1
\$\begingroup\$

These are what I know and what I've concluded about transistors. Correct me if I'm wrong:

  1. It's all about current. When we have a determined voltage differential source, BJTs are used to boost the available current. That's if the load needs to draw more curent, the BJT makes it possible.
  2. If 1 is correct then why one would ever buy an expensive voltage regulator with a high current rating? Can't we use a low-current regulator in small package and increase the current using BJTs?
  3. In datasheets, the DC gain (hfe) is given based on Ic current. (Example: 2SC3650) does it mean I should calculate and adjust the available current at base pin according to the Ic and hfe? (i.e. same 2SC3650 datasheet, if the load draws 500 mA, hence Ic = 500 mA, let's assume in a particular temperature hfe is exactly 1000. So what? does it mean the current at base should be at least 0.5 mA?
  4. hfe are given in range. But it is actually a single exact value. Right? i.e. for a single part, a constant temperature, and a constant Ic and Vce, the hfe is @some value and not changing.
  5. The only limiting factor at this current-increasing procedure is the max Ic rating of the part. (e.g 2 Ampere for the 2SC3650) that means even cascading two of these BJTs won't let me have more than 2A available.
  6. But if I add a higher current rated BJT after this 2A one, I can reach higher currents. I feel it's too theoretical to be of any use in real world.
  7. Hfe of some BJTs increases as Ic increases, but some are decreasing as Ic increases. How do I consider this specification when implementing a BJT in a circuit?
\$\endgroup\$
11
  • 1
    \$\begingroup\$ "Can't we use a low-current regulator in small package and increase the current using BJTs?" Yes, but you will get a voltage drop of Vbe and end up having to add further circuitry to compensate for this. \$\endgroup\$
    – Finbarr
    Jul 6, 2021 at 22:55
  • \$\begingroup\$ @Elementronics So, you mean why doesn't everyone just use a low-compliance voltage regulator (tiny thing in a TO-92, for example) and then layer on top of that BJTs or FETs for whatever is needed for a larger current compliance?? \$\endgroup\$
    – jonk
    Jul 6, 2021 at 22:59
  • \$\begingroup\$ right. That's wrong procedure. I should have said a voltage reference rather than a regulator. What I mean is : if I have a high ampere 12V source, I can use a 5v voltage reference chip (which is rated only @100mA ) and a BJT to gain current of 2A. \$\endgroup\$ Jul 6, 2021 at 23:01
  • 1
    \$\begingroup\$ The TL431 datasheet has example circuits combining it with one or more BJTs to make a voltage regulator. So it's perfectly possible, it's just not always the best solution. \$\endgroup\$
    – Finbarr
    Jul 6, 2021 at 23:35
  • 1
    \$\begingroup\$ @Elementronics Sure, you can do things that way. And it's simple and cheap, too. Finally, it relies upon parts that will probably exist almost forever. It's just going to be wasteful and you won't get many of the protections that ICs offer, today. I've done it the way you suggest, so I'm not sitting on a high horse here. If this is what you want to do, just do it. \$\endgroup\$
    – jonk
    Jul 7, 2021 at 1:30

2 Answers 2

2
\$\begingroup\$

I'm answering just a few of your points and comments here, in no particular order:

if I have a high ampere 12V source, I can use a 5v voltage reference chip (which is rated only @100mA ) and a BJT to gain current of 2A.

Yes, you can indeed. In fact, this is essentially what's inside a voltage regulator like your classic 7805 or LM1117. An op amp (or equivalent) is used in these regulators to increase the gain and lower the current draw on the reference, so that the reference itself only needs to supply a few microamps (or whatever the input bias current of the amplifier is).

But if I add a higher current rated BJT after this 2A one, I can reach higher currents. I feel it's too theoretical to be of any use in real world.

You feel wrongly; you've reinvented the Darlington pair (or Sziklai pair, if you mix npn and pnp), a very common way of getting immense current gain at the cost of increased saturation voltage and (in the case of the Darlington) increased base-emitter voltage. This is such a useful configuration that you can get any number of such pairs, conveniently packaged together in a single package that you can treat essentially like a single transistor with much higher gain than normal.

A Darlington pair can be made with two identical transistors, and frequently is when all you care about is getting high gain for a small signal, but when using a power transistor, you usually don't use an expensive high-power part for the input transistor, which only has to carry a fraction of the current of the output one--just get one of those high current parts and use it as the output transistor, and use a cheaper one for the input.

hfe are given in range. But it is actually a single exact value. Right? i.e. for a single part, a constant temperature, and a constant Ic and Vce, the hfe is @some value and not changing.

Not necessarily. Thermal cycles may change the characteristics of a part slowly over time, for instance. When you look at very long time scales, all sorts of weird things can happen to a part--I wouldn't be surprised if you could get electromigration of contaminants within a part, for instance, though I've never looked into it.

If 1 is correct then why one would ever buy an expensive voltage regulator with a high current rating? Can't we use a low-current regulator in small package and increase the current using BJTs?

This is very common and is, in fact, a suggested circuit application in numerous linear regulator datasheets. Have a look at page 21, figure 11 of this datasheet for instance.

\$\endgroup\$
3
  • \$\begingroup\$ Thanks. What I mean is from the given range for Hfe in datasheet, the gain is actually one value.( some unknown value in that range. But only one exact value) so the theoretical calculations are based one and only one value for hfe. (i.e do we calculate the minimum or maximum gain of a Darlington pair? Or we know for a given part, a temp and Ic etc. the hfe is a specific value?) \$\endgroup\$ Jul 7, 2021 at 6:21
  • 1
    \$\begingroup\$ @Elementronics Measuring beta is more trouble than it's worth in most cases--don't design circuits in a way that depends on the exact value of beta. \$\endgroup\$
    – Hearth
    Jul 7, 2021 at 13:28
  • \$\begingroup\$ @Elementronics And again, the exact value of beta for a given part may change with time, even if it's the same part at the same temperature and same collector current and everything--so you can't really rely on it to be stable over time either. I'm not sure just how much it drifts with time, as it's not usually a concern because circuits are usually designed in such a way that the exact value of beta doesn't matter. \$\endgroup\$
    – Hearth
    Jul 7, 2021 at 14:44
1
\$\begingroup\$

If 1 is correct then why one would ever buy an expensive voltage regulator with a high current rating? Can't we use a low-current regulator in small package and increase the current using BJTs?

Voltage regulators include a lot of stuff. They're more than just a voltage reference in a closed loop with a power amplifier. They have thermal and short-circuit protection which is nice. But they can also have things like sequencing and soft start which are really important and not so easy to just add on.

And what you would actually need is an op-amp, a voltage reference, and a power transistor. Just a reference an a transistor won't really be enough to close the negative feedback loop, at least not without the transistor introducing some offset that might vary with temperature and other conditions.

That's three components when you could just have one. Three times the complexity without the features listed above, and you might already be dealing with dozens of other ICs in your design.

Also, the performance of a regulator has been specified so you know how it's going to behave. You might be walking into the design knowing specific voltage tolerance, noise, and transient response requiresments. If you're whipping up your own you might not necessarily know that. In any case, it takes time to calculate it and verify it and you may already have your hands full dealing with other stuff.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.