NPN Transistor Base Resistor Calculations - Page 1
Forum section: electronics / beginners

author: TerraByter
author_rank: Contributor
timestamp: January 03, 2021, 12:32:18 pm
content: Hi all and Happy New Year,i want to learn how to find the Base Resistor for a NPN Transistor, i search and i found a lot of information but also found different methods to find the Base Resistor with different results.i make 3 excel with a 3 different methods how to calculate the Base Resistor.Can you help me to find the right one?

author: Heisen
author_rank: Contributor
timestamp: January 03, 2021, 12:49:40 pm
content: You need to know the minimum beta value of the transistor you are using (available in the datasheet), as beta changes with temperature that's why we look at the minimum beta value to make the circuit conservative.Assume the minimum beta value to be 30. Current at base times the beta value is the current that is allowed through collector to emitter. With the VCC 12v a 10k resistor at the base is enough to flow roughly 1mA into base, which will allow 1x30 = 30mA to flow between Collector and emitter in the worst case, which is higher than your 20mA requirement for the led.

author: Peabody
author_rank: Super Contributor
timestamp: January 03, 2021, 04:03:35 pm
content: Method C gives a much different result than the other two, apparently because of the "overdrive factor". I'm not sure what that is. The other two methods are ok provided the gain is actually 100 at the current you will be drawing. Gain can vary with temperature, and tends to be lower at high current levels. Generally, you take the worst case based on the datasheet, then give yourself a little leeway. That means you may be wasting a little base current, but for a product you need to be sure you've covered the individual variation among parts. For a single project, you can just test it, and find the highest base resistor that still works.I think Method B is technically correct. The calculation of collector current in Method A appears to be wrong. The collector current is (Vcc - Vf - VceSat) / 480. That gives 20mA, and at a hFE of 100 that's a 57K base resistor. But that's before you give yourself any room, so you'd probably end up with 47K.The other thing you have to consider is speed. The faster you have to switch the LED on and off, the higher the base current you may need.

author: TerraByter
author_rank: Contributor
timestamp: January 03, 2021, 04:43:58 pm
content: Thanks Peabody for the detail reply,Quote from: Peabody on January 03, 2021, 04:03:35 pmI read that articlehttps://www.nutsvolts.com/magazine/article/may2015_Securaand make the calculations for Method C. If you read it in the middle say :QuoteQuote from: Peabody on January 03, 2021, 04:03:35 pmI make the correction and now Method A give the same Result with method B.Quote from: Peabody on January 03, 2021, 04:03:35 pmCan you please explain little more that?

author: Peabody
author_rank: Super Contributor
timestamp: January 03, 2021, 05:46:56 pm
content: There are capacitances involved with both bipolar transistors and mosfets. To change the state of something, you have to charge or discharge that capacitance, and to do that more rapidly, you have to source or sink more current. We think of things turning on or off instantaneously, but that's really not the case. Every change takes time. For a simple power switch, that's not something you have to worry about, but if the transistor is part of an oscillator, then switching time comes into play. So if you want something to switch fast, you have to drive it hard.

author: exe
author_rank: Unknown
timestamp: January 03, 2021, 07:21:04 pm
content: If it's a switching application, then the value is not that important, just add a bit of "overdrive" current. Like, two-three times more base current than, say, minimum beta requires. If it's an amplifier, well, then things become more complicated, but in any case a good bjt circuit should work in with a wide range of betas.It's also easy to breadboard and test it.

author: srb1954
author_rank: Super Contributor
timestamp: January 04, 2021, 03:41:45 am
content: Quote from: exe on January 03, 2021, 07:21:04 pmTypically, using bipolar transistors in a simple LED driver or relay driver circuit, the transistor is over-driven with excess base current to ensure that the transistor is fully turned on (saturated) to achieve the minimum possible voltage drop across the transistor (VCEsat). The overdrive of the base current is usually determined as a ratio; Ic=Ib*20 for small signal transistors or Ic=Ib*10 for larger transistors. If you check your transistor data sheet it will typically include various specification figures or graphs for the saturated switching parameters e.g. Vce(sat), Vbe(sat) at specified collector currents and a particular Ic/Ib ratio.Knowing your required Ic you can quickly determine the Ib from the specified Ic/Ib ratio. The base resistor can then be easily calculated Rb <= (Vin - Vbesat)/Ib.

author: gcewing
author_rank: Regular Contributor
timestamp: January 04, 2021, 06:01:50 am
content: The symbol Ω on it own already means "ohm". If you're spelling out in full, it's "ohm" (with a small o), not "Ωhm".

author: emece67
author_rank: Frequent Contributor
timestamp: January 04, 2021, 11:07:41 am
content: .

author: Zero999
author_rank: Super Contributor
timestamp: January 04, 2021, 01:37:51 pm
content: Quote from: srb1954 on January 04, 2021, 03:41:45 amYes, the base current is often selected to drive the transistor hard on, into saturation, but that's normally unnecessary. Quite often, a considerable power saving can be made by not driving it so hard. This of course has to be ballanced with the the slightly higher voltage drop across the transistor. For example, the BC337 is specified with a minimum hFEof 60, when VCE= 1V & IC=300mA and 100, when IC= 100mA, so I normally use those figures when choosing a base resistor, for that transistor. Another good thing is hFEhas a positive temperature coefficient, so the the voltage drop will reduce, as it warms up. Relays will normally activate at slightly below the voltage rating, so a voltage drop of 1V is no problem and drivng it hard, will only save a few hundred mV, so it generally isn't worth it.https://www.onsemi.com/pub/Collateral/BC337-D.PDFIn this case it's an LED, which has a votlage drop of around 2V and the supply voltage is 12V, which drives both the base and LED, I'd set RBto roughlly 100 times the current limiting resistor value. There's no need to be accurate. Use the nearest standard E24 value.What is the LED being used for and what type is it? If it's used as an indicator, under normal office lighting conditions and it's a modern high efficiency red LED, then 20mA is probably overkill and 5mA will probably be bright enough. If it's a crappy old, orange/yellow/green LED, then you probably will need to drive it with 20mA.