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Phototransistor with three pins

 
 
Geir Klemetsen
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      10-27-2005, 08:27 PM
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/manufa...ight/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


 
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Tim Williams
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      10-27-2005, 09:53 PM
"Geir Klemetsen" <(E-Mail Removed)> wrote in message
news:djrd72$khn$(E-Mail Removed)...
> 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


 
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mike
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      10-28-2005, 02:10 AM
Tim Williams wrote:
> "Geir Klemetsen" <(E-Mail Removed)> wrote in message
> news:djrd72$khn$(E-Mail Removed)...
>
>>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.


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
>
>




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Watson A.Name - \Watt Sun, the Dark Remover\
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      10-28-2005, 04:59 AM

"Geir Klemetsen" <(E-Mail Removed)> wrote in message
news:djrd72$khn$(E-Mail Removed)...
> 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/manufa...ight/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


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.



 
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Robert Baer
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      10-28-2005, 07:14 AM
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/manufa...ight/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,
making it look like a diode, but still act like a transistor:

+----+---o +
| |/
+--|
|\
v
|
+----o -
 
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Robert Baer
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      10-28-2005, 07:16 AM
Tim Williams wrote:

> "Geir Klemetsen" <(E-Mail Removed)> wrote in message
> news:djrd72$khn$(E-Mail Removed)...
>
>>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
2) base tied to emitter: less sensitivity than if base tied to collector.
 
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Tim Williams
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      10-28-2005, 09:46 AM
"Robert Baer" <(E-Mail Removed)> wrote in message
news:Odk8f.2916$(E-Mail Removed) ink.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


 
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Robert Baer
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      10-29-2005, 06:22 AM
Tim Williams wrote:

> "Robert Baer" <(E-Mail Removed)> wrote in message
> news:Odk8f.2916$(E-Mail Removed) ink.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.
 
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Tim Williams
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      10-29-2005, 10:58 AM
"Robert Baer" <(E-Mail Removed)> wrote in message
news:_yE8f.1180$(E-Mail Removed) nk.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

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


 
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Keyser Soze
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Posts: n/a
 
      10-30-2005, 02:26 AM
"Tim Williams" <(E-Mail Removed)> wrote in message
news:dCI8f.3990$(E-Mail Removed)...
> "Robert Baer" <(E-Mail Removed)> wrote in message
> news:_yE8f.1180$(E-Mail Removed) nk.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.



 
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