Micro Phototransistor datasheet.

[cowboybob]

We were also taught about signals passing through the ether I just could not figure what the either was and how it worked. That's because the either does not exist. There is only air or a vacuum

Slight deviation from the thread's theme, but...

The definition of the "ether" is often nebulous at best, if not downright drifting into the realm of "magic smoke". So don't feel bad. I grappled with the concept as well .

But to add to your concept, RF wave propagation (of any type) is also affected, to a greater and lesser degree, by solar "winds", as they exist outside of the Earth's atmosphere. A pretty comprehensive piece on the subject can be found here: https://deepspace.jpl.nasa.gov/dsndocs/810-005/106/106B.pdf

 

[kal.a]

​

As you quite rightly say R2 (2K2) serves no purpose in this circuit configuration. The current through the opto transistor and Q1 is already limited by R3 (10K). But R2 will do no harm and will not stop the circuit from working.

The PMOSFET has a few problems:
(1) It is a 20V max VSD device and it is being used with 24V VSD
(2) The source and drain connections are swapped
(3) The IRF7404 PMOSFET is not suitable for driving a motor- it is not man enough

Once the above points are fixed the circuit will work OK but it would be operating the opto receiver transistor with a VCE lower than is specified in the characteristics defined by the data sheet.

Sorry. Much of the schematics I posted did not reflect what I actually did so I have to start again.

I picked up an IRF9540N P-Channel Mosfet and wired it like **broken link removed**with 24V directly to the source and the drain to the load (24V relay). The result was that the Mosfet got quite hot and the relay was on all the time though the voltage at the gate was changing (with the dark and light operation of the photo transistor) from 17.35V to somewhere around 2V if my memory serves me right. I believe that mosfet is damaged .
So I put a new Mosfet and wired it the same way with the addition of a 10K resistor between the 24V and the source then connected an LED to the drain. The LED is on all the time with high voltage of 17.3 and low of 0.033. Which I don't get. I find it so confusing that I thought I got the pinout wrong but all my search resulted in the same pinout configuration.
To your points:

1) I put anything in the schematic that was P-Channel just to work on the wiring but I hope that the IRF9540 is man enough
2) You mean connecting the 24V to the drain instead of the source and the load to the source? I just tried that too and the LED is on all the time.
But I don;t get the last line about the opto receiver. The opto receiver is connected to the BC547 and that did not change when I replaced the N-Channel with the P-Channel. How is it being affected by the Mosfet?

 

[kal.a]

​

This is a light on function. ie when light is shining on the opto receiver transistor, the motor will run.

The reason why Q1 (BC546B) is overheating is that there is nothing to limit its collector current and hence the base current of Q2 (BD140).
The opto reciever transistor can conduct 18ma for a really good sample. Q1 (BC546B) can have an hFE of 300 so potentially Q1 collector current could be 18mA * 300= 5.4 Amps. The voltage across Q1 collector emitter is 24V, so Q1 could possibly be dissipating, 5.4A * 24V = 129.6W. If you had a power supply with a current capability of 5.4 A and the Opto reciever transistor were particularly good and Q1 had an hFE within its data sheet specification there would be an explosion and all that would be left of Q1 would be the collector, emitter, and base leads.

To fix the problem all you need to do is put a 2K2 resistor between Q1 collector and Q2 base. This will limit Q1 collector current to around 24V/2K2 Ohms= 10.9mA, slightly different to 5.4A

This circuit will work with the addition of the 2K2 resistor, but once again it is relying on parameters not given in the dara sheet.

I think this reply is the one I really need to understand so I'm still reading it........I'm a bit slow.

 

[spec]

Sorry. Much of the schematics I posted did not reflect what I actually did so I have to start again.

I picked up an IRF9540N P-Channel Mosfet and wired it like **broken link removed**with 24V directly to the source and the drain to the load (24V relay). The result was that the Mosfet got quite hot and the relay was on all the time though the voltage at the gate was changing (with the dark and light operation of the photo transistor) from 17.35V to somewhere around 2V if my memory serves me right. I believe that mosfet is damaged .
So I put a new Mosfet and wired it the same way with the addition of a 10K resistor between the 24V and the source then connected an LED to the drain. The LED is on all the time with high voltage of 17.3 and low of 0.033. Which I don't get. I find it so confusing that I thought I got the pinout wrong but all my search resulted in the same pinout configuration.
To your points:

1) I put anything in the schematic that was P-Channel just to work on the wiring but I hope that the IRF9540 is man enough
2) You mean connecting the 24V to the drain instead of the source and the load to the source? I just tried that too and the LED is on all the time.
But I don;t get the last line about the opto receiver. The opto receiver is connected to the BC547 and that did not change when I replaced the N-Channel with the P-Channel. How is it being affected by the Mosfet?

**broken link removed**
​

Hi Kal,

Your circuit is reproduced above.
Yes, the PMOSFET is correctly connected as you say.
But the 18V Zener is in the wrong position. Its purpose is to ensure that the gate/source voltage of the PMOSFET never exceeds 20V. To do this the Zener needs to be connected across the gate/source terminals rather than the gate/drain terminals as in the above schematic (consider R1 to be a conductor).

At the moment, when Q1 is conducting, its collector will be at 0V, so the gate of the PMOSFET (call it Q2) will also be at 0V. The source of Q2 is connected to 24V, so this means the there will be 24V across the gate/source terminals of Q2. This will turn Q2 hard on and will probably damage Q2 because the limit of 20V across the gate/source voltage will have been exceeded by 4V.

When Q1 is turned off, R5 (2K2) will try to drag the collector of Q1 to 24V. It will not be able to do this because D1 (18V Zener) will start conducting at 18V and clamp Q1 collector to 18V. Thus Q2 will have 18V on its gate. But as Q2 source is connected to 24V this means that Q2 gate is 6V more negative than Q2 source. This means that Q2 will still be turned on but not as hard as with 18V across Q2 gate source.

You cannot just connect the Zenner diode across the gate/source terminal of Q2, because then when Q1 conducts there will be nothing to limit Q1 collector current or the current through the Zener, with the result that one or both components will be destroyed.

The solution is to use one of the circuits that I have suggested, which also use the opto sensor as defined in the data sheet.

 

[MikeMl]

You are exceeding the max allowed Gate to Source voltage for the PFet

 

[kal.a]

Thanks Mike. I'm still trying to figure out how i'm doing that. The voltage at the base never exceeds 18V.

 

[kal.a]

**broken link removed**
​

Hi Kal,


The solution is to use one of the circuits that I have suggested, which also use the opto sensor as defined in the data sheet.

Ok I finally got it. I will retry today.
Although I'm looking to build this circuit my goal is to learn so that's why keep messing around with different versions of things.

If I may ask you again and I do appreciate your patience, how's the opto sensor is affected by the way I wired it. I understand it has an effect on the mosfet but not clear on how it will affect the opto.

Thanks a lot
Kal

 

[kal.a]

**broken link removed**
​

At the moment, when Q1 is conducting, its collector will be at 0V, so the gate of the PMOSFET (call it Q2) will also be at 0V. The source of Q2 is connected to 24V, so this means the there will be 24V across the gate/source terminals of Q2. This will turn Q2 hard on and will probably damage Q2 because the limit of 20V across the gate/source voltage will have been exceeded by 4V.


The solution is to use one of the circuits that I have suggested, which also use the opto sensor as defined in the data sheet.

I think I understand a bit of this too and how it affects the opto receive. So as you stated because of the misplacement of the zener diode no I'm getting 24V across the gate/source of Q2 and the gate of Q2 is connected to the Collector of Q1 which at that same moment is also conducting so the current will flow through Q1's base which is connected to the collector of the opto receiver. I'm actually going to recreate that for fun and to see the effect in real time. There will be lots of smoke

But the opto at that moment is off with no path to ground so it should not be affected except for the fraction of time as the light is turning on and it starts to conduct it will be getting current from across the path I described above which will be soon turned off by the receiver conducting. Am I on the right track?

Thanks
Kal

 

[spec]

Ok I finally got it. I will retry today.
Although I'm looking to build this circuit my goal is to learn so that's why keep messing around with different versions of things.

If I may ask you again and I do appreciate your patience, how's the opto sensor is affected by the way I wired it. I understand it has an effect on the mosfet but not clear on how it will affect the opto.

Thanks a lot
Kal

**broken link removed**
​

Hy Kal,

Assume that the opto sensor data sheet reproduced above applies to your opto sensor: it won't be too far out, if not identical.

From the Absolute Maximum Ratings (AMR) you can see that the maximum VCE for the opto receiving transistor (ORT) is 30V so, with a supply of 24V, no protection measures will be required in this respect.

From the Electrical and Optical Characteristics (EOC) the ORT light current is 500uA to 14mA at an ORT VCE of 5V.

Your circuit, reproduced above, requires Q1 to be turned on and off to turn the MOSFET (call it Q1) to be turned on and off. I have already said in a previous post the the Zener diode is in the wrong place, so assume it is removed. I have also said that the PMOSFET VGS maximum rating will be exceeded, but ignore this for the time being.

With no light there will be no ORT collector current, but current will flow into the base of Q1 via R3 (10K) and R2 (2K2). Q1 base current will be (24V-0.6V)/(10K+2K2) = 1.91 mA. The hFE of Q1 will be around 300, so Q1 collector will be fully saturated and its collector will be around 0V.

The collector voltage of the ORT will thus be 0.6V (VCB Q1) + (1.91mA * 2K2) = 4.78V, so the ORT VCE is already below the data sheet operating condition of 5V.

With light on, current will flow through the collector of the ORT, but you will have no idea how much because the data sheet does not tell you.

In order to turn Q1 off the VCE of the ORT will need to drop to below about 300mV (nominal on voltage for Q1 is 600mv) No where on the data sheet does it say that the ORT will pass any current with a VCE as low as 300mV.

So your circuit has two problems:
(1) not quite enough ORT VCE in the dark state.
(2) virtually 0V ORT VCE in the light state. This is the killer.

I have been playing devil's advocate because, in practice, the circuit would probably work with most samples of the opto coupler, but it is not a worst case design that could go into production. The other thing is that temperature effects have not been taken into account.

I hope this explains why the above circuit does not use the opto sensor within the conditions specified on the data sheet.

By contrast, my circuits keep the ORT VCE well above 5V under all circumstances- aren't I a clever boy

 

[spec]

I think I understand a bit of this too and how it affects the opto receive. So as you stated because of the misplacement of the zener diode no I'm getting 24V across the gate/source of Q2 and the gate of Q2 is connected to the Collector of Q1 which at that same moment is also conducting so the current will flow through Q1's base which is connected to the collector of the opto receiver. I'm actually going to recreate that for fun and to see the effect in real time. There will be lots of smoke

But the opto at that moment is off with no path to ground so it should not be affected except for the fraction of time as the light is turning on and it starts to conduct it will be getting current from across the path I described above which will be soon turned off by the receiver conducting. Am I on the right track?

Thanks
Kal

Hi again kal,

A fraction of a time to you is very short. To a transistor a human fraction of time is an eternity and more than sufficient time for it to pass all the current that is possible and thus burn out. To give you a clue a BC546/556 will turn on in around 1uS (yes, one millionth of a second).

 

[kal.a]

View attachment 97181
​

Hy kal,

Firstly let me assure you that the above circuit is correct and will work as intended. I suspect that you have the PMOSFET incorrectly wired. I also suspect that that your PMOSFET is blown. The IRF7404 type is not suitable for this application anyway.

P Type MOSFET Operation

A PMOSFET is the compliment of an NMOSFET. This means that all the operating voltages are reversed. Under normal operating conditions the drain of a PMOSFET is more negative than the source. If the gate and drain are at the same potential no drain current will flow. If the gate is more negative than the source drain current will flow. So in the opto sensor the PMOSFET source would be connected to 24V and the drain would be connected to the top of the motor.

Circuit Function light on

Light impinges on the Opto Receiving Transistor (ORT) this causes an ORT collector current to flow from the 24V supply line through the emitter base junction of Q17 (consider R7 to be a short circuit at the currents concerned).

The current flowing through Q17 base emitter will generate a Q17 collector current of hFE Q17 * Ib to flow in the collector of Q17, but this current is limited by R5 (2K2) to 24V/2K2 = 10.9 mA.

As the whole supply line voltage is dropped across R5, the collector of Q17 will be at 24V. This voltage will also be on the gate of PMOSFET Q16 (R8 has no effect). As the drain of Q16 is also at 24V there is OV between the gate and source, so Q16 is turned off.

Circuit Function light off

No light impinges on the ORT so only a small dark current flows in the ORT collector. This small current flows through R24 (10K) and does not generate enough voltage (600mV) to turn Q17 on.

As Q17 is off R5 tries to drag Q17 collector down to 0V but Zener D8 prevents this and starts conducting at 24V-18V = 6V. The net result is that gate of PMOSFET Q16 is at 6V and its source is at 24V so the gate is 18V more negative than the drain. This means that Q16 is fully turned on.

Zener Diode

You ask what the Zener diode is for: simply to protect Q16. It serves no other purpose. The data sheet for the PMOSFET says that the gate should never be more than +-20V with respect to the source. Without the Zener the gate could be 24V more negative than the source which could potentially destroy the PMOSFET.

What does R7 (2K2) do?

R7 in the base of Q7 probably looks a bit odd but it is there to protect the collector of the ORT and Q17 base from destructive current flow from the 24V supply rail to 0V. You might say but the data sheet says only a maximum of 18 mA will flow in the ORT and you would be right, except:

(1) Radio frequency interference could cause the ORT to conduct heavily

(2) Electrostatic discharge could cause the ORT to conduct heavily.

(3) High energy radiation could cause the ORT to conduct heavily (Gamma, X, etc)

Also an arrangement like that is like a loaded gun waiting to go off and even if an excess current could not possibly flow in theory, you would still protect against it as a matter of good design.

This explanation is just brilliant.

 

[spec]

Slight deviation from the thread's theme, but...

The definition of the "ether" is often nebulous at best, if not downright drifting into the realm of "magic smoke". So don't feel bad. I grappled with the concept as well .

But to add to your concept, RF wave propagation (of any type) is also affected, to a greater and lesser degree, by solar "winds", as they exist outside of the Earth's atmosphere. A pretty comprehensive piece on the subject can be found here: https://deepspace.jpl.nasa.gov/dsndocs/810-005/106/106B.pdf

Back on how electronics was taught, and a big deviation from thread:

In the 1950s and early 1960s you knew where you were. The main active device was a valve (tube). Effectively, it only had three terminals: cathode, grid, and anode. The anode was pretty much always positive with respect to the cathode and the grid was normally a bit negative with respect to the cathode.

Electrons flowed from the hot cathode to the anode and the number of electrons flowing was controlled by the grid/cathode voltage: the more negative the grid was with respect to the cathode the less anode current flowed. The grid was an infinite impedance so no grid current flowed. OK, you had to use your imagination a bit to sort out the electron flow/conventional flow cock-up so that you had to assume that current flowed from the positive rail into the anode but you soon came to terms with that. I knew most of this when I was about 12. By the age of 14, I also knew that the voltage gain of a valve was gm * RA (mutual conductance times anode load). By age 17, I had some idea about the Miller feedback capacitor, between the anode and grid. And that was pretty much a valve's characteristics in a nut-shell. Armed with that basic information it was not too difficult to understand a valve operating in common cathode, common anode (cathode follower), and common grid (cascode) configurations.

Then these funny transistor things came on the scene. For a start there was PNP types and also the very rare NPN types. Then there was holes and electrons flowing in the transistor. Then there was leakage current and this terrible thing called thermal run-away. There was also the view, probably quite rightly in the early days, that transistors were not reliable and would blow at the slightest provocation, especially if they got warm, so only girls used transistors- men on the other hand used valves.

In any electronics community there was always someone who knew about transistors, there was a similar situation with binary arithmetic and computer programming at one time. These people had the mantra of a shaman in a Red Indian tribe. They were also mysterious and kept their secrets to themselves.

It was in this atmosphere that I tried to learn about semiconductor diodes and transistors. I read book after book, but just could not get it. Soon I came to the conclusion that semiconductors were just too complicated for me and perhaps so for the whole of electronics- we had just started learning about AC circuits with capacitors and inductors and phasor diagrams and all that kind of stuff and the whole thing was a bit overwhelming. Then they hit us with j (square root of minus one) and that did it for me. For about a month I thought seriously of going into another line of business. But the AC stuff started to gel and soon became second nature with a bit of practice. The big breakthrough was learning the formula: one, over, two pi root LC equals the resonant frequency. One of our group was an ace musician, pianist, and singer and he even wrote a song with that line. The rhyme continued through the song.

The big day came; we were going to have our first lesson on semiconductors from the expert in the field. I couldn't wait. The lecturer kicked off with a long description about crystal structures, minority and majority carriers, valency co-valent bonds, doping, leakage current, and, of course, the much feared thermal run away. That took the first couple of lessons. Then we moved on to the operation of a diode with more physics and a load of exponential equations involving Boltzmans constant and degrees Kelvin. It seemed that leakage current flowed one way and that normal current flowed the other way. By now, I decided that I would never understand how a semiconductor diode worked, even though I knew how a thermionic diode worked. The next week we were going to cover the operation of a transistors. I thought perhaps I will get on better with that.

On Monday morning, when we went in the class room, the instructor had already put a model of a transistor on the blackboard- I just could not make head nor tail of it. I know now that he had drawn an h parameter model of a transistor in the grounded base configuration, complete with re, rb, and rc. Off he went with blackboard after blackboard of formula. I realized then that I would never understand semiconductors and that is how it stayed. In our theory exams about 10% of the mark was allocated to semiconductors and the whole class decided that they would just forget about semiconductors and take the 10% hit. I did try to get a simple explanation of how a transistor worked from the instructor. He was most helpful. Once again he drew a grounded base configuration on the blackboard and than explained that a transistor has current gain. The input current into the emitter is Ib + Ic and the output current is Ic. He said that typically a transistor had a current gain of 0.95. I said that is not a gain that is a loss. He said that the gain was realized because the collector resistance is very high, in the order of 30K Ohms. That did it for me. Like all of us, I learned by rote a lot of standard questions and answers about semiconductors, sufficient to get exam marks, but did not have a clue about how semiconductors worked.

It wasn't until four years later, when I was in a design and development environment, that I got the hang of it all. An engineer explained, in about half an hour, how a semiconductor diode and a transistor worked in practical terms. Also, the common base transistor configuration was rarely used in real world circuits, mainly common emitter, similar to a valve, and common collector, emitter follower or cathode follower.

This is what I learnt:

Diode
If the anode is positive with respect to the cathode current will flow and there will be a forward drop of around 200mV for a germanium diode and 600mV for a silicon diode.
If the anode is negative with respect to the cathode no current will flow, except a very small leakage current of perhaps 1uA to 10uA for a germanium diode or 10nA to 100nA for a silicon diode.

Transistor
The base/emitter junction of a transistor is a diode with the same characteristics. The base current will be amplified by the hFE (Beta or current gain) of the transistor, so that the collector current will be IB * hFE and the emitter current will be, IB + (IB *hFE).

Armed with that simple information it was possible to design a whole raft of circuits, and once that basic information was assimilated, it was a progressive and simple matter to learn about the three transistor configurations: common emitter, common collector, and common base. It was even relatively easy to learn the h parameters and, more importantly, to understand their significance.

I realise now that our instructor, in spite of his expert reputation, did nor know a thing about designing a transistor circuit and probably didn't really know how a transistor functioned. Worse still, he did not know how to teach.

 

[kal.a]

Hello Spec, Hope you are well and don't mind me picking your brains a little more on the same topic.

I put the circuit together on a bread board and of course it worked like a charm. Then I got to thinking that I'm actually going to have two of those sensors positioned at 90 degrees out of phase. Only one will be on at a time.
So I **broken link removed**one of your circuits (the one with the BD140 I was using for an LED as I decided I don't need a Mosfet after all because the PLC input requires only 4mA to turn fully on) by adding a another BC557 PNP.
I tried it on a bread board and nothing burnt but buy now I learned that doesn't mean it's correct so I thought I'd double check.

Another thing I was thinking of is that although the BD140 preformed wonderfully and did not burn like the Mosfets with all my abuse I could also just use another BC557 as it will be required to conduct a few milliamps. What do you think?

Oh and one more thing what happens if I use the BD140 almost as a **broken link removed** so that it would switch the PLC input and shares only ground with the PLC . I'm not quite sure if current would flow the same way form V+ of one supply into the other.
I'm not inclined to do it this way but curious as to the effect.


Cheers

Kal

 

[spec]

Hello Spec, Hope you are well and don't mind me picking your brains a little more on the same topic.

I put the circuit together on a bread board and of course it worked like a charm. Then I got to thinking that I'm actually going to have two of those sensors positioned at 90 degrees out of phase. Only one will be on at a time.
So I **broken link removed**one of your circuits (the one with the BD140 I was using for an LED as I decided I don't need a Mosfet after all because the PLC input requires only 4mA to turn fully on) by adding a another BC557 PNP.
I tried it on a bread board and nothing burnt but buy now I learned that doesn't mean it's correct so I thought I'd double check.

Another thing I was thinking of is that although the BD140 preformed wonderfully and did not burn like the Mosfets with all my abuse I could also just use another BC557 as it will be required to conduct a few milliamps. What do you think?

Oh and one more thing what happens if I use the BD140 almost as a **broken link removed** so that it would switch the PLC input and shares only ground with the PLC . I'm not quite sure if current would flow the same way form V+ of one supply into the other.
I'm not inclined to do it this way but curious as to the effect.


Cheers

Kal

Hy kal,

I'm fine thanks and hope you are too. I don't mind answering any questions at all.

**broken link removed**

​

Your latest circuit is reproduced above:

First let me say that there is some very good thinking in this adaptation to a twin opto sensor circuit. It is quite clear that you know what you are doing judging by your approach. There are though two detailed issues:

(1) LED Current
You have connected both opto-coupler LEDs in parallel and fed them with a current via a single 1K resistor from 24V. As the individual LED forward voltages (VFs) will vary due to manufacturing tolerances, the LED with the lowest VF will hog all the current and the other LED will be starved. The unbalance might be so bad that one LED may not illuminate at all. In practice, both LEDs would probably illuminate, but this is not a worst case toleranced design. The solution is simple: use individual 1K resistors connected to 24V, one for each LED.

(2) BD140 Base Drive (call it Q2)
When U1 and/or U2 opto receiver transistors (ORTs) receive light they will turn on their respective PNP BC556 transistors which means that Q2 (BD140) will always be turned off.
The resolution for this problem is quite simple as shown in the schematic below.

To answer your question about interfacing to a PLC: yes that would be perfectly feasible. If you post details of the PLC input, I will post a simple interface circuit. Yes, the two circuits could use a common 0V, or if if you wanted two different 0Vs you could use an optocoupler to totally isolate the optosensor circuit from the PLC circuit.

​

ERRATA
(1) C18 should read 47uF or larger (as no motor is involved less current is conducted from the power line so the low frequency decoupling capacitor value can be reduced from 1mF)
(2) C4 can be deleted (same reason as (1) above)

 

[kal.a]

This is just fantastic. I didn't know diodes are used as switches. I already have ideas for using them that way.
And of course my schematic wouldn't work and I should've seen it. I had the sensors in my hand activating one at a time, no wonder I got false results.
I'm still reading on diodes work as switches and ecstatic to learn more electronics

You did it again spec, much appreciated.

The PLC wiring is pretty straight forward but I will post a picture of drawing later.

 

[spec]

Hi kal,
Thanks again for your kind words.
Yes, diodes can be used as switches and they can also be configured as logic functions. The two diodes in my circuit are configured as a TWO INPUT AND GATE with the following truth table (by convention, the inputs to logic gates are lettered a, b, c, etc and the output of the logic gates are labeled y):
If a=0 & b=0 then y=0
If a=0 & b=1 then y=0
If a=1 & b=0, then y=0
If a=1 & b=1, then y=1
(0=0V, 1=24V)

You can make any logic function with resistors and diodes. In fact one of the early logic families was Resistor Diode Logic (RDL).

As an aside, you may find this hard to believe, but you can make any logic function, including entire computers, with just TWO INPUT AND gates and INVERTERS. In, fact you can make all logic functions with just TWO INPUT NAND gates.

 

[kal.a]

Thanks for pointing out both my mistake and the function of diodes as logic gates. I had a great read and will continue to play with them.
But...... I decided to wire two sets of circuits and will try to fit them on the same board for practical reasons.

Here's what I'm doing. I have a motor which will be powered and controlled by a PLC through logic controlling electronic circuitry and relays. I will position on the motor shaft a slotted disc and two optical sensors.
The idea is to use those sensors as an optical encoder though a primitive one. The output of the Mosfet (now a BC556) has to change state from on to off or in other words it has to pulse.
With one sensor I would have let's say five pulses per revolution (the disc has five notches on it) with the addition of a sensor out of phase with the first I will have 10 pulses per revolution thereby doubling the resolution. I will also be using the rising and falling edges of each pulse which means that I will quadruple the resolution. Another reason they should not be on at the same time is that by pulsing one after the other I can tell direction.
I looked at the idea of having a logical wiring of the transistors to insure they provide a pulse the way I described it and I'm thinking they may screw up the timing of the Ons and Offs that they will not reflect the RPM of the motor although the idea intrigued me. But it's beyond my pay grade so to speak. I will keep it in the back of my mind though as I learn more electronics and digital circuitry and keep as a future project.

Thanks a lot
Kal

 

[kal.a]

The final result of the project is complete, tested and in action. I made an **broken link removed**with dual connection to connect two sensors and there by have a "Quadrature AB encoder"

 

[spec]

The final result of the project is complete, tested and in action. I made an **broken link removed**with dual connection to connect two sensors and there by have a "Quadrature AB encoder"

Excellent,

Well done

spec

 

[kal.a]

Hello spec,

The 1K resistor on the Anode of the photo transistor gets a bit warm and I'm starting to wonder whether I should replace it with a 2K2 to reduce the current or is better to replace the 1K- 1/2 Watt with a 1Watt resistor.
If my calculation is correct I have a 24V supply and a typical Vf of 1.2V 24-1.2 = 22.8 / 1000Ohms = .0228 Amps X 0.51 Watt. I think my 1/2 Watt resistor is too small.

Cheers
Kal