current for hard switching operation. In one of the examples given, the writer bases his calculation of base current on a measured beta of 318. Your aplication requires hard on-off switching which means that the transistor will be in saturation (i.e., at the lowest Vce possible) when it's on. Therefore, wherever possible, it's a good idea to use a transistor with a maximum current rating much higher than that required by the application.Īnother thing I'd like to point out is that the online references suggested before do not take into account the large reduction in current gain (or beta) at low Vce. The nature of silicon transistors is such that the gain falls off substantially long before the maximum current rating is reached. ![]()
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