Author Topic: Some noob questions  (Read 3072 times)

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Offline Brumby

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Re: Some noob questions
« Reply #125 on: August 18, 2018, 09:34:32 pm »
It is unfortunate that terms like "Earth", "Ground" and "Chassis" are all too often used to simply make reference to a common connection.  When the familiar definition of these terms is not applicable, it really can be confusing.

This common connection is also frequently given the role as the reference point for other measurements.  While this is not a guaranteed relationship, it would be rare to come across a schematic which did not do this.

The only time you can really be certain of a common connection being your reference point is when you have something described as 0V - such as a 0V power rail.

In practical terms, this reference point is (as you have worked out) where you would put the black lead from your DMM.

As you become more familiar with schematics and circuit topologies you will be able to make better judgements about reference points and measurements - but the first steps are understanding the simple things.  Each of these can be mysterious and confusing, until it 'clicks'.  Those are wonderful moments.
 

Offline Mr D

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Re: Some noob questions
« Reply #126 on: August 19, 2018, 09:04:21 am »
Thanks!

I feel like i've got just enough understanding to start looking at some real circuits.

Could anyone tell me: in the circuit below, how come i see no power source?

Or is that triangular element in the middle the power source? But it's not a battery, so what is it?!

Yeah yeah, this is a super-nooby question!!

 

Offline rstofer

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Re: Some noob questions
« Reply #127 on: August 19, 2018, 09:10:34 am »
The triangle is an op amp and the power supplies (probably +-15V) are not shown.  This is typical of all op amp circuits used in tutorials.  There might be an exception when the circuit uses a single rail op amp.  Then the + power input will have some voltage and the - power input will have a ground symbol.

That' an awfully advanced circuit considering...

Here's a circuit that shows the power supplies:

http://www.learningaboutelectronics.com/Articles/How-to-connect-a-LM741-op-amp-chip-to-a-circuit
 

Offline Mr D

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Re: Some noob questions
« Reply #128 on: August 19, 2018, 09:15:02 am »
OK!

I'm not trying to understand the circuit, i'm trying to understand and make sense of the elements and general structure!

So is the circuit only missing the power supply to make it a complete, viable circuit?
 

Offline rstofer

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Re: Some noob questions
« Reply #129 on: August 19, 2018, 09:25:49 am »
It looks complete.  The tricky part is that incandescent lamp in the upper left corner.  That device is used as a non-linear resistor.  When cold, it has very low resistance which means the op amp has very high closed loop gain.  When the circuit is oscillating normally, the resistance will increase (because the current increased) until everything balances out with a value of Rf/2 so the op amp closed loop gain should be around 2.
 

Offline Mr. Scram

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Re: Some noob questions
« Reply #130 on: August 19, 2018, 09:29:24 am »
OK!

I'm not trying to understand the circuit, i'm trying to understand and make sense of the elements and general structure!

So is the circuit only missing the power supply to make it a complete, viable circuit?
Power supplies, plural. Most op-amps require a postive and negative supply. Have you checked out Dave's op-amp video? You might not understand all or even most, but it helped me when I was completely unfamiliar with what they are and how they work. Knowing the common configurations helps a lot when looking at schematics.

 

Offline Mr D

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Re: Some noob questions
« Reply #131 on: August 19, 2018, 09:40:56 am »
Haha, you're getting way ahead of me there! ;) (Mr Rstofer)

I'll check of Dave's vid.

OK, i need some more fundamentals filled out:

I understand how a simple circuit with a bulb works. It's a closed loop with a resistor (the bulb) converting energy into light, so the circuit is useful.

But if i imagine an oscillator, how does the oscillating voltage get on to the next part of the circuit? How does it move on to an amplifier, while still keeping the closed loop back to the negative battery terminal?

Or is induction used to have several closed loops of circuit communicate information (voltage difference) between each other while being part of a bigger schematic consisting of several such closed loops?
 

Offline rstofer

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Re: Some noob questions
« Reply #132 on: August 19, 2018, 09:54:59 am »
Vout is shown on your schematic and while I don't want to even think about the filters (the Z equations), at the end of the day, Vout is produced from the op amp  power supplies through a bunch of transistors inside the op amp.

So, Vout and Gnd (part of power supplies not shown) goes on to whatever comes next.

When I said +15 and -15V, those voltages are relative to something.  That something is the ground created where the two supplies get connected together to essentially create a 30V supply with a center tap to ground.

Vout is relative to ground (not shown).   But the op amp itself doesn't care about ground, doesn't need to know about ground and isn't connected to ground.  That's why op amp videos spend so much time on this fact.

 

Offline Mr D

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Re: Some noob questions
« Reply #133 on: August 19, 2018, 09:58:58 am »
Ok, so I guess a circuit is always a single interconnected network, with multiple ground points? So the goal is to start with a power source with a steady voltage difference and eventually output either light through a bulb, or audio via an oscillating voltage difference to a speaker, or information through a 8 seg LED, or whatever. But always there's the current running to ground after the bulb or speaker or 8seg display?

But then i don't see how such a complicated scheme could work. You change one resistor value somewhere and won't that have a knock-on effect thoughout the rest of the circuit?? How can this be manageable in a complicated circuit? Or are there ways to sandbox parts of the circuit against this sort of chaos?
« Last Edit: August 19, 2018, 10:02:59 am by Mr D »
 

Offline Brumby

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Re: Some noob questions
« Reply #134 on: August 19, 2018, 10:41:21 am »
OK!

I'm not trying to understand the circuit, i'm trying to understand and make sense of the elements and general structure!

So is the circuit only missing the power supply to make it a complete, viable circuit?

To answer this question directly: Yes.

The actual power situation is also a bit obscure - with this describing what's around, but not shown:
So, Vout and Gnd (part of power supplies not shown) goes on to whatever comes next.

When I said +15 and -15V, those voltages are relative to something.  That something is the ground created where the two supplies get connected together to essentially create a 30V supply with a center tap to ground.

Vout is relative to ground (not shown).

The only hint is the symbol at bottom left which indicates a connection to the "ground" point - which is the 0V point of the +15V / 0V / -15V power supply.  The +15V and -15V power connections to the op-amp are implied - because it is known that an op-amp will need some sort of power supply.  Also, the output (voltage) will be referenced to something - because it has to be referenced to something - and this is commonly the 0V point.


These connections are omitted in many cases to simplify the schematic, so that the important aspects of it - the signal path, biasing and so on - aren't cluttered by connections we know must exist.

Sometimes you will see the full set of connections for everything; sometimes you will see nothing of the power connections ... and sometimes you might have short lines going nowhere with labels, such as this:



While not explicitly shown those +12V points are connected together and go to the +12V point of the power supply.  Same for the -12V points.

Don't worry about the VSRC "battery".  That is just a way of showing an input voltage, often used in teaching situations.
 

Offline rstofer

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Re: Some noob questions
« Reply #135 on: August 19, 2018, 10:49:27 am »
Yes, making a component change will change how the circuit operates, to some extent.  The thing is, real circuits are made up of independent modules interconnected in some way

Take LEDs connected to a uC.  The dropping (ballast) resistor is calculated at some point in the process and it is based on two factoids:  The Vf of the diode at the desired If (essentially intensity).  Then we calculate the series resistor to drop Vcc the proper amount (Vcc - Vf) when If is flowing.  If we want to be pedantic, there is a bit of voltage drop inside the uC to account for.

Having done all that, nothing changes even if I change clock frequency or the input voltage to the analog to digital converter (ADC).  They are in different modules.  Now, if I multiplex the displays (common with 7 segment displays), the calculation is a lot more involved because each segment is only illuminated for some percentage of time.  So the current needs to be higher and the resistor needs to be lower.  And so it goes.  Ohm's Law at work.

We don't tend to make things interdependent by having series resistors if we can avoid it.  Sometimes they are necessary and we know when they are.  Watch w2aew's transistor videos.  Pay attention to how he biases the base in the Common Emitter Amplifier.  And how gain is controlled by the ratio of the collector and emitter resistor.  Of course, the collector resistor also has to represent the load being driven - a subject for later on.
 

Offline Brumby

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Re: Some noob questions
« Reply #136 on: August 19, 2018, 12:13:01 pm »
But always there's the current running to ground after the bulb or speaker or 8seg display?

I'm not sure what you mean by "after".  Electricity (except for static discharge) flows through a circuit.  Any output to a bulb, speaker or display will put a current through that circuit element (typically) to ground.  It's not really "after" - it's an essential part of the single circuit function of making the speaker cone move or the bulb to light up.


Quote
But then i don't see how such a complicated scheme could work. You change one resistor value somewhere and won't that have a knock-on effect thoughout the rest of the circuit?? How can this be manageable in a complicated circuit? Or are there ways to sandbox parts of the circuit against this sort of chaos?

This is a good question - and leads us into the topic of impedance.  Actually working through this in a "complicated circuit" can become a bit of a rabbit hole - so be careful about jumping in too deep!  However, you don't need a bagful of maths to understand the principles.

Interaction between stages (where a "stage" is a group of components that do a specific function within a larger circuit) is, indeed, a consideration in circuit design and the primary factor involved is impedance.

Let's look at a circuit that processes a signal.  It will have an input and it will have an output.  That output will, in turn, be used as an input to a following circuit.  Now the question is - how much will the following circuit affect the circuit we are looking at when that input is connected to our output?

The answer is impedance.

If the output of our circuit has a *"high" impedance, then the signal it produces will look like it is coming through a *high value resistor and it will not be capable of providing much current, before the voltage drops down to an *"unusable" level.  If it has a *"low" impedance, then the signal will look like it is coming through a *low value resistor and it will be able to provide a *fair bit of current with the voltage dropping slightly.

If the input of the following circuit has a *"low" impedance then it is going to look like a *low value resistor that is going to want to pull a *significant amount of current, but if it has a *high impedance, then it will not need much current for it to be able to process the signal.


The simplest way to look at this is with the following diagram.  It is very simplistic and will not reflect what you will actually see in schematics - but it demonstrates the principle.



Here, Z1 will be the output impedance of the previous circuit and Z2 the input impedance of the circuit that follows.  How these impedances actually come about are the subject of some maths - maths that can be bewildering - but, for the sake of simplicity, just think of it like this and you will get the "feel".


Plugging in High/Low Input/Output combinations, you should notice the preferred approach is for the output impedance of one circuit be low compared to the input impedance of the following circuit as this will minimise the effect of connecting the following circuit to the one we are looking at.  The greater this ratio, the less the interaction will be.

This principle applies not only to electronics, but to an extremely wide range of physics as well.  However, sticking to electronics, this matter of impedance is pervasive.  It is everywhere.  If we take a simple example of a PA system, we have to consider the output impedance of the microphone and the input impedance of the pre-amp; the output impedance of the pre-amp and the input impedance of the power amp ... and then the output impedance of the power amp and the speaker impedance.

But this issue of impedances also exists within any circuit.  Each section of circuit that performs a particular job will have an input impedance and an output impedance.  For example, inside the pre-amplifier, you might have a transistor or two to boost the signal from the microphone to a level that makes it easier to work with.  This signal might then go through some tone control circuitry.  Then it might need a circuit that looks like an amplifier, but doesn't increase the signal level.  Such circuits are often used as a "buffer" where its input impedance is such that it does not affect the previous circuit (which may have a *high impedance) but provides a *low impedance output for whatever is subsequently connected.  (It is not uncommon to see unity gain buffer amplifiers - and that is their specific purpose.)

As these chain along the circuit, these impedances must be taken into consideration to ensure the signal is passed along in a useful manner.

* These terms are relative.  (This is important!)  What's painfully high in one situation might be insanely low in another.


This issue of impedances also extends into digital circuits and is the effective issue behind them as well.  We see them expressed more as the current capability of an output or the specification for an input which makes life easier as it hides away the internals (unless you really want to go deep) and just provides the numbers you need to know for a practical circuit.
« Last Edit: August 19, 2018, 12:36:43 pm by Brumby »
 


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