To drive 24v relay with arduino, please look at the schematic and help

[Zero999]

I_B = I_C/10 is just a general rule of thumb to ensure the transistor fully saturates, i.e. passes the desired collector current, with a very low collector-emitter voltage. Quite often the figures for the saturation voltage on the data sheet are quoted for a forced h_FE of 10. This isn't always the case. Note that the BC546's saturation region is specified with I_B = I_C/20.

Data sheets often specify h_FE, with much higher collector-emitter voltages, normally 1V, 5V, or 10V, so beware, if you don't supply enough base current the voltage drop can be excessive. Fortunately the BC337's h_FE is specified with a V_CE of 1V, which is low enough in this application, so all you need to do is divide the load current, by the minimum h_FE figure, to give the minimum base current. Unfortunately the PN2222A and BC546 h_FE ratings are specified with higher collector-emitter voltages, which is why I recommended the BC337. You can use the other transistors, but need to multiply the minimum base current, calculated by their hFE ratings by a factor of two or three, otherwise the voltage drop might be too high. Look at the V_CE vs I_B curves to give more of an idea of the minimum base current required.

[gcewing]

As what I understand hFE is a ration of current flowing between C E and B E but when i look at hFE section in datasheet it shows different unmarked (without units) ranges as MIN, MAX I do not understand,

hFE is a ratio of two currents, so it has no units.

Yes, it varies wildly from one specimen of transistor to another. One copes with this by designing circuits so that they don't depend on hFE very much. When using a transistor as a switch, this means overestimating the required base current to ensure the transistor its turned on hard. A rule of thumb often used is to take the minimum hFE from the datasheet and divide it by 10. Use common sense, though -- don't exceed the maximum allowable base current.
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[Zero999]

As stated above the h_FE varies widely, from part to part, even of the same batch and changes with the operating conditions such as: current, collector-emitter voltage and temperature. It should not be relied upon.

A BJT has three operating regions.

Cut-off. no base current is supplied and the collector current is tiny, theoretically zero, but there will be some leakage.

Active. The collector current varies, depending on the base-emitter voltage and current. Increasing the base current, results in a proportionate rise in collector current, depending on the h_FE. For example, if a part has an h_FE of 500, when V_CE = 10V, then when I_C = 10mA, I_B = IC/h_FE = 10mA/500 = 0.02mA. If the base current is increased, the collector current will increase by the same factor, so setting I_B to 0.04mA, will result in a collector current of 20mA.

Saturation. Beyond a certain point, increasing the base current further, no longer results in an increase in collector current, which is now mostly set by the collector load resistance. The transistor will still stop a small voltage, known as the saturation voltage. To achieve this, the base current needs to be much higher, than that in the active region.

In reality, there's some crossover between one operating region and the next, as the h_FE varies. When in cut-off there should be no current, apart from a small leakage, which increases, as the base-emitter junction is forward biased. The h_FE is low, at tiny currents, but increases rapidly, in the active region. Past a certain point, the h_FE falls, especially as the collector-emitter voltage  drops below, the base-emitter voltage, when the collector-base diode junction starts diverting some of the base current to the collector.

In short. You ideally want the transistor to be in the saturation region, when it's turned on. Unfortunately to get a really low, saturation voltage, a lot of base drive is required. I_B = I_C/10 is often specified on data sheets, but in reality, you can often get away with much less, in many applications. Your relay is run off 24V, so a voltage drop of 1V wouldn't be a big deal, which is why I suggested the BC337, because its h_FE is specified when V_CE = 1V.