I've been struggling with details of building accurate (\$\pm1\%\$ tolerance maximum) beta meter. The simplest idea I came with went to a diagram as shown below, with R set to a value providing a \$0.01\ \mathrm{mA}\$ current and a digital ICL7107 meter put to measure \$I_C\$. Obviously, with \$I_B\$ set to a constant value, \$I_C\$ could be easily transposed as a beta meter...


After building this simple circuit in a simulator, I've noticed that the actual \$I_B\$ shifts a little bit as a function of given beta, with about \$10\%\$ difference between \$\beta \rightarrow 0\$ and \$\beta \rightarrow \infty\$; I provided actual test case betas below (\$\beta=10\$ for a low-current transistor and \$\beta=1\mathrm{M}\$ for a high-current Darlington pair, the difference below goes to approximately \${}8\%\$). Of course, I could say this shift is small enough to be ignored - sadly, it's not (at least not in my case).

  • What is the source of this difference?
  • Do actual transistors behave similarly (possibly with recalling actual test results)?

And yes, I'm coming to a conclusion that in this case \$V_{CC}+R\$ can't be treated as a stable base current source. And I'm aware that \$V_{drop}\$ on \$R\$ changes with \$\beta\$, and that \$V_{BE} \mathbin{/} V_{BC}\$ distribution changes as well... but I don't understand how this affects \$I_B\$.


The circuit simulator I used.

  • \$\begingroup\$ In comments you refer to making Ib constant. But idea isn't to make the base current Ib constant! In the simplified model of the transistor, it is a current amplifier which sources a collector current Ic proportional to Ib times HFE. So, if you build a very stable circuit with a stiff bias and degenerated emitter, then when you can substitute many different bipolar transistors and get about the same Ic. What will vary is the base current Ib. High HFE BJTs will require less base drive. Lower HFE require more. So, measure Ib, and its reciprocal gives you HFE (for the given Ic, temperature). \$\endgroup\$ – Kaz May 13 '13 at 22:57
  • \$\begingroup\$ So, basically, Ib isn't a quantity that will be easy to keep constant in BJT circuits, and it's not something that is done in ordinary, everyday circuits. We care about the voltages on the emitter or collector, the collector current, the DC input resistance or AC input impedance, and outward-facing things of that sort which impact surrounding circuits. Base current is more of an internal matter in the circuit. (If someone has a counterexample, please ...). BJT op-amps expose their base currents to the user, but those are in nano or micro amps. They lead to small offsets with large resistors. \$\endgroup\$ – Kaz May 13 '13 at 23:07
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    \$\begingroup\$ sigh It's true, but it doesn't matter for this specific application. The question was not about T circuits, it was about the reason for Ib fluctuations. The idea to measure beta simply is to make Ib constant. My physics lab teacher (PhD on Gdansk University of Technology, about 10 years of expertise in electronics) told me when asked that my idea for measurement is valid and gives correct results within +-10% threshold. An actual application built yesterday proved to be working to the specs - thus I'm sorry, but I find your comments irrelevant to the subject in question and misguiding. \$\endgroup\$ – vaxquis May 14 '13 at 10:39

It is not clear why the experimenter expects \$I_B\$ to be constant in the face of changing transistors, including substitutions of Darlingtons for regular NPN's.

We can assume that the voltage source is ideal (valid: since these results are from a simulation), then the magnitude of \$I_C\$ does not disturb the voltage. (Indeed in the circuit on the right, we have the simulated transistor cheerfully passing 39000 Amperes, yet the source delivers!)

Even two different diodes will not pass exactly the same current if they are hooked up to the same voltage source and same resistor, because they have different curves.

In the case of a Darlington versus NPN base junction, we are looking at two diodes versus one. Plus, the Darlington may incorporate an internal bypass resistor.

The base current can be approximated using simple diode arithmetic: subtract the forward \$V_{BE}\$ drop from 5V, and divide that by the 100 ohm base resistance:

  • Regular NPN: \$(5 - 0.7)/100 = 0.043\$
  • Darlington: \$(5 - 1.4)/100 = 0.036\$.

If we use the exact \$V_{BE}\$ figures given in the diagram, we get the exact \$I_B\$ current values:

  • Regular NPN: \$(5 - 0.72945)/100 = 0.0427055\approx 0.0427\$
  • Darlington: \$(5 - 1.01)/100 = 0.0399\$.

The base current is simply that: application of Ohm's Law to the base resistor, subject to to voltage which remains when \$V_{BE}\$ is subtracted from 5V.

It is not a variation of base current with \$beta\$. That is to say, of course it varies with beta, but beta is perhaps not the relevant parameter to choose as an independent variable for understanding the variation. \$beta\$ is a high level summary of the characteristics of a transistor, connected with the simplified model.

If you wish to hold \$I_B\$ absolutely constant, then you have to drive the base with a current source. (Your circuit simulation software surely has an ideal current source component that you can immediately plant into the circuit.)

Take the most sensitive transistor that you want to measure, and choose the base current so that this transistor is just barely saturated. Less sensitive transistors will then derate the collector current from there. Include a collector resistor to protect the transistors, and as a basis for measuring current.

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    \$\begingroup\$ 0. "It is not clear why the experimenter expects IB to be constant in the face of changing transistors, including substitutions of Darlingtons for regular NPN's." I'm not asking it to be const, I'm asking what's the main factor influencing its change. Again, you're seeing things that aren't there. 1. The 39kA is just figurative. Up the R value and it goes down to reasonable values, this one was just for comparison. 2. Yes, I know about the 'current source solution'. Yet I chose Vcc+R... you still don't get why, despite a couple thousand chars answering your question? \$\endgroup\$ – vaxquis May 14 '13 at 16:49
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    \$\begingroup\$ 3. Using resistors on C/E makes sense only in actual circuits and depends on biasing mode AFAIK. "Protecting" Ts in this case is homogenous with "protecting" green 2,1V LED powered by 1,6V source with a series R... serve no actual purpose. The current gets limited by beta, no need to limit it further. 4. Variations on Ib happened on the same Vd. Your calculations are right but completely irrelevant... you still don't get it, don't you? 5. "It is not a variation of base current with beta. That is to say, of course it varies with beta,"... etc makes no real sense. \$\endgroup\$ – vaxquis May 14 '13 at 16:58
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    \$\begingroup\$ <cont> If Ib changes with beta change, either we have Ib(beta), Ib(something(beta)), or we come up with Ib = const. 3 logical options. There is no other possibility. Lastly, 6, the application worked exactly as expected, with A-meter on Ic and R set to tens of kOhm range (YMMV) it displayed beta within +-10%, mainly due to A-meter being not calibrated well enough yet (damn hard to finetune Vref with those 10/1/0.1 Ohm shunts.) I consider the question answered then and require no further input. (btw, "But idea isn't to make the base current Ib constant!" ... yet this time, it was.) \$\endgroup\$ – vaxquis May 14 '13 at 17:03
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    \$\begingroup\$ * yes, I know that actual Vd varies with I etc. Still, if you're able to find a LED that would get over its allowed I with actual Vd on it lower than actual Vd at allowed I, I'd be kind of surprised and amused and would finally stand corrected in my blissful ignorance. \$\endgroup\$ – vaxquis May 14 '13 at 17:17
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    \$\begingroup\$ Yeah, only it wasn't an actual Darlington pair T. HEAVY SIGH. In the simulation is a regular T with beta set to a Darlington range JUST FOR REFERENCE (FIGURATIVE VALUE). If you read the values *below the chart, you'd probably understood that. Drop doesn't go nowhere near double one... please, either start processing the input data properly or restrain yourself from further input on this subject, as I can clearly see you're neither getting the general concept nor the exact details provided. \$\endgroup\$ – vaxquis May 14 '13 at 18:32

If I understand the data provided by Danny & Kaz correctly, the strict beta measurement by simple current/voltage measurement is neither possible nor sensible due to changes in operating conditions (thermal emissions etc), reverse current gain and other variables related to specific transistors. Ib itself is varied based on the Vbe/Vbc voltages and although the coefficient of change is low, higher precision is not actually possible by simple, one-point measurement.

To sum up, I'll stick with approximations and up the error tolerance by one order of magnitude, to +-10%, leaving a simple bias with known Vcc/R values as a base of beta calculation.

By the way, it seems that increasing Vcc actually makes Ib(beta) more constant. With Vcc in orders of tens of V, Ib ~ const. The actual drift occurs heavily on low (<5V) voltages though, at least in the sim; I suppose that thermal emission on higher voltages would cancel this const-ness though.


Welcome to the art of electronics, your design will have to account for the non-ideal in the real world, most practical circuits will not depend on the exact gain, but will constrain the application to a gain less than the minimum expected of any given transistor type and for lot to lot variations. The beta of the device isn’t a concern so long as it’s higher than the minimum you require. The best way to test an active device is by building a set of curves with a source measure unit or curve tracer. Such data is usually provided in the data to save you the effort. Transistors are active devices that are dynamic (I realize this is obvious, but should be stated regardless) and should be tested as such. If you require matching for your application, then matched devices are available that share the same substrate. You’ll pay a fortune for this of course if you need very tight matching.

  • \$\begingroup\$ Please format your question, I am also a little confused as to how this answers the original posters questions., maybe you could clear that up? \$\endgroup\$ – Voltage Spike Jan 17 at 6:07

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