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How to interpret the hFE of a datasheet? How should I interpret this data, the image is extracted from a PN2222A datasheet

With Vce = 10v and Ic = 150ma I can obtain an hfe between 100 and 300 I understand that depending on the base current that I apply, this is correct?

But with Vce = 1v and Ic = 150ma I only to obtain an hfe of 50 regardless of the base current that I apply, is this correct?

But with Vce = 1v and Ic = 10ma, I only get an hfe of 75 regardless of the base current that I apply, is this correct?

Thanks

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    \$\begingroup\$ The variation you see is largely because of process variations between one unit and another. You need to design your circuit so it can cope with an hFE anywhere in that range. \$\endgroup\$ Feb 25 at 18:59
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How to interpret the hFE of a datasheet?

\$h_{FE}\$ (capital FE) is the DC gain. It's used when figuring out how to bias the amplifier, it's used in \$I_C = h_{FE} I_B\$. Because of that, the datasheet focuses on the information designers need to know when figuring out biasing. What bias \$I_C\$ current should I use? That will tell me what all my bias resistor values should be. This says that no matter how bad or good my transistor is, having \$I_C\$ around 10 mA will give me the highest gain. It also tells me what happens if I don't pick the "optimal" value for whatever reason. It does tell me the worst case scenario so I can make sure it still works in that case (e.g., make sure my circuit's gain < hFE). It gives me the best best case scenario too. To designers, we just want to know what the "best" case scenario is (just to make sure that, no matter what, we don't exceed some limit somewhere - put too much current through a feedback resistor or something like that) - so that's why we only get the one value of 300. Design for that, and you're safe.

Some important notes: this is different than if you were using this BJT as a switch! The gain in that case is very different. See the Collector-Emitter Saturation Voltage and Base-Emitter Saturation Voltage lines on the datasheet. That's for when the BJT is in saturation (i.e., switch is closed) - and you'll see that the \$I_C / I_B=10\$ in those cases - much smaller than the 100 you'd expect!

And to confuse things a little further, BJTs also have an \$h_{fe}\$ (lower case fe, also known as \$\beta\$) corresponding to the small signal gain. That's used for small signal analysis at higher frequencies. It's typically around a similar value of \$h_{FE}\$, so sometimes it doesn't show up (and you, as a designer, can just design for the worst-case scenario using \$h_{FE}\$.

With Vce = 10v and Ic = 150ma I can obtain an hfe between 100 and 300 I understand that depending on the base current that I apply, this is correct?

But with Vce = 1v and Ic = 150ma I only to obtain an hfe of 50 regardless of the base current that I apply, is this correct?

But with Vce = 1v and Ic = 10ma, I only get an hfe of 75 regardless of the base current that I apply, is this correct?

All of that data is just a measurement at a single Vce and Ic. In fact, your questions answer themselves! Check this out: Vce = 1v and Ic = 150ma I only to obtain an hfe of 50 that means the \$I_B\$ was 3 mA. Vce = 1v and Ic = 10ma, I only get an hfe of 75 that means \$I_B\$ was 0.13 mA. so there you have it, if Vce = 1v, then hFE changes with base current (using the calculations above)!

So how are you supposed to use this datasheet? Design your circuit to have the Vce and IC needed for best gain, if you can. If you can't, see which one you are closest to and design for the worst case scenario. Sometimes datasheets have charts showing hfe vs IC and/or Vce. You can use that to get a better value.

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  • \$\begingroup\$ >> so there you have it, if Vce = 1v, then hFE changes with base current (using the calculations above)! If vce = 10v hfe also changes with the base current, what I don't understand is why for vce = 10 they express a range of 100 - 300 and yet for 1 v they express a fixed hfe \$\endgroup\$
    – Mario
    Feb 25 at 21:26
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    \$\begingroup\$ The value is not fixed, it's just the minimum "worst case scenario." It can be larger than that, but it won't be smaller than that. They didn't tell us the range because it kind of doesn't matter, only the bottom of the range matters. We just need to know the real hfe will be greater than the value they gave us and something less than 300. \$\endgroup\$
    – KD9PDP
    Feb 25 at 21:52
  • \$\begingroup\$ Just one more thing ..., With Vce = 10v and Ic = 150ma that means IB was 1.5 mA and that if I increase the base current, I will reach 300 hFE? \$\endgroup\$
    – Mario
    Feb 26 at 6:13
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    \$\begingroup\$ Most likely, no. If you see cdn-shop.adafruit.com/datasheets/PN2222A.pdf, if you have Vce = 10 V and IC = 500 mA, the minimum hFE is 40. At IC=150 mA, hFE was 100. So IC=100 mA is the sweet spot. Any larger and hFE goes down, any lower and hFE goes down. You're right to try to get your mind around this, it's not as easy as you'd learn from a book! hFE changes with IC and VCE. There is a point that is "best," and everything that's not that combination will have a lower hFE. Here, VCE=10V, IC=150 mA is "best." \$\endgroup\$
    – KD9PDP
    Feb 26 at 19:54
  • \$\begingroup\$ >> There is a point that is "best," and everything that's not that combination will have a lower hFE. Here, VCE=10V, IC=150 mA is "best." Sorry for my English and for abusing so many questions, I have read your answer several times and now I understand it much better, just one more doubt, according to the datasheet the maximum hFE for Vce = 10v and Ic = 150ma is 300, it is correct then that the base current that gives me 300 hFE for Vce = 10v and Ic = 150ma is 0,5ma? \$\endgroup\$
    – Mario
    Feb 27 at 8:26
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But with Vce = 1v and Ic = 150ma I only to obtain an hfe of 50 regardless of the base current that I apply, is this correct?

No, the \$h_{fe}\$ will still vary. Depending on bias current, temperature, manufacturing variation, or phase of the moon.

They just aren't giving any maximum limit on the parameter. This keeps prices low because it's one less way the part could fail a test and need to be discarded rather than sold to you.

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