Let us analyse this design together, I have several points to clarify with you so I will make a bulleted list.
Tell me if I am wrong.

1) On the left, if I'm not mistaken, Vdd takes on the value of Va or Vb depending on which is present, right? So it is as if they are in OR?
If both Va and Vb are ON ... Vcc is equal to the higher value of the two?

Now let's focus on the right side.
Let's assume that DCX_OK and DCY_OK are "status" signals, i.e. they tell me whether +3.3_X and +3.3V_Y are good or not (whether they work or not)

2) The central bjts (BJT1 and BJT2) will be closed or opened depending on whether +3.3_X and +3.3V_Y on their base are present or not, right?

3) If +3.3_X is present, that opto-bjt1 (suggest name) will close and then C and B of BJT1 will short? I don't quite understand how OBJT1 works ... of

4) If the answer is yes, how does BJT1 close when C=B? .. usually closure implies that E and C are ideally a "wire"




Please, if possible, answer in order, perhaps respecting the bulleted list

Thanks

 

you did not label your circuit.
if light from led hits transistor OBJT1 then Vce of that transistor will be some 0.1V or pretty close to "touching wires". so Vce of BJT1 will be about 0.8V and current through 1k resistor will be (Vdd-0.8V)/1k or about 8.5mA. though, not sure for what purpose...

 

you did not label your circuit.
if light from led hits transistor OBJT1 then Vce of that transistor will be some 0.1V or pretty close to "touching wires". so Vce of BJT1 will be about 0.8V and current through 1k resistor will be (Vdd-0.8V)/1k or about 8.5mA. though, not sure for what purpose...

Thank you for your reply.
I agree with the calculation you made, but unfortunately I too do not understand the point of the circuit

It doesn't look like a simple ON/OFF switch of the middle BJT, does it?

 

Hi 74,
What is the original source of the circuit design, who gave it to You?
E

 

This looks like homework or a school assignment. Please provide your best understanding of the circuit so far.

Also, like Panic Mode mentioned, you have to label the components, D1, D2, R1 and so on. Saying “the middle BJT” could create confusion to exactly what you meant.

 

1. D1,D2 and R5 form an OR circuit. Higher one wins. Not sure what way the right side "tells" is something is ok or not - there are no indicators or outputs - the circuit is a dead end. are you SURE that DCX_OK and DCY_OK is not on the other side of R1/R3 (on collector of Q1/Q2)?
2. 3.3V is never on the base. signal from 3.3V may reach base if X1/X2 is powered and depending if OBJT is an optocoupler or photointerrupter. later one would work if light path is clear.
3. already explained
4. already explained

 

Let us analyse this design together, I have several points to clarify with you so I will make a bulleted list.
Tell me if I am wrong.

1) On the left, if I'm not mistaken, Vdd takes on the value of Va or Vb depending on which is present, right? So it is as if they are in OR?
If both Va and Vb are ON ... Vcc is equal to the higher value of the two?

Now let's focus on the right side.
Let's assume that DCX_OK and DCY_OK are "status" signals, i.e. they tell me whether +3.3_X and +3.3V_Y are good or not (whether they work or not)

2) The central bjts (BJT1 and BJT2) will be closed or opened depending on whether +3.3_X and +3.3V_Y on their base are present or not, right?

3) If +3.3_X is present, that opto-bjt1 (suggest name) will close and then C and B of BJT1 will short? I don't quite understand how OBJT1 works ... of

4) If the answer is yes, how does BJT1 close when C=B? .. usually closure implies that E and C are ideally a "wire"


View attachment 355792

Please, if possible, answer in order, perhaps respecting the bulleted list

Thanks

Hi,

The short answer to all of this is that when you see something that looks obviously wrong, it means you have to do a more careful analysis. That sometimes means you have to modify the assumptions or at least go over them to see if they are not truly compatible.

In this case, the assumption that vC=vB is not good, and neither is vC=vE when the transistor is 'on'. It is not only rare to see those conditions is it impossible. That's because when a BJT turns 'on' as much as possible, there is always some voltage drop. It can vary from roughly 0.1 volts up to almost anything even 5 volts or even higher (fully supply voltage). That's because the voltage drop C to E depends on the CE current and that depends a lot on the BE current.
Looking at the BE current first, the BE current will not be very high because the opto-couplers we commonly see do not have a very high current transfer ratio and that is usually less than 1. That means if you apply 5ma (for long opto life) then you might expect 5ma output. That means the BJT being driven will have less than 5ma BE current because the BE resistor will shunt some of that current, and we don't want to get rid of that resistor entirely. But let's say it gets 4ma and that BE resistor only conducts 1ma. If the gain of the BJT is 50 then the CE current will be around 200ma (if possible) and that would be enough to drive the 1k resistor (1k at 10v will never draw that much). However, the CE voltage will not go to zero it will stay at some level like 0.1 to 0.8v or about that. A general-purpose transistor might go as low as 0.5v, and a more expensive BJT might get to 0.2 volts, but that is about the best you can get. That in turn means the 1k resistor will get a voltage max of about 9.5 to 9.8 volts.

We could also talk about starving the BJT for BE current but I'll assume for now that you drive the opto with enough current. A flaw in the diagram though is that the cathode of the opto LEDs do not have a return path. If the return is ground then that's good, but if the cathode is at 3.3 volts then it will never turn on. Thus, I have to assume for now that the cathodes are at ground potential.

In the end your analysis is not really that bad, but your assumptions about the CE voltage drop for any of the transistors (including the internal ones for the opto's) is just not good enough to fully understand how this circuit works. Once you modify that though you will come up with good results I'm sure.

As a side thought, if MOSFETs were being used instead of BJT's then we would get different results if the MOSFETs were chosen to have low drain to source resistance. Even a 1 Ohm DS resistance would only drop about 0.001 volts with a 10v source and 1k resistor as load.

 

View attachment 355797
1. D1,D2 and R5 form an OR circuit. Higher one wins. Not sure what way the right side "tells" is something is ok or not - there are no indicators or outputs - the circuit is a dead end. are you SURE that DCX_OK and DCY_OK is not on the other side of R1/R3 (on collector of Q1/Q2)?
2. 3.3V is never on the base. signal from 3.3V may reach base if X1/X2 is powered and depending if OBJT is an optocoupler or photointerrupter. later one would work if light path is clear.
3. already explained
4. already explained

@MrAl @schmitt trigger @ericgibbs @panic mode

I was looking through some old notes on my tablet (probably from my last job, I did fimrware not hardware) when I found the circuit I sent in the main post (I send it back to you below).
After a bit of research to get and extra information, I found that the part on the right with the bjts is an internal circuit of the power supply, probably used in one of the machines some colleagues were working on.
In the schematic I don't know why it says DC and not AC (as stated on the datasheet) but it doesn't matter, I'm interested in the logic.

I unfortunately I do NOT have any other details on the schematic, so I would like to analyse this (which you have done, but maybe with this new important information there are other things to add).

In the image below I enclose the same schematic of the main post ... and a second circuit that is the same original schematic but with the addition of a pull-up (circled and dashed), because I vaguely remember that there was a pull-up at Vcc ... but I'm not sure, so maybe we'll analyse that case too.



Unfortunately, I have no more information than these.
I am quite convinced, however, that those DCX_OK and DCY_OK signals are output from the power supply, signalling whether it is correctly supplying voltage.

 

I am quite convinced, however, that those DCX_OK and DCY_OK signals are output from the power supply, signalling whether it is correctly supplying voltage.

what is receiving or turning those signals into visible information? there are no LEDs or connections to other circuits. you placed those two labels on supply voltage. in other words, state of those two does NOT depend on inputs +3.3_X or +3.3_Y

it should be something like this:

 

@MrAl @schmitt trigger @ericgibbs @panic mode

I was looking through some old notes on my tablet (probably from my last job, I did fimrware not hardware) when I found the circuit I sent in the main post (I send it back to you below).
After a bit of research to get and extra information, I found that the part on the right with the bjts is an internal circuit from this power supply (page 57), probably used in one of the machines some colleagues were working on.
In the schematic I don't know why it says DC and not AC (as stated on the datasheet) but it doesn't matter, I'm interested in the logic.

I unfortunately I do NOT have any other details on the schematic, so I would like to analyse this (which you have done, but maybe with this new important information there are other things to add).

In the image below I enclose the same schematic of the main post ... and a second circuit that is the same original schematic but with the addition of a pull-up (circled and dashed), because I vaguely remember that there was a pull-up at Vcc ... but I'm not sure, so maybe we'll analyse that case too.

View attachment 355958

Unfortunately, I have no more information than these.
I am quite convinced, however, that those DCX_OK and DCY_OK signals are output from the power supply, signalling whether it is correctly supplying voltage.

Hi,

Since DCx and DCy are connected directly together, I would think that the better place to see the outputs would be on the collectors of Q1 and Q2.

 

what is receiving or turning those signals into visible information? there are no LEDs or connections to other circuits. you placed those two labels on supply voltage. in other words, state of those two does NOT depend on inputs +3.3_X or +3.3_Y

it should be something like this:
View attachment 355961

"The transistor shall turn on when input mains level is good>85Vac"
Sorry, I do not understand your doubt

 

Sorry, I do not understand your doubt

i have no doubts - i know exactly what i stated. it is you who does not understand the circuit...

you should think of J1-2 and J1-1 as a simple switch contact. you should also see Vdd as a fixed voltage.
this is one way to use that status - turn on an LED:


another possibility is like this. note that it requires external series resistor (R2). and in general, R2 should be larger value than R1 (the internal resistor of the PSU). This R2 is the Rp resistor that I show in post #9



but this is what your idea in post #8 was:

 

i have no doubts - i know exactly what i stated. it is you who does not understand the circuit...

you should think of J1-2 and J1-1 as a simple switch contact. you should also see Vdd as a fixed voltage.
this is one way to use that status - turn on an LED:
View attachment 356056

another possibility is like this. note that it requires external series resistor (R2). and in general, R2 should be larger value than R1 (the internal resistor of the PSU). This R2 is the Rp resistor that I show in post #9

View attachment 356057

but this is what your idea in post #8 was:
View attachment 356058

View attachment 356059

Thanks for your reply! I will analyse the circuit on LTSpice that you made.

Going back to what is written in the power supply datasheet, I don't understand the operation of the transistor:


I read from your reply that J1-2 and J1-1 are switch contacts, but that bjt configuration I still don't understand (electrically speaking) .. same for the right first opto-bjt (I don'tknow the name)
Should I also see pins 3 and 4 as switch contacts?
Can we analyze the behavior in function of S3V3D?

 

Thanks for your reply! I will analyse the circuit on LTSpice that you made.

Going back to what is written in the power supply datasheet, I don't understand the operation of the transistor:
View attachment 356087

I read from your reply that J1-2 and J1-1 are switch contacts, but that bjt configuration I still don't understand (electrically speaking) .. same for the right first opto-bjt (I don'tknow the name)
Should I also see pins 3 and 4 as switch contacts?
Can we analyze the behavior in function of S3V3D?

(See attachment)

When we apply 0v to the input side (Red) we see 0v at the base of the transistor and +10v at the collector (Blue is output) of the transistor.
When we apply 3.3v to the input side (Red) we see about 0.7v at the base of the transistor and about 1v at the collector (Blue is output) of the transistor.

So, the input determines the voltage that appears at the collector of the transistor. The 1k0 resistor draws current from the +10v supply but the +10v supply does not change in voltage.
Also note that the Red GND ground does not connect to the Blue GND ground that's how we get galvanic isolation.

Note the original diagram that first appears in this thread may not be draw correctly.

 

and all of that is inside the PSU... inaccessible. there is nothing external to react to signal. the 1k0 internal resistor is not the load, it is current limiter to protect the PSU output. the internal detail was shown only as a simplification to inform you of polarity, isolation, and limited protection.

they could have go more abstract and show it just as two terminals and text showing polarity and voltage/current limit. then you would interpret it as a weak contact and try to think of a circuit. pretty sure you would not want to wire it the way you did because this is what you actually had - switch connected to power source - without load. You are lucky that PSU designer added some protection or that output would be toast.