Same concept as current from photovoltaic cells?

Of course.
But a solar cell has two terminals for the current to flow, as does any source of electricity.
It's not generated out of thin air at a single terminal.

The other terminal for the base current in a phototransistor has to be the collector.
You cannot have more current coming out of the emitter of a phototransistor, then goes into the collector.

 

It’s not nothing. It’s like current generated by solar power. It’s created when the photon are excite the electrons to jump the band gap. Otherwise wouldn’t there be no current from photovoltaic cells?

Current generated for solar power is from existing charge carriers moving in response to a e-field created from charge separation from photon energy. Current is not created, it's a rate of existing charge moving.

There has to be a source of the electrons for the current. The photon energy is large enough to raise an electron from the valence band into an empty conduction band state, thereby generating one electron-hole pair.


That doesn't create an extra charge carrier. It creates a 'hole' that we simplify to a +e charge. This charge separation generates an electric field that in essence sucks charge from the collector and sends charge into the base as a bias current.

 

Ah Kirchoffs Law... got it makes sense now

 

I think I've got it now, my mistake was mixing up standard BJT calculations with a 2 terminal phototransistor.

In the phototransistor the base current is "induced" by light. The confusion arises having a base current value which you use with Beta to calculate gain. Unlike a standard transistor there is no current flow from the base to emitter, it's only changing the impedence of the Collector/Base so current flows. So base and emitter current is the same, but it's regulated with the "base current" generated by detection of light. Dark currrent is noise, any current above that generated by the photons can be attributed to base current.

I'm sure it's not a perfect model but it's better than my original confusion.

 

Unlike a standard transistor there is no current flow from the base to emitter, it's only changing the impedence of the Collector/Base so current flows.

This isn't correct; in a phototransistor, incident light causes a current to flow from collector to base (in effect, a light-induced increase in Icbo), and that current flows in turn through the base-emitter junction, causing an increased collector-emitter current (that is, the collector-base photocurrent is amplified by the transistor's β).

The behavior is the same as if a PN photodiode were connected between the collector and base of an ordinary transistor. The only difference is that in a phototransistor, the collector-base junction functions as the photodiode.

 

Thanks @OBW0549, that is what I meant to say.

 

Thanks @OBW0549, that is what I meant to say.

So, is the book right?

 

I would say yes, but again does not go into detail, I really wish we were spending more time on diodes.

 

I would say yes, but again does not go into detail, I really wish we were spending more time on diodes.

It's a lifetime of learning based on having the fundamental physics background to see behind the curtain. Keep asking questions but don't assume the book is wrong without knowing what's really right as you learn.

 

Hello,

Do the following pages shed some light?
https://circuitglobe.com/phototransistor.html
https://www.electrical4u.com/phototransistor/

Bertus

 

There are errors in the book, so I'm always questioning... some of it is due to over simplifying to get a point across for instance take this circuit below... it doesn't work at the very least it needs the RC filter at the emitter. But the common emitter voltage amplifier concept works... again, just wary.

 

There are errors in the book, so I'm always questioning... some of it is due to over simplifying to get a point across for instance take this circuit below... it doesn't work at the very least it needs the RC filter at the emitter. But a common emitter concept works... again, just wary.

View attachment 186106

You don't need an RC filter at the emitter but it help to stabilize the quiescent state bias of the circuit with DC negative feedback while allowing full gain of the signal without negative feedback because of the capacitor bypass.
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/feedback/

 

@nsaspook i tried the circuit. It does not work without the high pass filter as given in the example. I haven’t tried the feedback. The link explains very clearly the issues.

 

i tried the circuit. It does not work without the high pass filter as given in the example.

Are you referring to the high pass filter (dc block) at the input, since there is no filter at the emitter?

 

We need the DC bias at the base to keep the input from cutoff... Can't get output waveform as the design below... if you can make it work that's amazing. I had to put a emitter resistor and a bypass cap in parallel to get the output to work correctly. The gain isn't incredible but multiple stages required to get substantial voltage gain.

book example:

example that I got to work from elsewhere: I put an output cap at Vout to get rid of the DC bias.


Did I miss something?

 

But do you know why it does not work?

 

I think it has to do with the DC biasing at the base. It just goes into saturation without the divider.

Vbb acts like a giant capacitor.

the base needs a consistent bias current. The battery and the signal mixing in that resistor doesn't work. If you move the resistor in series with the DC bias source, now you are able to set the bias (quiescent current) while seeing the signal at the base. It's done best with the resistor divider in the other example. Unfortunately there is distortion of the bottom half of the input signal on the "battery source" example because the current is not consistent on the negative half of the input. Ah... so the divider keeps the total resistance consistent.

Then the issue of the gain is next.

 

Okay, there is an error in that simple circuit, as the Vbb source will look like a short circuit to the AC input signal.
Below is the LTspice simulation with the resistor in series with the bias voltage, as it needs to be.
It shows an AC gain of more than 100.
Note that the gain is very sensitive to bias current.

 

Okay, there is an error in that simple circuit, as the Vbb source will look like a short circuit to the AC input signal.
Below is the LTspice simulation with the resistor in series with the bias voltage, as it needs to be.
It shows an AC gain of more than 100.

Nice one... it looks clean... mine is similar but looks like you nailed it.

 

I would say yes, but again does not go into detail, I really wish we were spending more time on diodes.

Does this help?

HP Optoelectronics Applications Manual, first edition: