hFE of a NPN transistor - Page 1
Forum section: electronics / beginners

author: Fuldup
author_rank: Contributor
timestamp: April 18, 2024, 11:15:27 am
content: Hi i bought a new transistor and i need to know the hFE value of the transistor, but in the datasheet there are 3 different values, can someone explain me why and how should understand it?

author: wasedadoc
author_rank: Super Contributor
timestamp: April 18, 2024, 11:23:05 am
content: The datasheet is only telling you the minimum value it will have at 10 Volts between collector and emitter and at three specific values of collector current.What voltage and collector current will you be using the transistor at? Also in most cases if you need to know the actual hfe, the circuit design is not good.

author: wraper
author_rank: Unknown
timestamp: April 18, 2024, 11:26:25 am
content: There is no single hFE value, it changes depending on conditions transistor is used at.

author: tggzzz
author_rank: Super Contributor
timestamp: April 18, 2024, 11:33:00 am
content: hFEvaries between transistors; manufacturers usually specify minimum and typical values.For any given transistor, hFEvaries with IC, frequency, and probably other parameters such as temperature.Hence circuits are designed so they are tolerant of variations in hFE.Identical transistors, either manually selected and thermally connected, or on the same silicon die will have hFEthattendto track each other.

author: Fuldup
author_rank: Contributor
timestamp: April 18, 2024, 11:44:33 am
content: i want to use the transistor to turn an LED on and off, the transistor will be connected to a 9v 200mah battery and i think that LED can have maximum input of 5v and 25ma.atteched you can find the a screenshot of the datasheet of the LED, i have a green one.

author: wasedadoc
author_rank: Super Contributor
timestamp: April 18, 2024, 11:48:22 am
content: 5 Volt is the maximum the LED will survive if you connect it the wrong way round.25 mA is the maximum continuous current that will not kill the LED. Modern LEDs are quite visible with as little as 5 mA.You should use a series resistor to control the current. There are lots of tutorials on the internet which cover this. Find some and read them.

author: Fuldup
author_rank: Contributor
timestamp: April 18, 2024, 12:03:56 pm
content: ok thanks, but i need to know the hFE of the transistor, if i'm going to give 1ma and 10v to the base of the transistor is the hFE going to be greater than 25 or what to they mean with >25?

author: wasedadoc
author_rank: Super Contributor
timestamp: April 18, 2024, 12:45:35 pm
content: Quote from: Fuldup on April 18, 2024, 12:03:56 pmYou only need to know the minimum hFE. The transistor acts as a switch and goes into saturation meaning it has only a vey small (around 0.4 Volts) between collector and emitter. The current is then determined by the resistor in series with the LED and transistor, NOT by the base current and hFE. You only need to know the minimum hFE so that you can provide sufficient base current. Have you done any background reading on this?

author: TimFox
author_rank: Super Contributor
timestamp: April 18, 2024, 02:34:43 pm
content: There is a useless controversy about if BJTs are controlled by the voltage Vbe or the current Ib.Since Vbe is a monotonic function (approximately logarithmic) of Ib, either concept is valid.However, as a practical matter with a set of devices manufactured to a given part number, hFE varies from part to part (between the minimum and maximum datasheet values), but the Vbe required to obtain a given Ic shows a much smaller variation from part to part.Therefore, for a simple on-off switch circuit, you should ensure that the base current is sufficient (at minimum hFE) to obtain the collector current, taking the Vbe value into account for the drive to the BJT.At higher than minimum hFE, the extra base current will just drive the transistor into "heavier saturation", since the external collector circuit (resistor and LED) limits the collector current.

author: BrokenYugo
author_rank: Super Contributor
timestamp: April 18, 2024, 03:26:27 pm
content: It would be helpful if you provided the datasheet of the transistor in question, searchable PDF preferred.Also note that for indicator use a LED often does not require driving it anywhere near the limit, you'll find modern high brightness types blinding at the usual 20mA. I like to breadboard the LED in question and pick a pleasant looking drive current, then go from there.As a rule of thumb you can just assume beta=10 in this sort of situation and drive the BJT into saturation more often than not, the tradeoff is you waste a few mW driving the base harder than necessary. It is good to learn/know how to actually work it all out though.

author: Zenith
author_rank: Frequent Contributor
timestamp: April 19, 2024, 11:00:03 am
content: Too much analysis in my view.You need a suitable transistor with decent hFE. BC547 can be bought for pennies each if you buy 50 and they are always useful. There are loads of others. You need a ballast resistor to limit the LED current to somewhere between 5mA and 10mA. Since you are using a 9V 200mAh battery, you need no more current through the LED than necessary to make it comfortably visible. That's subjective and so you'll need to experiment. You need to limit the base current so the transistor is saturated, but not enough to damage it. 10K should be about right. Breadboard it to check that it works.You may find this worth watching. He goes into more detail than I just did.

author: CaptDon
author_rank: Super Contributor
timestamp: April 19, 2024, 01:41:23 pm
content: You don't use HFE/Base current to control LED current, far to many variables. You use the transistor as an on/off switch. Take the lowest shown HFE and use a value slightly less for your calculation. A simple calculation process here, assume the transistor has .4vdc across it when ON. Your supply is 9vdc and the LED will have a certain voltage drop also, perhaps 3.5vdc at 25ma. You need to choose a current limiting resistor. 9v - 3.5v = 5.5v Then 5.5v - .4v = 5.1v. A resistor that will provide 25ma, at 5.1v = 204 ohms. Now for your base drive, if the minimum HFE shown was let's say 50 then use a slightly lower value like maybe 40. The collector will pass 25ma and with a gain of 40 the base drive would have to be .625ma or 625 microamps. If your base drive voltage was 3.3vdc, subtract a typical base voltage of .7vdc leaving 2.6vdc you then need a base resistor that will provide 625 microamps at 2.6vdc which equals 4160 ohms. You could set up a spreadsheet calculator to play with the values I have given here.

author: MarkT
author_rank: Frequent Contributor
timestamp: April 19, 2024, 05:45:41 pm
content: Quote from: Fuldup on April 18, 2024, 11:44:33 amThen hFE is irrelevant, it doesn't apply to a transistor used as a switch. You need the Vsat entries in the datasheet. Typically this will be quoted at Ib = 0.1 * Ic, which is what's usually needed to saturate a transistor on. So 2.5mA or so for base current to turn it on.If the LED is 5V and 25mA you will need a current-limiting resistor of (9-5)/0.025 = 160 ohms in series with the LED.

author: mariush
author_rank: Super Contributor
timestamp: April 19, 2024, 06:16:13 pm
content: Quote from: Fuldup on April 18, 2024, 11:44:33 amA green led will have a forward voltage of around 2.2v to 2.4v , it would be silly to power it with a 9v battery because majority of the battery power would be wasted as heat in the resistor that limits the current.You limit the current going through the led using formula derived from Ohm's law : Input voltage - (number of leds in series x forward voltage of led) = Current x ResistanceYour input voltage will be 9v , minus around 0.2-0.4v in the transistor, between collector and emitter.If you want to limit the current to 20mA (0.02A) and you use 2.2v for the forward voltage of the led, then your formula becomes 9v - 0.4v - 1 led x 2.2v = 0.02A x R so R = 6.4/0.02 = 320 ohm ... I'd probably use 330 ohm and get a bit less than 20mA through the led.The problem is you're gonna waste a lot of energy in the resistor : Power = Current2xR = 0.02 x 0.02 x 320 = 0.128 watts .... while your led will consume 2.2v x 0.02A = 0.044 wattsChange your battery to 2 AA rechargeable batteries (2.4v) or 2 alkaline batteries (3v) and not only you'll have more energy, you'll also waste less in the resistor.With rechargeable batteries you'd have R = (2.4v - 2.2v ) / 0.02A = 10 ... so a 10 ohm resistor would be perfect, and the power wasted in resistor would be 0.004 wattsYou want to use the transistor as an on/off switch, you won't use it to limit the current. So even with a hFe of 25 you'd get the transistor going with as little as 1mA on the base.Assuming same batteries are used to turn transistor on or offInput voltage - ~0.6v drop on the base = Current x Resistance ... for 1mA you'd have R = (2.4v - 0.6v) / 0.001A = 1800 ohm ... so even with a 1kOhm, you'd get more than 1mA on the base.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: April 21, 2024, 11:13:55 am
content: I don't know which transistor you have, but it's a poor choice for controlling an LED. Transistors with an Hfe < 50 are usually high power transistors, while transistors for currents upto 100mA have an Hfe somewhere between 200 and 1000.And for controlling an LED, you always need a series resistor (or a complicated circuit) and the transistor is used only as a switch.So if you want 5mA though the LED(Plenty for signal LED's), then divide that by the minimum Hfe (25 in this case) and then calculate the base resistor to set the base current to 5mA / 25 = 200uA.