Interpretation of existing small circuit

[Adder_Noir]

Dear All,

Before I lose the link I'll post it up:

http://img290.imageshack.us/img290/9122/voltageregulatorlh7.jpg"

Can anyone help me identify the unknown component in this picture drawn in blue? I'm trying to come up with an interpretation and circuit diagram of this circuit. It's a voltage regulator from an alternator on a car. I presume it has the task of both rectification and voltage output limitation. It is externally linked to a capacitor before being transmitted to the battery.

My electronic engineering is not my strong point although I do have a strong electrical background (I/O board testing and programming) and I can understand most simple stuff with a nudge in the right direction.

So, does anyone have any ideas what these components are? Thanks in advance for the help and to anyone who even just takes the time to read the thread

 

[NoTime]

The image is not good enough to tell.

This unit does not do rectification.
The diodes (usually 6) for that are large and on heat sinks.

The regulator circuit is probably similar to the one shown here
http://www.1stconnect.com/anozira/SiteTops/energy/Alternator/alternator.htm

 

[Adder_Noir]

That's great thanks a lot mate. Just one thing to let you know that the image shown is a heat sink. The circuitry is glued onto the back of a heat sink (the metal behind the circuit board has fins on the other side).

I thought they would be too small to be doing rectification especially for charging a car battery. Only thing is I can't find any diodes anywhere which could be doing the rectification.

So can I ask exactly what does the voltage regulator do?

**Edit I've just had a look at that link you posted. It's superb, thanks alot.

 

[Adder_Noir]

Hi again. I've done a lot of reading and a lot of work on the information contained in that link (and what a link it is!).

Given that I have very little (virtually no) background in electronic schematics and only just figured out what a transistor does the other day, I could do with a little bit of advice. I've worked as hard as I can on the following diagram to come up with an interpretation of what it does and how the components work. All I need is someone to just confirm whether or not I'm right on the key points.

http://www.1stconnect.com/anozira/SiteTops/energy/Alternator/alt_06_1.gif"

Before I list my conclusions I should state that this is a voltage regulator circuit for an automobile alternator. It works by altering the current in the rotor to keep the heavy output current of the stator at 12V. It does this by using a set potential divider (the resistors labelled 'C' in the diagram) to deliver a pre-determined fraction of the output voltage back into the regulator circuit. This is then compared to a Zener diode which either boosts current into the rotor or dumps it out as heat instead.

My conclusions are as follows:

• The output voltage comes in and is divided by the potential divider.
• From there it attempts to drive a current through the Zener diode and its success depends upon the comparison between the voltage driving it and the breakdown voltage of the Zener diode.
• If unsuccessful the control transistor remains open, and vice versa.

So basically if the voltage is too high the control transistor closes, but if it's too low it remains open. Now looking at the right hand part of the circuit:

• The voltage drives current through the charge lamp and into the loop containing the two bias resistors, rotor coil, and bias diode.
• The right hand bias resistor has high resistance whilst the other has a low/medium resistance.
• When the voltage is too low the control transistor remains open and a small amount of current flows through the high bias resistor on the right and turns on the valve transistor closing it. This allows a virtually zero resistance (with the exception of the rotor coils internal resistance) path through the rotor coils, through the closed valve transistor to earth, so increasing current in the rotor. The bias diode being there to make certain current always flows in the desired direction.
• If the voltage is too high then the control transistor is turned on and closes. Consequently almost all the current on the right side of the circuit now flows through the low/medium bais resistor and not enough current reaches the valve transistor to turn it on, so it remains open. Current is consequently dumped to Earth and rotor coil current drops.
• The protection diode is there to allow current to be dumped to Earth in an emergency through a filter resistance (shown on the far right bottom part of the diagram) should the valve transistor overheat to a dangerous level.

And that's it. So how did I do? I guess this is the kind of control which is called 'hunting'. I heard a little about it in Uni from my old tutor. Also seeing as this circuit board is obviously non-programmable could it accurately be called a microprocessor?

I'd be as always most grateful for a response. Thank you in advance

 

[NoTime]

Before I list my conclusions I should state that this is a voltage regulator circuit for an automobile alternator. It works by altering the current in the rotor to keep the heavy output current of the stator at 12V.

Just be aware that there is some simplification in the website and that a car battery actually charges at 13.6v, not 12v.

My conclusions are as follows:

• The output voltage comes in and is divided by the potential divider.
• From there it attempts to drive a current through the Zener diode and its success depends upon the comparison between the voltage driving it and the breakdown voltage of the Zener diode.
• If unsuccessful the control transistor remains open, and vice versa.

So basically if the voltage is too high the control transistor closes, but if it's too low it remains open. Now looking at the right hand part of the circuit:

You seem to have this part correct.
Note that there is a breakdown voltaage on transistors as well. This is about 0.6v for silicon BJT as shown here.
For purposes of this schematic you can add the 0.6v to to the breakdown voltage of the Zener.

• The voltage drives current through the charge lamp and into the loop containing the two bias resistors, rotor coil, and bias diode.
• The right hand bias resistor has high resistance whilst the other has a low/medium resistance.
• When the voltage is too low the control transistor remains open and a small amount of current flows through the high bias resistor on the right and turns on the valve transistor closing it. This allows a virtually zero resistance (with the exception of the rotor coils internal resistance) path through the rotor coils, through the closed valve transistor to earth, so increasing current in the rotor. The bias diode being there to make certain current always flows in the desired direction.
• If the voltage is too high then the control transistor is turned on and closes. Consequently almost all the current on the right side of the circuit now flows through the low/medium bais resistor and not enough current reaches the valve transistor to turn it on, so it remains open. Current is consequently dumped to Earth and rotor coil current drops.
• The protection diode is there to allow current to be dumped to Earth in an emergency through a filter resistance (shown on the far right bottom part of the diagram) should the valve transistor overheat to a dangerous level.

A little less well on this part. All in all not a bad first try though.

Take the coil, right hand bias resistor and valve transistor.
All coil current flows thru the valve transistor.
The bias resistor is there to always keep the valve transistor turned on.
Partly this provides the same function as the protection diode.
It also functions to provide a minimum charge current, by always keeping some current flowing in the rotor coil.

The bias diode keeps the control transistor from being able to turn the valve transistor completely off.

The left hand bias resistor sets the maximum current to flow thru the rotor coil
You can determine the amount of current from the gain of the valve transistor and value of the bias resistor.

When the control transistor turns on because of overvoltage it shorts the left hand bias resistor to ground. Effectively removing it from the circuit as the bias diode becomes reversed biased.
Note that the control transistor will normally reach an intermediate state between off and on.

Protection Diode: Current will keep flowing in a coil after power is removed. This can generate a very high voltage destroying the transistor. The diode shorts this current in a loop with the coil eventually dissipating it in the coil resistance.

"Filter" resistance provides a bias current on the charge lamp. It has no other function.

Hope this helps.

 

[Adder_Noir]

That's great mate. It's amazing the amount of help you can get on here if you're prepared to do some work

I'll have a good read through your post today and find out where my key mis-conceptions are. Thanks again

 

[Adder_Noir]

I've read through your reply and cross-referenced it with the link material and I understand exactly what you are saying as you explained things really well.

It has raised another question though which I hope is the last but somehow suspect not

What kind of current would we usually be talking about to turn on a small transistor such as this one? I get the impression transistors are slightly analogue in nature and can have an intermediate stage between on and off. So what kind of current values are we looking at to saturate it on, and what current values would just maintain it partially on?

Thanks in advance to anyone whom answers or just reads the thread

 

[berkeman]

The beta (current gain from Ib to Ic) for small transistors is in the range of 100-300, so for a collector current Ic of 100mA or so, you will need to supply a base current of about 1mA. For larger power transistors, the beta is smaller, say 50 or so, which means that for an Ic of a couple amps, you will need to supply an Ib of a few 10s of mA or more.

I'd recommend that you check out the book "The Art of Electronics" by Horowitz and Hill. You are at the point in your curiosity and learning where that book will help you a tremendous amount. Read it cover to cover, and you will have a very good understanding of all this basic electronics stuff.

 

[Adder_Noir]

I'd recommend that you check out the book "The Art of Electronics" by Horowitz and Hill. You are at the point in your curiosity and learning where that book will help you a tremendous amount. Read it cover to cover, and you will have a very good understanding of all this basic electronics stuff.

As soon as I next get paid that's the first thing I'll do! I was hoping someone would step forward and recommend a good book on the subject pitched at beginners level.

Thanks for the reply

 

[Adder_Noir]

I've done some more work on this. Quite a bit actually. Haven't had the money to buy the book yet (things really are that tight). Got some great info from Wiki on transistors though and spent a week or so understanding as much about them as I could. I see them more as amplifiers now than switches!

I've measured the actual resistance of my alternator's rotor coil and it came back at 4.9 ohms. I've come up with two suggested figures for minimum charge current and maximum rotor current so see if I could evaluate the bias resistors values when desired current values had been stated. It worked out ok:

I reduced the driving voltage to 13.0V from 13.6V to account for the overcoming of the majority carrier depletion region in the valve transistor base-emitter junction. I assumed the transistor gain beta to be 100.

I aimed to achieve a minimum charge current of 0.5A. Whether or not this is a realistic figure doesn't matter for the moment.

Putting x1 (the right hand bias resistor) in a circuit with the rotor coil with a driving voltage of 13.0 linked to ground produced a value for x1 of 2595.1 ohms given that beta = 100.

Again given that a maximum charge current of 2A means a base-emitter current of 0.02A then the current contribution from x2 (the left hand bias resistor) when the control transistor is firmly closed is 0.015A because the x1 route is already supplying 0.005A regardless of the control transistor's behaviour.

Pumping it through came out with a value for x2 of 866.667 ohms.

So I think if I could acquire real world values for maximum and minimum rotor current I could now make a suitable choice of bias resistors and valve transistor.

I have one important question though.

If I chose a control transistor with a gain that was too high could I produce an unstable system or would it just produce a much more responsive system? Likewise could I overdamp the system by choosing a control transistor with a gain too low? Is there a realistic way I can model the circuit to come up with an ideal choice of reaction gain?

One more question also. I notice on my existing board there is a large transistor located in the middle. Is it there to make the best use of the heat sink (by placing it right in the middle)?

Thanks again in advance for the help

 

[NoTime]

The values look reasonable, but I haven't checked your math.

The values of the resistors in the voltage divider determine the required gain or conversly you pick the resistor values for the gain you have.
The allowable output error determines how much gain you need.
More gain less error.
Generally, the output voltage will rise as the final output load decreases, so enough current has to flow from the divider multiplied by the gain to divirt the x2 current and shut down drive to the field coil.

The filter cap damps the system.
If its too small the system could indeed oscilate.
If its too big then the system will not accurately follow rapid output load changes.

Yes, the large transistor needs a lot of heat sinking and centered is the best place.

 

[Adder_Noir]

Ah, thank you kindly again for the help NoTime. It's most appreciated. Just a shame that with the exception of only a few others you're the only one who's taken the time to assist

Sometimes I have to print off your posts and do some work on them for a while before I fully understand them but I got the gist of that one quite well. I'll pick through all the concepts you raised next week. The one that's surprised me the most is that it is the capacitor which is responsible for keeping the system stable. The original one in my car was 2.2 mu farads.

I'll be back in touch again if I have anymore questions, but in the mean time thanks for the help

 

[Adder_Noir]

More work done!

I've been bashing away at my notebook hard again and have reached the next question. It's capacitor related. Perhaps someone other than poor old NoTime can answer on this occasion

I've been doing yet more reading and happened across a very nice linear way to model a capacitor's behaviour regardless of applied voltage by using the time constant equation. Only question is that the time constant equation is:

time constant = resistance * capacitance

But there is no resistance in the circuit with the cap in the voltage regulator circuit. There is a resistor there but it's supplying the voltage across its terminals to the cap so is not counted. Therefore what rough figure could I use for the resistance of the loop? I'm trying to get as accurate as I can here and I was wondering would something like 0.001 ohms be realistic for an electronic pathway with no resistor?

Also is there any significant time delay for transistor behaviour I need to worry about with regards to collector/emitter current lagging base/emitter current?

As always I never ask a question I can find the answer to easily myself. I always crunch everything through before pestering people. Once I've got this sorted out I'll attempt to create a graphed model of it's behaviour in excel. I expect I'd be able to come up with a reasonably accurate behavioural model given that the circuit is not too complex and I don't have to get involved with Laplace stuff thankfully

All I need to do now once the above problems have been solved is to try and find out how quickly the alternator output reacts to changes in rotor current. I suspect I may need a new but brief topic started on that one

 

[berkeman]

I've kind of lost track of which circuit you are talking about now. Can you point me back to it, or re-post it?

 

[Adder_Noir]

I've kind of lost track of which circuit you are talking about now. Can you point me back to it, or re-post it?

I decided to forget about the actual one in my car, and work on producing a new design based on this template here:

http://www.1stconnect.com/anozira/SiteTops/energy/Alternator/alt_06_1.gif"

That's what all the discussion has been about in the recent posts

 

[berkeman]

Then the diagram is mislabeled with the two resistors and the potentiometer all called "C". Maybe that's where the confusion is coming from. The R for the RC would be the parallel combination of those three resistances (and also the parallel impedance presented by the zener feeding the Vbe junction), and the C is the capacitance of the capacitor. Or else I'm still not understanding your question...

 

[Adder_Noir]

Thanks for the reply. I'll make it a bit clearer below as it's got a bit confusing.

This is already established and I'm okay on it

The use of the letter 'C' is purely arbitrary in the diagram.

Just in case it is causing confusion, I decided to label the two bias resistors x1 and x2 respectively for indentification purposes.

The potential divider made up of the three resistors labelled 'C' (why they didn't label them as 'R' instead I really don't know ) is there to put a desired percentage of the feedback voltage across the zener diode.

Say for example the feedback voltage optimum was 12 volts. You could use the potential divider to split it so an optimum of 6 volts would be present across the Zener diode. You then obviously choose to use a Zener diode with a breakdown voltage of 6 volts.

That way if the output of the stator coils (and consequently the feedback voltage) goes above 12 volts, then the Zener diode conducts in the reverse bias direction and turns on the transistor.

If the feedback voltage is less than 12 volts the Zener stays non-conductive and the transistor stays off.

The next bit is the part I need help with

If you look at the diagram it shows that the potential divider is placing a voltage across the capacitor which is in parallel with the Zener diode.

On the diagram it shows this as being represented by part of one resistor, and all of another. Just ignore that and assume there were just two resistors in the potential divider, both of equal magnitude, splitting the feedback voltage 50/50 so that half of the feedback voltage was across the Zener diode.

What I want to know is:

The very bottom left loop in the diagram can be thought of as just one resistor with a voltage across it, in parallel with a capacitor.

Given that the time constant equation has a resistance and a capacitance in it, and I want to use the time constant equation - should I put the value of 'R' in the time constant equation t = RC equal to the resistor which has the voltage across it (i.e. the bottom half of the potential divider)...

..Or should I put it almost (but not actually) equal to zero.

The reason I ask is the voltage trying to drive current round that loop exists across the terminals of the resistor. So does that mean I can remove the value of that resistor from the circuit when using the time constant equation or should I just include it as if the resistor were in series with the capacitor and the voltage were coming from a battery?

Here's a picture to help explain what I mean:

http://img329.imageshack.us/img329/254/picrcdc3.jpg"

 

[NoTime]

I resemble that remark

time constant = resistance * capacitance

But there is no resistance in the circuit with the cap in the voltage regulator circuit. There is a resistor there but it's supplying the voltage across its terminals to the cap so is not counted. Therefore what rough figure could I use for the resistance of the loop? I'm trying to get as accurate as I can here and I was wondering would something like 0.001 ohms be realistic for an electronic pathway with no resistor?

Without the zener connected.
If the cap is charging then R is the upper divider resistance.
When discharging R is the lower divider resistance.

Also is there any significant time delay for transistor behaviour I need to worry about with regards to collector/emitter current lagging base/emitter current?

Not a consideration here, but yes this can be important.
This time is related to the cutoff frequency.
T_on/T_off is another way of presenting this.

 

[Adder_Noir]

Before I begin a new topic on working out input/output voltage relationship on the windings I just wanted to thank those who replied in this topic for the help they gave me