Phototransistor with three pins

G

Geir Klemetsen

Guest
I have a set of one sender (led) and one retriever (phototransistor) from a
disassembled old printer.

The phototransistor is quite similar to
http://www.allproducts.com.tw/manufacture1/everlight/p18-s.jpg the left one
apart from it has three pins, not two.

How to get the light signal from the diode into an electric analog signal
from the transistor?


With only two pins, it's wery easy:

VCC
+
|
|
.-.
| |
| |
'-' Output
| |\
o--------| >--o
\ | |/
\ -
^
|
|
===
GND
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

But with a 3 pins phototransistor, I don't know how to assemble it to do
the same as in the chematic above. I have googled for a tutorial, but
couldn't find any.

TIA
 
"Geir Klemetsen" <regeirgarbagespam@start.no> wrote in message
news:djrd72$khn$1@services.kq.no...
But with a 3 pins phototransistor, I don't know how to assemble it
to do the same as in the schematic above.
Click to expand...
If the third pin is base, leave it floating, or connect it to emitter. If
for some reason you wanted to operate the phototransistor as a regular
transistor as well, that's what it's there for.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
 
Tim Williams wrote:
"Geir Klemetsen" <regeirgarbagespam@start.no> wrote in message
news:djrd72$khn$1@services.kq.no...

But with a 3 pins phototransistor, I don't know how to assemble it
to do the same as in the schematic above.


If the third pin is base, leave it floating, or connect it to emitter.
Click to expand...
Interesting. I always thought that connecting the emitter to the base
would effectively turn off the photo transistor??? Learn something new
every day.

Sometimes you can do special things using the base in the feedback
loop, but for most applications it's just a high impedance place for
extra noise/leakage to mess things up.

If
for some reason you wanted to operate the phototransistor as a regular
transistor as well, that's what it's there for.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
Click to expand...


--
Wanted, Serial cable for Dell Axim X5 PDA.
Return address is VALID but some sites block emails
with links. Delete this sig when replying.
FS 500MHz Tek DSOscilloscope TDS540 Make Offer
Bunch of stuff For Sale and Wanted at the link below.
MAKE THE OBVIOUS CHANGES TO THE LINK
ht<removethis>tp://www.geocities.com/SiliconValley/Monitor/4710/
 
"Geir Klemetsen" <regeirgarbagespam@start.no> wrote in message
news:djrd72$khn$1@services.kq.no...
I have a set of one sender (led) and one retriever (phototransistor)
from a
disassembled old printer.

The phototransistor is quite similar to
http://www.allproducts.com.tw/manufacture1/everlight/p18-s.jpg the
left one
apart from it has three pins, not two.

How to get the light signal from the diode into an electric analog
signal
from the transistor?


With only two pins, it's wery easy:

VCC
+
|
|
.-.
| |
| |
'-' Output
| |\
o--------| >--o
\ | |/
\ -
^
|
|
===
GND
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

But with a 3 pins phototransistor, I don't know how to assemble it to
do
the same as in the chematic above. I have googled for a tutorial, but
couldn't find any.

TIA
Click to expand...
In the above circuit you show no value for the resistor. You don't show
what the inverter is, but it should have a very high input impedance,
for the following reason. Also, the diode is used in the reverse biased
mode, so the current will be very low, microamps if that.

With the three lead device, it would seem that it's a phototransistor.
If so, then the third or base lead doesn't need to be connected. But if
it's a phototransistor, then you shouldn't use it like the diode above,
reverse biased.
 
Geir Klemetsen wrote:

I have a set of one sender (led) and one retriever (phototransistor) from a
disassembled old printer.

The phototransistor is quite similar to
http://www.allproducts.com.tw/manufacture1/everlight/p18-s.jpg the left one
apart from it has three pins, not two.

How to get the light signal from the diode into an electric analog signal
from the transistor?


With only two pins, it's wery easy:

VCC
+
|
|
.-.
| |
| |
'-' Output
| |\
o--------| >--o
\ | |/
\ -
^
|
|
===
GND
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

But with a 3 pins phototransistor, I don't know how to assemble it to do
the same as in the chematic above. I have googled for a tutorial, but
couldn't find any.

TIA


A simple-minded way is to connect the transistor in the DCT mode,
Click to expand...
making it look like a diode, but still act like a transistor:

+----+---o +
| |/
+--|
|\
v
|
+----o -
 
Tim Williams wrote:

"Geir Klemetsen" <regeirgarbagespam@start.no> wrote in message
news:djrd72$khn$1@services.kq.no...

But with a 3 pins phototransistor, I don't know how to assemble it
to do the same as in the schematic above.


If the third pin is base, leave it floating, or connect it to emitter. If
for some reason you wanted to operate the phototransistor as a regular
transistor as well, that's what it's there for.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms


Poor choices.
Click to expand...
1) floating base: NO GO
2) base tied to emitter: less sensitivity than if base tied to collector.
 
Tim Williams wrote:

"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:Odk8f.2916$yX2.2374@newsread2.news.pas.earthlink.net...

A simple-minded way is to connect the transistor in the DCT mode,
making it look like a diode, but still act like a transistor:


No?? If it's NPN, it's going to be forward-biased! I don't see how that
could ever work.

My understanding is the transistor has an additive (OR) choice between base
current and light turning on the collector.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms


Start with a transistor.
Click to expand...
If the base is connected to the collector, it is still technically
and pracitcally an active device, with the base current controlling the
collector current.
Take one step backwards, close your eyes to that.
Shine a light on a silicon PN junction and notice that an electrical
voltage is produced if open circuit (or hi Z load) and that an
electrical current is produced if shorted circuit (or low Z load).
Partly open eyes, take one-half step forward.
Shine a light on a transistor die (that is how the vast majority of
phototransistors were made; a lens that focused the light on a 2N2222 or
equivalent die).
In effect, charge is being injected in the base.
You now may open the eyes all the way and finish stepping forward.
 
"Tim Williams" <tmoranwms@charter.net> wrote in message
news:dCI8f.3990$7s1.1612@fe04.lga...
"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:_yE8f.1180$8c5.717@newsread3.news.pas.earthlink.net...
In effect, charge is being injected in the base.
You now may open the eyes all the way and finish stepping forward.

Ok, so you get a variable knee (i.e., variable Vbe) diode? Kinda spoils
the
useful switching character of a transistor.

Tim

There are a few things that have not yet been mentioned.
Click to expand...

Most phototransistors are junction transistor, not field effect transistors.

This means that they operate on current flow not voltage levels as are
common with FETs.

The rise and fall times of junction phototransistors are determined by the
rate of change of the current through the base-emitter junction.

The collector current of phototransistors is usually very low, in the range
of 400 to 800 microamps with full illumination. Darlington phototransistors
can switch more collector current but rise and fall times increase from a
few microseconds to several milliseconds.

The larger the area of the base is the more sensitive the phototransistors
is. A lager base area means a large base-emitter capacitance. This sets up
two competing goals. Where one would like fast turn on and off times the
emitter impedance should be low, but to get a large output voltage swing
with only 400 microamps of collector current the emitter impedance needs to
be quite large.

One way to manage all these thing is to use a three terminal phototransistor
and establish the base impedance independent of the emitter circuit.

Keeping the voltage drop across the phototransistor (Vce) low can also
improve rise and fall times and sensitivity to light.
 
Keyser Soze wrote:

"Tim Williams" <tmoranwms@charter.net> wrote in message
news:dCI8f.3990$7s1.1612@fe04.lga...

"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:_yE8f.1180$8c5.717@newsread3.news.pas.earthlink.net...

In effect, charge is being injected in the base.
You now may open the eyes all the way and finish stepping forward.

Ok, so you get a variable knee (i.e., variable Vbe) diode? Kinda spoils
the
useful switching character of a transistor.

Tim


There are a few things that have not yet been mentioned.

Most phototransistors are junction transistor, not field effect transistors.

This means that they operate on current flow not voltage levels as are
common with FETs.

The rise and fall times of junction phototransistors are determined by the
rate of change of the current through the base-emitter junction.

The collector current of phototransistors is usually very low, in the range
of 400 to 800 microamps with full illumination. Darlington phototransistors
can switch more collector current but rise and fall times increase from a
few microseconds to several milliseconds.

The larger the area of the base is the more sensitive the phototransistors
is. A lager base area means a large base-emitter capacitance. This sets up
two competing goals. Where one would like fast turn on and off times the
emitter impedance should be low, but to get a large output voltage swing
with only 400 microamps of collector current the emitter impedance needs to
be quite large.

One way to manage all these thing is to use a three terminal phototransistor
and establish the base impedance independent of the emitter circuit.

Keeping the voltage drop across the phototransistor (Vce) low can also
improve rise and fall times and sensitivity to light.



....for example, keep the Vcb zero with an opamp would improve the
Click to expand...
risetime by a significant amount.
 
Robert Baer wrote:
Keyser Soze wrote:

"Tim Williams" <tmoranwms@charter.net> wrote in message
news:dCI8f.3990$7s1.1612@fe04.lga...

"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:_yE8f.1180$8c5.717@newsread3.news.pas.earthlink.net...

In effect, charge is being injected in the base.
You now may open the eyes all the way and finish stepping forward.


Ok, so you get a variable knee (i.e., variable Vbe) diode? Kinda
spoils the
useful switching character of a transistor.

Tim


There are a few things that have not yet been mentioned.

Most phototransistors are junction transistor, not field effect
transistors.

This means that they operate on current flow not voltage levels as are
common with FETs.

The rise and fall times of junction phototransistors are determined by
the rate of change of the current through the base-emitter junction.

The collector current of phototransistors is usually very low, in the
range of 400 to 800 microamps with full illumination. Darlington
phototransistors can switch more collector current but rise and fall
times increase from a few microseconds to several milliseconds.

The larger the area of the base is the more sensitive the
phototransistors is. A lager base area means a large base-emitter
capacitance. This sets up two competing goals. Where one would like
fast turn on and off times the emitter impedance should be low, but to
get a large output voltage swing with only 400 microamps of collector
current the emitter impedance needs to be quite large.

One way to manage all these thing is to use a three terminal
phototransistor and establish the base impedance independent of the
emitter circuit.

Keeping the voltage drop across the phototransistor (Vce) low can also
improve rise and fall times and sensitivity to light.



...for example, keep the Vcb zero with an opamp would improve the
risetime by a significant amount.
Click to expand...
Would there be any advantage to keeping the Vcb high, smaller Ccb
but let the dVcb/dT be zero??
mike

--
Wanted, Serial cable for Dell Axim X5 PDA.
Return address is VALID but some sites block emails
with links. Delete this sig when replying.
FS 500MHz Tek DSOscilloscope TDS540 Make Offer
Bunch of stuff For Sale and Wanted at the link below.
MAKE THE OBVIOUS CHANGES TO THE LINK
ht<removethis>tp://www.geocities.com/SiliconValley/Monitor/4710/
 
"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:Ifk8f.2917$yX2.1069@newsread2.news.pas.earthlink.net...
Tim Williams wrote:

"Geir Klemetsen" <regeirgarbagespam@start.no> wrote in message
news:djrd72$khn$1@services.kq.no...

But with a 3 pins phototransistor, I don't know how to assemble it
to do the same as in the schematic above.


If the third pin is base, leave it floating, or connect it to
emitter. If
for some reason you wanted to operate the phototransistor as a
regular
transistor as well, that's what it's there for.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms


Poor choices.
1) floating base: NO GO
Click to expand...
No go electrically, but that's not what it's used for. It's used for a
phototransistor, and normally the base is left floating.

2) base tied to emitter: less sensitivity than if base tied to
collector.
Click to expand...

Then it will have no photo sensitivity. It'll just be a forward biased
diode.
 
"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:_yE8f.1180$8c5.717@newsread3.news.pas.earthlink.net...
Tim Williams wrote:

"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:Odk8f.2916$yX2.2374@newsread2.news.pas.earthlink.net...

A simple-minded way is to connect the transistor in the DCT mode,
making it look like a diode, but still act like a transistor:


No?? If it's NPN, it's going to be forward-biased! I don't see how
that
could ever work.

My understanding is the transistor has an additive (OR) choice
between base
current and light turning on the collector.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms


Start with a transistor.
If the base is connected to the collector, it is still technically
and pracitcally an active device, with the base current controlling
the
collector current.
Take one step backwards, close your eyes to that.
Shine a light on a silicon PN junction and notice that an
electrical
voltage is produced if open circuit (or hi Z load) and that an
electrical current is produced if shorted circuit (or low Z load).
Partly open eyes, take one-half step forward.
Shine a light on a transistor die (that is how the vast majority of
phototransistors were made; a lens that focused the light on a 2N2222
or
equivalent die).
In effect, charge is being injected in the base.
You now may open the eyes all the way and finish stepping forward.
Click to expand...
My understanding is that phototransistors are not used as photovoltaic
devices. They are used just as a PIN photodiode is used, reverse
biased, but with current amplification.
 
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