Early Voltage problems

[Nik_2213]

I'm trying to model a 6n139 split-darlington opto-coupler in Spice. I've collected hFE data on the Q1 (~45) and Q2 (~100) of my tube of 25 chips, but I've run into a problem with the Early voltage of Q2.

I set base currents at 5~~100 uA using an LM334 and 15 resistor chain, used a 3V~~12V switched PSU for Vcc.

When you plot iCollector against Vcc for different values of iBase, the tangents are supposed to converge, albeit waaay off to the left.

Well, mine don't. At 10 uA (μA), Early voltage is ~200. At 20 uA ~160, 50 uA ~100, 70 uA ~85 and 100 uA ~65. I've tried ignoring the 3 & 4.5V data-points, but the tangents still don't converge...

Any ideas where I've gone wrong ??
(I'm a hobbyist, not a student.)

 

[FvM]

Apart from the fact, that the Early voltage model is a simplification that won't be exactly met by a real
transistor, I wonder how you measure hFE of the first transistor of 6N139, as the base terminal isn't accessible?

 

[Nik_2213]

"as the Q1 base terminal isn't accessible..."

I connected LED cathode to Q2 emitter, commoned Q1 & Q2 collectors. That let my Peak Atlas component tester recognise entire device as a low-gain Darlington, with hFE mean 7.3, range 6~~9, SD(n-1) 0.7

Also, in H-P Optoelectronics Application Manual (c)1977, p3.7, I read that the LED==> Photodiode CTR is ~0.0015 (0.15%) across a range of devices. This matches the ~0.16% quoted by 6n136 datasheet. Calculating from individual chips' Device & Q2 hFEs gave me approximate figures for Q1.

I've since e-mailed Avago (was Agilent, was H-P) to ask about Early voltages as, IIRC, those may be derived from details of the production process. Sadly, I can't expect an answer as I'm a hobbyist rather than a DotCom.

 

[Nik_2213]

I got non-convergent tangents, giving VAF ranging from 70~~170. After much hair-tearing and deep searching, I've learned that this is a 'well-known' issue in devices with small areas and/or self-heating.
;-(

 

[Nik_2213]

The basic problem with modelling the 6n139 is that Q1's base is not accessible, so *only* the product LED==> photo-diode CTR x Q1 hFE is available...

(CTR = Current Transfer Ratio ;-)

Or so I thought...

Last night, I had a rush of blood to the brain: What if I fed the LED a few milliamps and put my DVM across Q1's emitter and collector ? No Vcc, just the photo-diode's output ? Would its photo-current flow through Q1's base/emitter PN ?

So, today, I tried it. I bridged Q1 emitter to Q2 emitter as latter is connected to substrate. I put an LED in series with the opto's LED to see current was flowing. A 9V battery and 680R, 1K0 or 1K8 fed 9.05, 6.21 or 3.49 mA by DVM. I measured 53.4, 35.4 and 18.6 uA photo-current across Q1 emitter and collector.

These gave CTR of 0.0059, 0.0057 & 0.0053, equivalent to ~0.5 %.

This is approx three (3) times the CTR from the old H-P reference, thrice the quoted figure for the data-sheet and model of 6n136...

If this ratio is representative, it means that a Q1's hFE is only ~15, in marked contrast to the Q2s' measured hFE of ~100...

D'uh, can you see any hole in my methodology or logic ??

 

[FvM]

I understand, that you are operating the photodiode in photoelemnt mode? Then the measurement is incorrect, because the photodiode juntion is connected in parallel to "Q1" BC and BE junction. But you are measuring only the the current amount through BE. To measure the full photocurrent, you must bypass a photodiode to junction voltage zero. Unfortunately, it's impossible in this case.

The basic problem with modelling the 6n139 is that Q1's base is not accessible

Very true. And you can't overcome it.

 

[Nik_2213]

D'uh, yes, of course !!

Thank you.

Why so much of the photo-current found its way to my DVM is another question, probably related to the 'small' geometry of device...

 

[pratchaya.aut]

use resister limit current or calculation new resister

 

[Nik_2213]

"use resistor limit current or calculation new resistor"

Thank you for replying to my post !

Do you mean to add series resistors, measure resulting photo-current and extrapolate back, as you would measure the internal resistance of a cell or battery ?? Neat idea !! I don't know if it will give the equivalent E+B//C series resistors, though...

D'uh, I have considered biasing the device with only ~0.6 V, just enough to prevent current through the BC PN junction and, hopefully, steering most of the photo-current to the emitter. Of course, this just adds another layer of potential confusion...

FWIW, I'm now in a quandry: Avago will *not* supply a Spice model for 6n139. The ~0.16% photo-diode CTR from 6n136 is a factor of three (3) less than even the partial figure I've just calculated for the '139. Darlington transistors *should* have similar gains, but my measurements imply otherwise. I cannot get Q2's Early Voltages to converge because of self-heating and small geometry. I can't yet understand Avago's spice models for the related 6n136, hcpl 4504 or hcpl 4506...

I'm just glad I'm a hobbyist, not a student...

 

[mribble]

Using a resistor to limit current means putting a resistor in series. Then use Ohms law to figure out the current. For instance if you have a 5V batter and put a 1K resistor to limit current you'd have a current of 5mA.

 

[Nik_2213]

moRe: Early Voltage problems

Hi ! I used 1% 680 R, 1k0 then 1k8 with a fresh 9V(nominal) PP3 battery minus a pilot LED's drop to deliver approximately 10, 6 and 3.5 mA, then measured the actual current through 6n139's LED with my DVM as 9.05, 6.21 and 3.49 mA.

For lower currents, I used LM334 and 1% resistor chain ( mA = 67.7/ R Ohms ) but measured current to be sure as an LM334 has nominal ~3% error.

FWIW, I've just found a 3rd-party Spice model of related 6n140. I'm currently evaluating it. However, it uses 2n2222 as its Q1 & Q2 rather than defining NPN models, and seems to use arbitrary CTR etc. Looks like a 'curve matcher' rather than a 'physical' model. D'uh, having exhausted other options, I can live with that...

 

[Nik_2213]

Because it is a 'monolithic' Darlington, BF for Q1 must equal Q2. Only the areas may differ. Given that, I can calculate the CTR...