Electronics > Beginners

Talk me through this simple circuit?

(1/11) > >>

Mr D:
Hi folks,

I'm really a beginner!

I'm trying to get a foothold, and i found this simple looking circuit, and i was wondering if anyone would like to talk me through it.

So let's start at the beginning with the power supply. The placement of the ground means that the voltage rails to the OSC section are +-4.5V.

But does this matter? What would have been wrong with 0-9v?

Could R2 & R4 have been replaced with one 20k resistor?

Also, the LED will get more or less bright depending on the load downstream? How does one think about this or calculate it?

Thanks in advance!



rstofer:

--- Quote from: Mr D on May 30, 2019, 08:24:54 pm ---Hi folks,

I'm really a beginner!

I'm trying to get a foothold, and i found this simple looking circuit, and i was wondering if anyone would like to talk me through it.

So let's start at the beginning with the power supply. The placement of the ground means that the voltage rails to the OSC section are +-4.5V.

But does this matter? What would have been wrong with 0-9v?


--- End quote ---

It matters if the intent is that some signals swing + and - around ground.  An audio signal, for example.  The signal could as well swing around 4.5V and use blocking capacitors to remove the DC offset.  Either way...

ETA:  I didn't see the Op Amp on the first go-around.   The op amp doesn't care whether you power it with +-4.5V or 0-9V but the datasheet for the 741 shows a minimum voltage of +-10V (20V total) on the power pins.  I'm wondering...


--- Quote ---
Also, the LED will get more or less bright depending on the load downstream? How does one think about this or calculate it?


--- End quote ---

Well, sure, the brightness can change depending on load.  The LED is across the battery and we assume, when first looking at the circuit, that it is a perfect voltage source.  It puts out 9V no matter the load.

But it isn't a perfect source, it has internal resistance and when the load increases, the internal resistance drop more voltage.  Also, the battery voltage falls as it is discharged.  The arithmetic for determining voltage versus load as impacted by source resistance (inside the battery) is a little more involved than just talking about it.


The entire purpose of R2 and R4 is to create an artificial ground so that signals revolve around 0V.

rstofer:
If you were going to go with 0-9V, you wouldn't use R2 and R4 at all.  You would just call the bottom rail ground and the top rail +9V.  That may or may not be helpful.

ETA:  You would need to think through what happens if you don't create the +-4.5V/Gnd reference.  I'm not convinced the circuit will work as drawn but I think the artificial ground is required.  This kind of thing (rail splitter) is done all the time with single supply op amps and that is exactly what is happening here.  The author is taking a bipolar op amp and running it on a single supply.

Mr D:
Many thanks, there's a lot for me to mull over there.

But first this concept of ground, I'm still really struggling with.

So from my other thread from a few days ago i think i understand what ground conceptually is: basically a single wire (or node), and wherever a ground is indicated on the circuit, there's a connection to this wire or node. (just to be sure, all grounds in all 3 pix are attached to the same node, right?)

But how does one decide: "ok, we need to connect to ground here, here, here and here"? And surely doesn't adding one more ground connection totally mess up the whole circuit? Because aren't you adding a new path down which the current can flow, having a knock on effect around your whole circuit?!

And to zoom out even more: is there some sort of technique one can use to design or analyze a circuit, working backwards?
I mean: you start with the end goal (in this case to output a square wave at variable audio rate frequency),  and then work back from there, designing (or analyzing) what components go where and why?

(i still want to come back to the specifics of your reply, but these are the things that are bugging me the most right now!)

EDIT: i've been playing with EveryCircuit for a few minutes, and i think something just clicked: you use all those grounds in your circuit to persuade current to flow where it otherwise wouldn't want to flow? And then you use resistors in the circuit to balance the whole circuit out, speeding or slowing this flow back to ground?


 

virtualparticles:
A ground is a common node in the circuit as you surmised. If your circuit is driven with a single +9V supply, the ground could attached to the negative terminal of the battery and we would call it 0V for reference purposes. We often then make the rash assumption that all connections to that ground and the implied connections between those grounds are all superconductors. In actual fact, there will always be some small voltages between any two ground nodes depending on the resistance and the amount of current flowing into each ground node. Frankly, it's just cleaner on a schematic to use a ground symbol rather than showing all the connections that go to a single wire which goes back to the negative side of the battery. But you can do that if you like and never use the ground symbol.

As a practical matter, it is common to build a circuit on a PC board that has all copper on one side. If you use that as your ground, you can attach things to it and "skywire" a circuit together.

You asked how to work backwards from a requirement to an implementation. This is what electronic design is all about and it takes a little experience and analyzing other people's circuits to learn the ropes. To generate a square wave, you would need a switch. That could be a biploar transistor or it could be a FET. Either one could do the job. I would suggest you load "LTSpice" from the Analog web site and experiment with transistor circuits to get the hang of it.

Navigation

[0] Message Index

[#] Next page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod