Help understanding Darlington CB leakage current and its ramifications

Those resistor values sound a little high to me, for no extremely rational reason.

All sounds right to me.

Well, if the leakage is there, it's real collector current, but it's likely small. If your base-ground resistor diverts most of the leakage to ground, the collector leakage will *not* be multiplied by beta, which could make it a lot bigger.

Measure a few parts?

Assume it doubles every 10 degrees C.

Most any handheld DVM can measure nanoamps... on its *voltage* range. Just measure the voltage drop across a 1M resistor (1 mv per na) or even across the meter's internal resistance, usually 10 megs. A Fluke and a 9-volt battery and a couple of clip leads will quickly measure Iceo, Icbo, and Icex.

He sounds OK to me. And he's likely aware that some darlingtons, especially the bigger ones, can be very leaky.

John

Some quick thoughts, without going in to any real depth.

I pulled the data sheet, and it's a little sparse - I assume (but do not know) that there is no internal resistor (pull from input device base to lower emitter) ; Assuming that to be so, then it won't affect other things.

The base in a darlington, is the base of the actual top collector device. As (when off, anyway) a reverse biased PN junction, there will be leakage current which increases linearly with temperature, and can usually be assumed to be roughly equal to that of an ordinary bipolar device (unless other guidance is given). (When on, it's more like a low quality current sink towards the emitter).

For typical PN reverse leakage, I suggest you look at some PN diode data sheets (they usually specify such things).

I would be wary of a 100k pull down on *any* bipolar device - you wouldn't need much leakage to give you significant problems. I usually reserve such values for FETs.

You may want to reduce that to more like 10k (or even less). That will still be the vast majority of the current drawn from your processor output pin (the minimum current transfer is 15,000 for the device).

Others here will no doubt weigh in :)

Cheers

PeteS

OK, took your advice and looked at diode leakage on a variety of diodes. Leakage ranges from the low single digit nA range clear up into the double digit uA range. I was dead in the water without knowing how to characterize this specific transistor in comparison to one of the numerous diode models.

So, I decided to run a experiment. I took a transistor and connected its collector to 13.50VDC. I connected a 979K (at 25C) resistor from the base to ground. I connected the emitter to ground. Measured leakage current at room temperature was 0.2nA. I next moved the experiment over to an environmental chamber that is running a test at

70C. I placed the test sample inside (wiring all connected). The resistor dropped in resistance due to the rise in temperature so I had to compensate for this. The end result was 4nA of leakage current. This seems awefully low but makes sense if you look at the levels at room temperature. If you keep halving leakage current for every 10C drop in temperature from 70C it comes out to roughly 0.18nA at room temperature. Close to what I measured. The un certainty of my measurement equipment could have had a role to play in this small discrepancy or perhaps a perfect doubling does not occur at exactly 10C rise. Whatever...

Do the levels of leakage I mention make sense?

At 120C leakage should increase to approximately 128nA. Across my 100K pulldown this results in a 12.8mV drop across the load resistor. IMO this is nothing to be concerned about (assuming my findings are valid).

You are absolutely right. He is OK. As a matter of fact, I have nothing but respect for the man. He challenges me to push myself to be better.

I'm sure you have all been there at one point or another. Trying to impress your mentors. You know, show them that you're approaching their skill base and soon may surpass them,,, At least that's how I feel.

George W. Marutz II

effective low state resistance may be well below that needed >to keep the leakage under control under all circumstances with no >shunt resistor across the darlington.

Unfortunately the Darlington always has voltage across it. The micro can turn on and off freely. So, I cannot count on the base resistor to be a stiffer pull down at all times.

George W. Marutz II

All three of you made mention that my base pull down resistor is too high of a value. Funny, so did the senior design engineer.

Why did I do this? Well, for one thing, I did not realize that using a high resistance value pull down could cause potential problems. I know better now.

My train of thought was to keep current being sourced by the micro pin as low as possible (around 500uA as an arbitrary target). Higher source current can cause other problems in my system that I will not go into now (A/D accuracy). I decided that the high current gain of the darlington over the temperature extremes I will be working with would aid in this effort. I first chose the base resistor from the parts list already on my BOM. I chose another resistor that was already on the BOM for the base pull down. I wanted this value to be greater than the base resistor so I would ensure I would be turning the transistor on. The 100K part fit the bill.

I might have to break down and add a new component to the BOM. Perhaps a 4.7K base resistor and 10K pull down?

Thank you for your valuable input.

Ge0rge

snipped-for-privacy@yahoo.com wrote: (snip)

Reading ASCII drawings requires that you set your newsreader to a fixed width (per character) font, like Courier. This puts the characters on a Cartesian grid.

Clear enough.

He probably got bit on the ass for neglecting this effect at one time.

If you have no data on the input transistor's collector to base leakage versus temperature, how do you know that your 100k resistor will drain the leakage without the voltage building up high enough to keep the darlington from being turned on by it?

There are two transistors in a darlington. The first one can be turned on by its collector to base leakage, and then the second one is driven by that leakage current times the beta of the first one. The second device can also be driven directly by its own collector to base leakage. Many darlingtons contain an internal resistor base to emitter of the second transistor to handle the second case, since you rarely have access to this base.

Almost all of the CB leakage will pass through the resistor, till the base emitter voltage of that device gets high enough to start pushing current through that junction. That voltage may be as low as .3 volts at high temperature. Then that base current is amplified by the transistor beta and is applied either to the second base emitter junction or to the parallel combination of that junction and any internal base emitter resistor.

Yes, it is the voltage that determines how much current passes through the junction.

I don't think so.

No.

(snip)

I would perform and experiment on several devices from different batches with no input resistor and a higher temperature than you expect the circuit to ever see (up to the rated temperature for the device) and the collector voltage at least equal to the supply voltage plus the zener voltage, to get an idea how bad the problem might be. If there is an internal base resistor on the second transistor, the problem may be slight. Then find out by experiment how low a base resistor is needed to bring the collector current down to an essentially fixed value (no amplification of leakage). Then use one

1/10th of that value.

Of course, if your logic signal pulls down to essentially zero volts (and there is no way the logic can become unpowered while the darlington is powered), the effective pull down resistance is the parallel combination of the base resistor and the series resistor to the logic signal. So by the time you get the series resistor low enough for good saturation at lowest temperature and device gain, the effective low state resistance may be well below that needed to keep the leakage under control under all circumstances with no shunt resistor across the darlington.

John - KD5YI (so many John's around here:) )

So what you are saying is that cut off current is the same as leakage current?

Hmm... I guess I had a misunderstanding of what cut-off current meant. THANKS. If I had known this I could have saved myself a lot of anguish. I learned a few valuable leasoons here today.

Ge0rge

to know it and I hope one of the regular >gurus will come along and correct me.

Sounds right to me. I checked your hypothesis before I responded the first time. Go to:

Ge0rge

You might consider using a fet instead of a darlington, especially if the micro is a 5-volt part. A 2N7002 costs a few cents, saves one resistor, and will sink over half an amp with +5 on the gate.

John

Hi, George -

The data sheet gives Icbo of 10 nA max at 25C. At 125C that would increase by 2^10 or to 10 uA. (Probably won't be this high since the spec is at 100V)

The Vbe(on) is given as .8V typical at 50 mA Ic and about 125C (Vbe vs Ic at temperature graph on page 2)

So, Icbo is about 10 uA and 8 uA goes through the 100k leaving 2 uA for Ib. At high temperature, your gain is over 65,000. This could lead to a collector current of 130 mA (if the relay coil permits). If it was my design, I would go as low as I could with the b-e resistance as others have recommended. It may help with noise as well.

I realize you effectively have a 10k shunt via the logic input signal, but what is the microcontroller low output voltage at 125C? And, as John Poplish pointed out, what if, somehow, your micro output tristates to floating? Go for the robustness.

Just some thoughts. Good luck.

John

Actually, George, I ASSUMED it was the same as leakage. If I'm wrong, I need to know it and I hope one of the regular gurus will come along and correct me.

John

to know it and I hope one of the regular >gurus will come along and correct me.

Even so, I would support John Larkin's suggestion that you use a FET. No drain on the micro and none of these other problems. It is a match made in heaven.

Cheers, John

on the micro and none of these other problems. It is >a match made in heaven.

The original design used a FET. That's why the resistor values are the way they are. The darlington was supposed to be a drop in replacement. FETs for this purpose cost 5x as much as the Darlington in my circuit now. Their are only a hand full of options for a N-Channel MOSFET with a 100V breakdown rating in a SOT-23 package. The limited number of competitive products on the market dictate higher prices. When you manufacture 2 million+ products a year this sort of thing adds up.

Thanks

Ge0rge

micro is a 5-volt part. A 2N7002 costs a few >cents, saves one resistor, and will sink over half an amp with >+5 on the gate.

Thanks for the input John. However, read my comment to John - KD5YI. I need a part with a higher VDSS rating. The 2N7002, and many other comparable devices, just would not cut it. I desire as small of package as is possible. The SOT-23 fits the bill for now. Unfortunately you don't find many SOT-23 N-Channel FET offerings with a VDSS of 100V or greater. The price takes a considerable jump over the

60V counterparts.

Oh, one last question. What resistor would you eliminate? The Gate pull down? Why?

Ge0rge

micro is a 5-volt part. A 2N7002 costs a few >cents, saves one resistor, and will sink over half an amp with >+5 on the gate.

Assuming your uP starts up with its output port pin tri-stated, you would still need a pulldown to ensure the fet is off. But you wouldn't need a series resistor.

You can, incidentally, leave the gate lead of a 2N7000/7002 disconnected, and turn it on/off by briefly charging or discharging the gate by touching V+ or ground with one finger and then tapping the gate lead with the other. Once on or off, it will stay that way for a while... about a week is the longest I've verified.

John

Snip:

Excellent. Never came across this one. I was using the $0.55 Siliconix part.

to operate a 12V relay, but I'm sure you >do.

Unsuppressed battery feed / automotive load dump.

Ge0rge

drain on the micro and none of these other problems. It is >a match made in heaven.

What's the actual load voltage?

John

Normal operation range 9VDC to 18VDC. Circuit must survive 100V transient for 400mS. Also 24VDC for 30 minutes.

Ge0rge