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Suppose I have this BC635 transistor, if I look at hFE then I'd say that the lowest value is 25 from that table in the Electrical Characteristics category, but if I look at the Figure 3. Base-Emitter Saturation Voltage from where I get the beta characteristic, on the right side of the plot beta is 10.

Now I only learned about the beta characteristic in school, not about hFE, but I read online that they are the same thing, although this is not true judging from the datasheet I linked.

The question is: What should I use in my calculations for the base current if I want to use a transistor as a simple DC switch, beta from the plot or hFE?

What's more troubling is that some datasheets do not have those plots, what should I do there?

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3 Answers 3

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If you are using the transistor as a saturated switch then you decide what Ic/Ib should be.

The saturated characteristics of the transistor are guaranteed at Ic/Ib = 10. Most often you'll use a bit less drive, maybe Ic/Ib = 20 unless you are very close to the limits of the transistor.

hFE is gain in an unsaturated condition (as an amplifier) with a relatively huge Vce (2V in this case). It is a poorly controlled parameter and varies a lot from part-to-part and with temperature.

You can say that you should have forced \$\beta\$ << hFE (specified at a high Vce) if you want consistent results.

Plots are only an indication of typical behavior, you need to look at the guaranteed limits for design. The plots may help you interpret the guaranteed results at intermediate operating conditions.

So, suppose you want to switch 100mA. hFE (Vce = 2V) is typically around 100, and does not tail off much as you go higher to 200mA (Figure 2) so we can be fairly sure it doesn't do anything weird, but the 100 is only typical. We can see that hFE is guaranteed to be 40 at room temperature and 150mA, so it should be at least 40 at room temperature and 100mA. It might drop 30% at low temperature, so we're left with a guarantee of 28 at low temperature and for a low gain unit. I think I would use Ic/Ib = 10 in this case, not 20.

Now that does not mean that you can't pull a random BC635 transistor off the shelf and use Ic/Ib = 50 and have it work most of the time, but that is not proper design. Don't be that guy.

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    \$\begingroup\$ Last sentence is a great tip. \$\endgroup\$
    – efox29
    Jun 28, 2015 at 19:08
  • \$\begingroup\$ I still don't understand how I decide what Ic/Ib should be, because that's a property of the transistor. I also don't understant this statement: Most often you'll use a bit less, maybe Ic/Ib = 20, how can 20 be less than 10? \$\endgroup\$
    – Paul
    Jun 28, 2015 at 20:33
  • \$\begingroup\$ @Paul If you (say) double the base drive current when the transistor has very low Vce the Vce hardly changes. So you could pick Ic/Ib < hFE @ 2V to whatever you like by allowing Vce to drop below 2V. At some point the Vce starts to rise again slightly. By the statement I mean a bit less base drive, not forced beta. I'll edit to make that more clear. \$\endgroup\$ Jun 28, 2015 at 20:45
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In the table, Hfe is specified at a collector-emitter voltage of 2V. In figure 3 the transistor is being forced into heavy saturation i.e. Vce is mainly below 0.1 volts. You usually find that driving a BJT into saturation causes Hfe to dramatically reduce. Bear also in mind that the graph is "typical" and it could be slightly worse.

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\$h_{FE}\$ and \$\beta\$ are two symbols for the same thing.

The reason the saturation plot shows \$I_C=10I_b\$ is because saturation is a behavior where the collector current drops below the level that would be predicted in the forward active region (\$I_C < \beta{}I_b\$). If the situation was \$I_C=\beta{}I_b\$, the part wouldn't be in saturation. In the plot, they are specifying how much lower they drove it before they characterized the voltages.

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