Phototransistor with three pins

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:

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

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from a

left one

signal

do

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.

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 -

Poor choices.

1) floating base: NO GO 2) base tied to emitter: less sensitivity than if base tied to collector.

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:

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.

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:

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.

Would there be any advantage to keeping the Vcb high, smaller Ccb but let the dVcb/dT be zero?? mike

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

Tim

-- Deep Fryer: a very philosophical monk. Website:

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 ?

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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).

The 4N35 is an optocoupler not just a phototransistor.

See:

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.

emitter. If

regular

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.

collector.

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

that

between base

the

electrical

or

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.