Measure base spreading resistance. - Page 1
Forum section: electronics / projects

author: orolo
author_rank: Frequent Contributor
timestamp: February 13, 2016, 09:57:56 pm
content: I have been reading about methods to measure rbb' for a BJT, and all involve complex procedures. Thesimplest oneI found involved detecting the base resistance noise, with a low noise preamplifier and a bandpass stage, besides a signal detector. After some thinking, I believe I've a simpler method, quite easy to implement, but I'd like some suggestions about how to realize it.The idea is as follows. It is clear that the simple method of feeding the base of the BJT with a constant current I0, and then measuring the Vbe of the transistor, doesn't work, because that voltage is:$$ Vbe = I_0\cdot R + V_T\cdot \ln\left(\frac{I_0}{I_s}\right) $$so one has to measure both intensity, voltage, and everything depends on Vt (controllable)andIs, which is quite difficult to measure and control.After some thinking, I derived a measurement that depends only on intensity, is independent of Is, and rather insensible to vt. The idea is to charge a big, low ESR, capacitor bank in parallel with the base of the transistor, and then letting it discharge. If the initial current is high enough, the dynamic resistance of the BE diode will be very low, and current decay will depend mostly on Rbb'.The setup is on the first attachment. The cap is charged until constant current I0 is reached, then a MOSFET cuts the power, and discharge starts. Current is measured, and instantaneous current decay dI/dt is measured.After some basic algebra, I found an expression for the base spreading resistance:$$ R_{bb'} \ = \ -\frac{I_0^2 + CV_T\dot{I}}{CI_0\dot{I}} \ \approx \ -\frac{I_0}{\dot{I}C} $$the second formula is roughly 5% worse, if my calculations were right, but it doesn't depend on VT.Okey, so I tested the spice models for two transistors, one with Rb=10 Ohm in is .model, the other with 61.02. The current vs time are on attachment 2. In both cases the decay is almost linear, so it was easy to compute the initial current I0 and I' = dI/dt.For the 2N4401, with exact Rb=10: I got I0=35.9mA, I' = -0.549 A/s, from the graph, which with C=6e-3 F and VT=26e-3 V, gives Rbb' = 10.17 (better formula), Rbb'=10.89 (worse formula). Not bad.For the 2N1711, with exact Rb=61.02: I got I0=36.3mA, I' = -0.01 A/s. Applying the formulas, Rbb' = 60.44 (better), Rbb' = 61.16 (worse).So the formula works, and the circuit seems quite simple. Do you think it is realizable on the cheap? I was thinking of the circuit attached, with a very small current sensing resistance in the emitter, connected to an OP07 for amplification, and a simple arduino ADC at the other side, measuring in well-defined intervals. One thing that bothers me is the huge base current: even if it doesn't destroy the BJT, could it get the base in the high injection area, and need an ideality factor of 2 in the 'more exact' formula? Lesser currents could be used after all.In any case, if I get the time, I'll try setting the circuit up, and doing some measurements.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 14, 2016, 01:06:29 am
content: Hmm, seems plausible, but beware that:- You're dividing by a derivative. In general, signals are noisy, and derivatives even more so (a derivative has the inverse frequency response to an integrator, so an ideal one has infinite gain at infinite frequency, and therefore infinite noise voltage!).- You're probably venturing into the high level injection regime, where the Ebers-Moll equation may not hold (or more precisely, needs to be modified; SPICE uses the Gummel-Poon model, which I forget how well it models this region). I don't know offhand if the difference can simply be taken up by a series resistor RS (and therefore measured as Rbb' -- though exactly which parts come from the SPICE model's RB, RC and RE, may vary?), or if it's more involved than that.- As shown, you're also heavily saturating it, which means a large fraction of base current will flow through the B-C junction as well (not that it has anywhere to go, but out the emitter, anyway). Which I assume will somewhat change the Rbb' figure (additional current paths?), or at least reduce Vbe (though that shouldn't matter, with Vbe factored out at least).- Avoiding saturation by bypassing the collector is probably a bad idea, at least for time constantsthislong. You don't want it heating up during the test, or Vbe drops precipitously without any relation to Rbb'. So that's a concern.- Just leaving the collector open altogether would be another option, so presumably you'd only be measuring it as a diode, where RS = RB + RE.- Or you can always diode-strap it (B+C together) and see what it looks like, that way. (I would guess the apparent value of Rbb' is reduced by hFE, in that case, at least as long as hFE is available -- which will drop at very high pulse currents, where Vce(sat) rises and hFE falls.)Should be easy enough to implement and test. Measure some waveforms, see what the slope is like, see if it's at least in the ballpark of the real thing?!FYI, Art Of Electronics 3 has extensive measurements on a whole page worth of parts! So for a common range of parts, anyway, you might not need to do this at all...Tim

author: uncle_bob
author_rank: Unknown
timestamp: February 14, 2016, 02:04:39 am
content: HiI think you will run into trouble with the fact that your base current is greater than your collector current. The transistor is in saturation and that does not help things much at all. All of these parameters are approximations that are valid over a certain range of conditions. When you go outside that range ... odd things happen. Since the SPICE analysis uses the same approximations, you will need to look at real parts. Running at a base current that is 6X the collector current definitely is past where the models were intended to work.Just in case I missed something:Your supply is 5V in the schematic.Your resistor to the collector shows as 1KThe max collector current will be 5maThe I0 number in your calculations (which I take to be base current) is ~ 35 ma.If I'm wrong and it's the emitter current, you have 5X the base current vs collector current. Still outside the bounds.Bob

author: Guitarbro
author_rank: Newbie
timestamp: February 14, 2016, 05:27:57 am
content: Page 6 of the MAT-02 datasheethttp://www.analog.com/media/en/technical-documentation/obsolete-data-sheets/MAT02.pdfhas a circuit to test log conformity and emitter resistance. I imagine putting the current though the base would yield the base + emitter resistance. you could also connect the collector to keep it out of saturation.

author: orolo
author_rank: Frequent Contributor
timestamp: February 14, 2016, 02:17:53 pm
content: Quote from: T3sl4co1l on February 14, 2016, 01:06:29 amThat's a very subtle observation, I didn't think about it. Fortunately, the big cap low-pass filters the current, so its derivative shouldn't be more noisy that I itself: the integrator cancels the differentiatior. But this is an afterthought: I didn't see the red flag of the derivative. Thank you for pointing it out.Quote from: T3sl4co1l on February 14, 2016, 01:06:29 amThat's my main concern, the high injection region. If beta decay at high currents is an indicator of high injection, a 2N3904 is well into injection at Ib=5mA, according to the datasheet. The lowest base current possible is given by the highest dynamic resistance for the Vbe diode we are willing to accept. At Ib = 1mA, it is 26 Ohm, unacceptable. At 10mA, 2.6, still to high for good resolution. But the 'better formula', if it were not bothered by high injection, takes into account the dynamic resistance, so it could be used at lower currents. I guess only with experimentation can these details be clarified.Quote from: T3sl4co1l on February 14, 2016, 01:06:29 amOf course, and having collector current also distorts the current meter! It is clear that the best option is to leave the collector floating. I am instinctively loath to do it, but in this case it clearly is the way to go. Diode connecting the transistor is also an intriguing option.Quote from: T3sl4co1l on February 14, 2016, 01:06:29 amOh yeah, I've been drooling over these pages since I got the book. But I don't trust my suppliers will give me transistors from the same manufacturer or process. I know 2N4401 have remarkably low rbb' for a jellybean, have heard the same about BC337s, but without my own measures I will have always that lingering doubt.Quote from: uncle_bob on February 14, 2016, 02:04:39 amHi. I agree: as Tim suggested, I think the best way to avoid collector interference is to leave it floating. And you are right, only experimentation will tell if this is a good measure or not. Thanks for the pointers.Quote from: Guitarbro on February 14, 2016, 05:27:57 amHm, they pass a ramp through a transdiode log amplifier, but use a pair of matched transistors though a diff. amp to cancel the Is term. With only a tranny, I think that term wouldn't go away. With only the log amplifier, the output would be:$$ V_T\cdot \ln\frac{I_c}{I_s} + I_c\cdot R_{bb'}$$we can guess Vt (26e-3) is at, or over, Ic in order of magnitude, and the logarithm should be about 10 or more, which would mask a great deal of the resistance. Again, great currents would be needed. The circuit is very interesting, though. Thanks.Quote from: Guitarbro on February 14, 2016, 05:27:57 amThat's the weak point of this measurement method, the emitter resistance is also computed. And the caps series resistance. I was hoping it wouldn't interfere too much.I will definitely implement this idea as soon as I can, and come back with the results. I can't thank you enough for the insightful corrections!

author: uncle_bob
author_rank: Unknown
timestamp: February 14, 2016, 04:52:40 pm
content: Quote from: orolo on February 14, 2016, 02:17:53 pmHiIf you hook collector to base, you effectively bypass Rbb and the measurement stops working. If you supply enough current to the collector to keep it out of saturation:Ib = 36 ma.Beta = 100Ic = 3.6AThat's going to take you off into a very unusual area (and a lot of heat).Emitter resistance is likely on the same rough order of magnitude as base spreading resistance. It also (like Rbb) has some dependance on current. It will be difficult to simply ignore it.Bob

author: sarepairman2
author_rank: Frequent Contributor
timestamp: February 15, 2016, 08:33:14 pm
content: what good is measuring this parameter?

author: dom0
author_rank: Super Contributor
timestamp: February 15, 2016, 08:35:00 pm
content: Contributes noise, defines bandwidth and switching speed (w/ other parameters).

author: uncle_bob
author_rank: Unknown
timestamp: February 15, 2016, 08:36:27 pm
content: Quote from: sarepairman2 on February 15, 2016, 08:33:14 pmHiIt comes into two areas:One is in modeling the noise of the device. It is a "real" resistor so it generates noise.The other is in modeling the device at RF. It is a significant part of the input impedance at higher frequencies.There may be other reasons as well ....Bob

author: orolo
author_rank: Frequent Contributor
timestamp: February 16, 2016, 10:10:24 am
content: Quote from: uncle_bob on February 14, 2016, 04:52:40 pmI don't think so. The base is long and thin, and the emitter is heavily doped, both things contribute to higher base resistance. Besides, if emitter resistance was of the typical order of rbb', 100 Ohms, it would add quite a lot of emitter degeneration in the typical transistor. To rephrase it: since emitter current is 100xbase current, one would expect that a good transistor should have about 1 ohm or less of emitter resistance. Anyway, I've been reviewing transistor measurement manuals, and computing emitter resistance is not particularly challenging: just the dVce/dIc slope at increasing DC collector currents. A few measurements and a linear interpolation.Again, I'm becoming more and more convinced that rbb' current dependence comes mostly from collector current, which floods the base with carriers. Since this measurement method has no collector current, current dependence should be greatly diminished. That is what I hope: of course, without measurements this is just conjecture.Anyway, I'm setting up a rough first version of the measuring circuit; hopefully I will have it working within days. If all goes well, I'll post the data.I will also test transistors for injection effects at high base currents, with collector floating: I don't expect any injection at all. High injection should come from the flooding of the base with collector carriers. I'm expecting that the resistive effects will take over much sooner than any injection is visible.Well, we'll see.

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 16, 2016, 10:38:08 pm
content: Doesn't rbb' depend on emitter current? That is: a single value of rbb' is only valid at one DC bias point.

author: uncle_bob
author_rank: Unknown
timestamp: February 17, 2016, 12:30:00 am
content: Quote from: orolo on February 16, 2016, 10:10:24 amHiWell in a typical RF bipolar, you get a Rbb of 10 to 20 ohms and the emitter (likely ballasted) is in the 5 to 10 ohms.Bob