Is there a device that can...

You didn't notice Icbo ??

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

I'm not qualified to evaluate your eyesight.

If you don't have data, or the gadget in question is not cost-sensitive, it won't hurt to always add b-e resistors. But in cost-sensitive situations where the numbers are known, one can omit them.

Beta tends to fall off severely at nanoamp currents, and many loads could tolerate microamps of leakage. The obligatory use of b-e resistors is probably a holdover from germanium days, where a base bias network was as likely to remove base current as it was to supply it. A decent silicon ss transistor doesn't leak much.

I admit I usually use b-e resistors, but my gear isn't cost-sensitive.

John

Icbo = Icb with emitter OPEN... note the _uA_ with applied heat.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

OK, a few more words: worst-case, Beta, tempco.

The worst-case Ic of the PNP will be the maximum Beta * the worst-case leakage current. With Beta's up to 1000 your (optimistic?) nanoamps are amplified to microamps, which might just become a problem.

I tried to make a real-life check for the common BC547

but I could not find a leackage figure, much less a tempco. Am I blind?

Wouter van Ooijen

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Yes, but I do not recognise it as the leakage current. Maybe my folly, I am basically a software guy. But it its conditions state Ie = 0 (no emitter current) and a certain Vcb. I would have expected the leakage to be specified for a Vcb = 0 and a certain Vec?

Wouter van Ooijen

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beta times higher, applicable if the base is open (a more common situation, especially without any base-emitter resistor). The resulting Ic can be surprisingly high, at high temperatures.

--
 Thanks,
    - Win

OK. But I seldom use a transistor with open emitter. So how does this Icbo figure relate to CE leakage in a realistic situation?

Wouter van Ooijen

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when you draw a simple model of a transistor it shows two diodes with kathodes or anodes together (PNP/NPN), just like a double diode. now you can leave one leg open without problems of understanding. The collector/base diode is biased inversly and the leakage current is measured against temperature and bias voltage. When you connect the emitter to gnd this current will flow into the other diode and forward bias it. As Win writes, now this current will be multiplied with hfe and cause a collector current that might be substancial. This happens when the base is isolated by hiZ.

--
ciao Ban
Bordighera, Italy

Thanks Wilfried and Ban.

So now for the real calculation: assume a BC547 and BC557. BC547 Icbo is 15nA at 30C, 5uA (!) at 150C. Bmax is 800. BC557 Bmax also 800. So at 30C the resulting leakage-caused Ic of the BC557 (PNP) would be

15nA * 800 * 800 = 9.6 mA. At 150C the BC557 would probably be in saturation.

Now some of you will probably rise up and shout that you have never seen this in real life. That is possible. If you can live with the 'statistical certianty' that you will never combine two of these transistors that happen to have the worst case leakage and the (for this case) worst case Beta, OK, your choice.

Wouter van Ooijen

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Kind of depends.

If it's a fish tank level monitor, immersed in the tank, performance at 150C may be slightly less than relevant. Many consumer goods will have largely melted at 150C.

Indeed, design to worst cases, but it's probably worth actually thinking about what the worst case might possibly be, rather than assuming 150C, and a million rads for everything.

The above calculation which results in 9.6mA worst case is for 30C.

Wouter van Ooijen

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There are other ways to save parts, without compromising reliability. In case you missed my post, in one of the other forks in this thread...

Here's a classic level-shifting pass-element switch, four resistors.

. hv in ---+--- E C ------- out . | B pnp switched . R2 | . | | . '------+ . | . R4 . | . C . logic -- R3 -+- B . control | E npn . R1 | . | gnd . gnd

Current-switched level shifting, by contrast, saves two resistors and also has a few other advantages (there is a requirement that the switched voltage be higher than the logic-control voltage).

. hv in ---+-- E C ------- out . | B Q2 switched . R2 2.7k | pnp . | | . '-----+ . | . C . logic ----- B Q1 . control E npn . 5.0V | . R1 4.3k . | . gnd

The current provided by R1 is the desired base-drive current, plus the base discharge current Vbe/R2. The base-drive current level is provided independently of any changes in the switched power voltage. The circuit is suitable for switching high voltages, for example if Q1 is a mpsA42, up to 300V can be switched. If Q2 is a MOSFET, chose R2/R1 to provide about 12V of gate voltage for the ON state.

. hv in ---+-- S D -------- out . | G Q2 switched . R2 12k | MOSFET . | | p-channel . '-----+ . | . C scale resistor values . logic ----- B Q1 according to desired . control E npn MOSFET switching speed . 5.0V | . R1 4.3k . | . gnd

Note, no Q1 base-emitter resistor is needed for the OFF condition, because the logic-control line is strongly driven to ground, which bypasses any leakage currents, and also rapidly turns off Q1.

--
 Thanks,
    - Win

Frank, that's where the thread is about, read Wins replies a bit back. Your resistor will eliminate the current by factor 800, but still the leakage from the c/b-diode of the 2nd transistor will cause some collector current. But now also a part will be leaking through the e/b-junction of the

1st transistor, so it will be even lower. It will need a second resistor to gnd from the base of T2 to eliminate this effect completely. Many darlingtons have it inbuilt..

--
ciao Ban
Bordighera, Italy

The whole discussion was about not using such resistors.

I guess using a BE resistor on the NPN would reduce the Ic of the NPN to the figure in the datasheet, so the Ic of the PNP would be reduced to 12uA at 30C, 5 mA at 150C. That might be acceptable in some (most?) situations. But do might want to check the Icbo / temperature relation.

Wouter van Ooijen

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"Wouter van Ooijen

" schreef in bericht news: snipped-for-privacy@news.xsall.nl...

With the bc547 base floating? What if there is a small base resistor to ground, on the first NPN, bc547?

--
Thanks, Frank.
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One thing I like to do in these circuits is add a resistor between the collector and the base/gate of the switch transistor to protect against cascading catastrophic failure if the bottom device fails - or more likely I slip with the scope probe!

kevin

the

to

Yes, but I was under the impression it was about leaving out the resistor between e-b of the top PNP transistor. Since the drive for the lower NPN has to come from somewhere, a push-pull logic gate perhaps, or something that has some resistance to ground anyway, when in off condition.

Wouter calculated what may happen with the base of the lower NPN floating, which is perhaps not something that will happen in practice.

Not all that important anyway.

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Thanks, Frank.
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"Wouter van Ooijen

" schreef in bericht news: snipped-for-privacy@news.xsall.nl...

I assumed leaving out only the resistor between e-b of the upper PNP. You would still need one resistor between collector of the lower NPN and the base of the upper PNP, to limit the base current for the PNP.

That's what I meant. In practice the circuit that drives the first NPN will have some resistance to ground. It can be a push-pull output of a logic gate or something. If the driver is another emitter of a NPN transistor, you get 800 * 800 * 800. You have to stop somewhere ;)

So, I figured the thread was only about leaving out the resistor between b-e of the upper PNP. Which does not seem too bad, if it is simple low speed switching.

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Thanks, Frank.
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>
> Wouter van Ooijen
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"Winfield Hill" schreef in bericht news: snipped-for-privacy@drn.newsguy.com...

Hey, I like that one. You just saved me 500 resistors per year ;) Now I have to find a way to buy the Guru a cup of coffee and a donut...

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Thanks, Frank.
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