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Phototransistor with three pins

G

Geir Klemetsen

Jan 1, 1970
0
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
 
T

Tim Williams

Jan 1, 1970
0
Geir Klemetsen said:
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
 
M

mike

Jan 1, 1970
0
Tim said:
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



--
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W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Geir Klemetsen said:
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

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.
 
R

Robert Baer

Jan 1, 1970
0
Geir said:
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,
making it look like a diode, but still act like a transistor:

+----+---o +
| |/
+--|
|\
v
|
+----o -
 
R

Robert Baer

Jan 1, 1970
0
Tim said:
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
Poor choices.
1) floating base: NO GO
2) base tied to emitter: less sensitivity than if base tied to collector.
 
T

Tim Williams

Jan 1, 1970
0
Robert Baer said:
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
 
R

Robert Baer

Jan 1, 1970
0
Tim said:
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
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.
 
T

Tim Williams

Jan 1, 1970
0
Robert Baer said:
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
 
K

Keyser Soze

Jan 1, 1970
0
Tim Williams said:
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.
 
R

Robert Baer

Jan 1, 1970
0
Keyser said:
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.
 
M

mike

Jan 1, 1970
0
Robert said:
...for example, keep the Vcb zero with an opamp would improve the
risetime by a significant amount.

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/
 
T

Tim Williams

Jan 1, 1970
0
Keyser Soze said:
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.

Odd, my 4N35 says 20mA minimum Ic for approximately 20mA in the LED.

Tim
 
J

Jamie

Jan 1, 1970
0
Tim said:
Odd, my 4N35 says 20mA minimum Ic for approximately 20mA in the LED.

Tim
Hmm, i always thought raise/fall (slew/skew) depending on
how you want to use the lingo? improves on more current
since the miller effects are reduced ?
 
R

Robert Baer

Jan 1, 1970
0
mike said:
Would there be any advantage to keeping the Vcb high, smaller Ccb
but let the dVcb/dT be zero??
mike
Zero Vcb tends to zero out a lot of terms; noise, miller effect,
leakage to name a few (i forgot all of them, but i think about 6 moer or
less).
 
K

Keyser Soze

Jan 1, 1970
0
Tim Williams said:
Odd, my 4N35 says 20mA minimum Ic for approximately 20mA in the LED.

Tim
The 4N35 is an optocoupler not just a phototransistor.

See: http://www.fairchildsemi.com/ds/4N/4N35.pdf for data sheet.

You may want to note that while the current transfer ratio for the 4N35 is specified to be 100% at 25C it drops to 40% at the high
and low temperature extremes.

The CTR performance of related devices (4N25-4N28) is so poor (10-20%) that they have no specification for temperature extremes at
all.

Optocouplers tend to have the optical design optimized for best performance.

When using phototransistors on the other hand the optical path does not lend itself to convent optimizations.

Manufactures of phototransistors seem to try very hard to obscure the specification for photon sensitivity of their products. This
makes it quite difficult for a designer to compare products from various vendors.
 
W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Robert Baer said:
Poor choices.
1) floating base: NO GO

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.

Then it will have no photo sensitivity. It'll just be a forward biased
diode.
 
W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Robert Baer said:
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.

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.
 
R

Robert Baer

Jan 1, 1970
0
Watson said:
between base



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.
...and tying the base to the emitter gives a sensitivity almost three
orders of magnitude less than with a floating or biased base.
 
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