Darlington reflections

[W Letendre]

Björn Söderholm wrote:
>
> Hi everybody!
>
> I've just built a high power power supply, and it works!!
> But there is a few things i always been wonder at, and that is the
> resistors you find across the base-emitter junction in a darlington
> arrangement with
> descrete transistors. I read something about preventing collector to base
> leakage current in 1st of the transistors from being amplified and
> turning on
> the transistor pair. I'am not sure i know exactley what they mean.
> Does any one know how to calculate those resistors?
> My power supply contain a darlington with 47ohm and 470ohm, first
> transistor is a BD135 and the other is the well known 2N3055.
> Please help!
>
> Thanks in advance.
>
> Björn

Haven't heard of the effect you describe, but have used resistor across
BE junction of output XSTR in Darlington pair in order to provide known,
constant bias current for input XSTR. Without such a resistor, input
XSTR collector current determined by hfe of output XSTR, and may vary
quite a lot. Formula is simple: pick desired bias current, divide into
0.7 volts.

W Letendre
Dir Eng
NEAT

[Björn Söderholm]

[Ted Davis]

In article <32D9DB...@boden.mail.telia.com>,
sm2gxn.s...@boden.mail.telia.com says...

Pick a nice, safe, definitely OFF voltage - whatever you are confortable with
(well under .6 V). Divide it by a few times the maximum leakage current of
the transistor - pick the next lower value. Multiply the voltage you chose
and the current value you used - that's the dissapation - use a resistor with
a wattage rating at least twice that.

--
T.E.D. (tda...@umr.edu)

[Douglas P. McNutt]

In Article <32D9DB...@boden.mail.telia.com>, Bjrn Sderholm
<sm2gxn.s...@boden.mail.telia.com> wrote:
>Hi everybody!

>
the
>resistors you find across the base-emitter junction in a darlington
>arrangement with
>descrete transistors.

I think the resistor is used in darlingtons used as switches to guarantee
that the switch is really off even if the driver transistor leaks a little
bit. In an analog configuration with overall feedback it isn't required.

-> From the USA. The only socialist country that refuses to admit it. <-

[w]

To take it a step further:

At very low collector currents, hfe drops significantly. With no
resistor across the base-emitter of the second transistor (or
alternatively, a current sink pulling current out of the emitter of
the first transistor), the collector current of the first transistor
can be quite low. Therefore its gain (hfe) is low and the benefit of
the Darlington connection is lost. See Gray and Meyer, pp. 220-224
(Third Edition) for a more complete discussion.

Bill Huber, D.Sc., P.E.
Electronics Consulting Engineers
w.h...@ieee.org

W Letendre <w...@ziplink.net> wrote:

>Björn Söderholm wrote:
>>
>> Hi everybody!
>>
>> I've just built a high power power supply, and it works!!

>> But there is a few things i always been wonder at, and that is the

>> resistors you find across the base-emitter junction in a darlington
>> arrangement with

>> descrete transistors. I read something about preventing collector to base
>> leakage current in 1st of the transistors from being amplified and
>> turning on
>> the transistor pair. I'am not sure i know exactley what they mean.
>> Does any one know how to calculate those resistors?
>> My power supply contain a darlington with 47ohm and 470ohm, first
>> transistor is a BD135 and the other is the well known 2N3055.
>> Please help!
>>
>> Thanks in advance.
>>
>> Björn

[Bob Wilson]

W Letendre (w...@ziplink.net) wrote:
: Björn Söderholm wrote:
: >
: > Hi everybody!
: >
: > I've just built a high power power supply, and it works!!
: > But there is a few things i always been wonder at, and that is the
: > resistors you find across the base-emitter junction in a darlington
: > arrangement with
: > descrete transistors. I read something about preventing collector to base
: > leakage current in 1st of the transistors from being amplified and
: > turning on
: > the transistor pair. I'am not sure i know exactley what they mean.
: > Does any one know how to calculate those resistors?
: > My power supply contain a darlington with 47ohm and 470ohm, first
: > transistor is a BD135 and the other is the well known 2N3055.
: > Please help!
: >
: > Thanks in advance.
: >
: > Björn

: Haven't heard of the effect you describe, but have used resistor across
: BE junction of output XSTR in Darlington pair in order to provide known,
: constant bias current for input XSTR. Without such a resistor, input
: XSTR collector current determined by hfe of output XSTR, and may vary
: quite a lot. Formula is simple: pick desired bias current, divide into
: 0.7 volts.

The original poster is correct. The purpose of the emitter base
resistors is described in any basic text on transistors. Whether it is a
darlington or just a single common emitter driver, adding a
base-to-emitter resistor is considered good design practice. Without it,
and especially at high temperature, the collector base leakage current
can prevent the transistor from turning fully off.

Bob.

[Robert H. Penoyer]

>I've just built a high power power supply, and it works!!
>But there is a few things i always been wonder at, and that is the
>resistors you find across the base-emitter junction in a darlington
>arrangement with
>descrete transistors. I read something about preventing collector to base
>leakage current in 1st of the transistors from being amplified and
>turning on
>the transistor pair. I'am not sure i know exactley what they mean.
>Does any one know how to calculate those resistors?
>My power supply contain a darlington with 47ohm and 470ohm, first
>transistor is a BD135 and the other is the well known 2N3055.

If a transistor's collector-base leakage isn't shunted away from the
base, it tends to turn the transistor on.

In a Darlington pair the second transistor is affected not only by its
own leakage current but by the current coming from the first
transistor. The current from the first transistor is its leakage
multiplied by its hFE. This multiplication of leakage currents causes
Darlington pairs to be "leaky" so that they pull a fairly large
current when no drive is applied to the base of the first transistor.

If the base of the first transistor is not connected to a
low-impedance driver, a base-emitter shunt resistor is needed to keep
the first transistor turned off. The base of the second transistor
should be shunted by a second resistor. The second resistor must shunt
the second transistor's leakage current plus any leakage current that
might still be coming from the first transistor (despite the presence
of a shunt resistor there).

If the Darlington pair is driven by a low-impedance source, no shunt
resistor is required at the first transistor.

The method for calculating the values of the shunt resistors is
explained in another of the posts in this article.

[Winfield Hill]

Robert H. Penoyer at rpen...@pacbell.net says...

>
>If the base of the first transistor is not connected to a
>low-impedance driver, a base-emitter shunt resistor is needed to keep
>the first transistor turned off. The base of the second transistor
>should be shunted by a second resistor. The second resistor must shunt
>the second transistor's leakage current plus any leakage current that
>might still be coming from the first transistor (despite the presence
>of a shunt resistor there).

An even more important role for the second resistor, the one across the
base-emitter of the large output transistor, is to provide a current to
discharge its collector-base capacitance and reduce shutoff time.

That's why those resistors have such a low value. For example, the
resistor in a MJ10000 family part is only 15 ohms. This wastes over
65mA during normal use, but provides about -50mA for shutoff.

In the old days many of us in high-speed power design wished we could
gain access to that internal point with a fourth lead, to further
improve performance. Now we just use large MOS FET or IGBT devices.
By using currents over an amp into and out of the gate, I've been able
to get these to switch in well under 10 nanoseconds.

--
Winfield Hill hi...@rowland.org
Rowland Institute for Science
Cambridge, MA 02142

[Bill Sloman]

In article <dmcnutt.1...@news-2.csn.net>, dmc...@macnauchtan.com (Douglas P. McNutt) says:
>
>In Article <32D9DB...@boden.mail.telia.com>, Bjrn Sderholm
><sm2gxn.s...@boden.mail.telia.com> wrote:
>>Hi everybody!
>>

>the
>>resistors you find across the base-emitter junction in a darlington
>>arrangement with
>>descrete transistors.
>

>I think the resistor is used in darlingtons used as switches to guarantee
>that the switch is really off even if the driver transistor leaks a little
>bit. In an analog configuration with overall feedback it isn't required.

Depends whether you want bandwidth. We used to put a resistor between base
and emitter of the output transistor (typically 6k8) in order to get a
defined collector current (typically 100uA) in the input transistor.

Without that resistor, it took a long time (even in analog terms!) for
output current to turn off.

Bill Sloman (slo...@sci.kun.nl) | Precision analog design
TZ/Electronics, Science Faculty, | Fast analog design and layout
Nijmegen University, The Netherlands | Very fast digital design/layout
| e-mail for rates and conditions.