What are the Functions of the Resistors and Diodes in this Circuit?

[cjm181]

Homework Statement

i am working through this question, the final bit i am struggling with, describing what each bit of the circuit does. tying to understand what is going on. Looking at fig 1b

I have identified a bridge rectifier, and the zener diode. I am reading up on these so i am not looking for help with these at the mo! What i can't seem to sus is what the two resistors are doing in the ac supply before the bridge rectifier. And also what is the resistor after the rectifier, the vertical one? The coupling is an optoisolator. Can anyone help me or point me to some further reading please?

Homework Equations

The Attempt at a Solution

Are the two resistors to limit the current going into the rectifer?

Thanks

 

[jaus tail]

I don't think the resistors would be to limit the current going into the rectifier. Since the resistors are permanently in the circuits there'd be power loss across them.
Also if you want to reduce current, maybe use one resistor in series why both?

 

[Numbskull]

Also if you want to reduce current, maybe use one resistor in series why both?

If you were using resistors to limit the current, you could use two smaller resistors to dissipate the power rather than one big one? Cost / size constraints?

 

[gneill]

It could be for current limiting safety. It would appear that the circuit is to be directly interfaced to the mains, and you definitely don't want any direct contact with a mains line to be made directly accessible to external contact as might occur at the coupling interface (say if there's a connector there). An isolation transformer would be preferable, but presumably is omitted to save size, weight, and cost.

 

[NascentOxygen]

I'd draw in the relevant side of the optocoupler circuit, so it can be considered in association with your 230V components.

 

[cjm181]

its says breifly explain! maybe the resistors are for something to the left not shown?

 

[Nidum]

It could be based on an old style line dropper circuit with bits added on to drive the opto coupler .

The two resistors to the left of the diode bridge drop the voltage across the diode bridge to a level much lower than mains level . The other resistor provides a ballast load to set the current flowing to a level which gives the right voltage drops across the line dropper resistors .

nb: Current taken by opto coupler << current in ballast resistor .

Not a good or an efficient arrangement but very low cost compared with a transformer circuit .

 

[cjm181]

Thanks everyone!

My interpretation of the circuit so far (bear in mind i am a mechanical guy!)

From left to right

The two resistors in the 230V ac supply - to reduce the voltage and current going into the next component

The diode bridge rectifier - to convert ac to dc

The parallel resistor - to further reduce the current and voltage going to the zener diode

The zener diode - this regulates and stabilises voltage going to the load (in this case the opto isolator). it does this by operating in the breakdown region, where voltage remains relatively constant regardless of applied current (within limits). Also keeps a consistent voltage if the load changes. However not sure how the load would change with an optoisolator?

the parallel diode - not sure about this one. I am going to guess if anything goes wrong with the rectifier, it allows current to flow thru it from bottom to top. but won't allow current to flow from top down, so in that instance all the current / voltage goes to the opto isolator.

Am i miles off?

Thanks
Craig

 

[NascentOxygen]

The zener diode - this regulates and stabilises voltage going to the load (in this case the opto isolator).

Your explanation is okay up until here. The zener here is not used how you may think, and it does not stabilise the voltage across any element (apart from itself!).

I conclude that you haven't followed my earlier advice, then?

 

[cjm181]

ok, so to draw in the optoisolator will be to draw in an LED on the left side of the coupler box, and a receiver on the right side of the coupler box.

can you expand on the comment 'so it can be considered with the 230v components'?

ummm

so my led knowledge is not great, but i know u need to limit the current going thru them, with a resistor!

 

[Nidum]

The zener here is not used how you may think, and it does not stabilise the voltage across any element

+1

 

[NascentOxygen]

can you expand on the comment 'so it can be considered with the 230v components'?

Everything between the 230VAC plug and the optoisolator LED/s is there for the purpose of safely providing an appropriate drive current to power those LEDS. You need to consider the whole lot in combination when trying to understand what purpose each element serves, and ask yourself questions about each.

Start by drawing a large sinusoid, superimposing on it the rectified current waveform, then for each element work out what its current waveform will be doing throughout the mains cycle.

 

[cjm181]

ok, not sure then, let me have a think...

the zener, are we calling this a unidirectional zener? based on its symbol.

So, i am calling the top horizontal line positive, and the bottom line negative.

lets say there is a 12v DC supply, and the coupler needs 5v supply, if the zener diode is 7v, won't 7v be lost across the zener as heat, and the coupler will receive 5v across it?

 

[NascentOxygen]

lets say there is a 12v DC supply, and the coupler needs 5v supply, if the zener diode is 7v, won't 7v be lost across the zener as heat, and the coupler will receive 5v across it?

In this hypothetical circuit you have thrown in for discussion...yes.

 

[cjm181]

ok, back up to post 5

 

[cjm181]

so, diode 1, the zener, allows current and voltage to flow from left to right once the voltage exceeds the knee voltage. an amount of voltage will be lossed as heat. However it does allow large changes in current to pass thru??

so, looking at the led diode in the coupler, and indeed diode 2, what is limiting the current going thru the circuit?

and what's D2 doing? that's going to allow voltage / current to travel upwards, but stop it traveling downwards

 

[Nidum]

Deleted

 

[NascentOxygen]

so, diode 1, the zener, allows current and voltage to flow from left to right once the voltage exceeds the knee voltage. an amount of voltage will be lossed as heat. However it does allow large changes in current to pass thru??

Yes. (Though don't overstate "voltage lost as heat" because heat is also a function of current; just need to explain it as a voltage drop across the zener.)

so, looking at the led diode in the coupler, and indeed diode 2, what is limiting the current going thru the circuit?

That's a good point---LEDs do need their current limited. Nothing obvious here in your sketch. Perhaps something you have omitted to draw, something between here and the 230V plug?

 

[cjm181]

ok, so to summarise there are 4 things,

1. what's limiting the current going through the circuits for the led and D2
2. Whats the zener doing?
3. what is r1 doing
4. what's d2 doing.

so, answering point 1 - well the bridge can't be limiting current, as that is made of diodes. the only other thing is the two resistors R2 and R3, in the ac line. So R2 and R3 limit the current going thru the circuit!

2, still no closer to this. If the two resistors in the ac side are handling current, then the zener can't be current related. (although i know they let large currents thru. How ever when i type zener diode into google, a lot of the returns are regarding voltage regulation. Further on this, the voltage in the circuit must be controlled, so i am thinking it is still involved in supply to the coupler. i think its working in conjuction with r1.

3?

4. the bridge rectifier is made up of diodes. diodes of this type let voltage / current flow one way, and not the other way (ideally). I know in reality there is a little bit of leakage. I also know that if enough current / voltage is applied in the wrong direction, the diode will start to conduct, and damage it. So if this was to happen, current could flow along the negative line, so what would happen then? Current would travel thru r1, some thru d2, as that's the direction it will work in, but the coupler will become open circuit. if r1 and d2 were not there, then the coupler would be damaged, as the same situation would apply as what happened to the rectifier diodes, the current would become so high the led would start to conduct.

Going the right way, r1 is still in use, the zenner is being used, but d2 (assuming the current / voltage is ok) is not being used.

 

[NascentOxygen]

That looks okay.

You should be able to determine ballpark values for R₂ and R₃ by looking at manufacturers' datasheets to find a typical value of current needed to drive an optoisolator LED.

Have you thought about the waveshape being provided to the logic circuits in the PLC? You know what the rectified sinewave looks like...and the circuit you are examining here manipulates that mains sinewave into becoming what the PLC needs. What waveform would that typically look like, in order to clock the PLC?

 

[cjm181]

ummm, ok, i got this, ac waveform in, dc waveform out.

assuming the required input signal to the plc needs to be constant, then the dc input will effectively be an on /off signal.

But surely this could be taken care of in the logic if that's an issue? not sure. i don't see this answers point 2, 3 and 4

thanks for ur patience so far

 

[cjm181]

as i am mechanical, i am loosly thinking of the zener as a pressure (voltage) releif valve. so i see it releasing a voltage once the input voltage is at a certain point. so this will 'open and close' still with a dc supply? for point 2

 

[NascentOxygen]

On the PLC side it is getting sent a logic signal from the 230V mains for some purpose. This won't be a fixed DC. What do you expect it will be? Once you figure out what it will probably look like, you can work backwards and deduce how the components on the "hot" side of the optocoupler work together to shape it thus.

Converting the sinusoid to a rectified sinusoid is only a preliminary step.

 

[cjm181]

Ok. I think I see where you are going! Does the Zener diode help to 'square off' the dc waveform. The 'squaring off' will be determined by the knee voltage of the zener.

 

[NascentOxygen]

Ok. I think I see where you are going! Does the Zener diode help to 'square off' the dc waveform. The 'squaring off' will be determined by the knee voltage of the zener.

If component values are well chosen, it should do so---though by "squaring off" you may be thinking of something quite different from what I mean!

You could take two widely-different values of V_z, say 5V and 50V, and see how things differ between the two cases just by rough graphical sketching. This is getting beyond the "brief outline" of operation, but to understand a circuit's design you really do need to understand what each component is doing and why its value has been chosen to be what it is.

 

[cjm181]

I am thinking the zener outputs this, graph c

 

[NascentOxygen]

I am thinking the zener outputs this, graph

Did you intend to erase the large capacitor from that diagram before posting? Those 2 waveform sketches (b) and (c) will not exist on that rectifier circuit when it has a capacitor, but they will be seen when it has no capacitor.

There is no capacitor in your PLC drive circuit.

If you imagine that basic rectifier/regulator (but without the filter capacitor) in Fig (a) modified by having your optoisolator LED connected in series with its zener, i.e., between zener and ground, then it models perfectly what is going on in your PLC drive.

 

[cjm181]

Yeah, i ment literally graph C, ignore everything else in the picture! Just the first picture i came to that looked similar.

Ok, back to post 19

1, The two resistors R2 and R3 are limiting the current coming into the rectifier, and hence the DC part of the circuit. This is necessary because diodes will let large currents flow through them, which if uncontrolled will damage the diodes. This protects the rectifier, D1, D2 and the optocoupler

2, The zener diode allows voltage and current to flow thru it in reverse (reverse bias) once the knee voltage is reached. Any voltage less than the knee voltage and the zener diode will resist flow. This gives the opti isolator and the PLC signal as shown in post 26 fig C

3, I am thinking, if the knee voltage has not been reached, R1 allows any residual current to pass thru it. If this resistor, if sized correctly, help to reduce leakage across the zener?

4, D2 is there for protection. If anything goes wrong with the rectifier, and current is allowed to pass along the negative line, this would damage the opto isolator. With D2, current would be able to flow across this diode, protecting the opto isolator. Also the zener would allow flow in this direction the same as a conventional diode.

How close am i now? any screamers?

 

[NascentOxygen]

That is very good. I think it covers most points.

Harking back to the circuit in your original post, you can see the three R's also form a resistive divider across the mains, so whatever happens, the potential difference where the zener connects in cannot exceed ⅓ of the mains' peak (if all R's are equal valued). This sounds like a safety advantage.

 

[cjm181]

Hey Nascent!

Thanks very much for your help. Certainly got the cogs turning. I think i have enough now to complete the question.

Thanks again for your help and patience!