Gnd vs common vs +- voltage source ?

Thread Starter

Hextejas

Joined Sep 29, 2017
187
Darn spell checker got the title wrong. Should be GND, not And.

As you can see from the title, I am mightily confused about the - (minus) side of things.
Here is the circuit that I am trying to build. In the circuit there are 3 references to 《》, the arrowhead symbol., or gnd.
So, in my simple way of thinking, can I wrap all 3 of those together ? Somehow that don't seem right. The negative of a 17v DC supply, one side of my audio signal, AC, and the common of the circuit. If the audio signal were not there I would feel comfortable wiring the other 2 together, but this is a puzzle. This seems like a basic understanding topic which I clearly lack, so could someone point me to a clear discussion or explanation of it ?
My next project has both a -- and a + 15v DC requirement. I bought a bench power supply that delivers Both, but I just know that I will be back here with lots of questions. Somehow the -- and ++ issue seems related to this thread which is why I put it here. Please ignore it if you think it belongs in a separate thread.
Thanks.
Modified simple amp.png
 
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Tonyr1084

Joined Sep 24, 2015
7,853
The downward pointed triangles are ground points. Yes, they can be tied together. As for the AC component, yes, AC can go below ground. A 5 mV signal isn't going to hurt anything but the negative component will be cut off. So your amplifier will only amplify your signal when it is going positive. The capacitor (C1) will block the DC component from coming back into your AC source, so I don't see any problems with tying all the grounds together.

If you had a ground point (as you do) and you had a dual supply (as you don't) you would also be amplifying the negative side of the signal. Of course your amp isn't built for amplifying the negative component so there'd be more circuitry if you were to, but for the sake of understanding, keep in mind that it's possible to have voltages below zero. A bit hard to understand at first, but if you look at an AC sine wave you see it has both a positive and a negative component to it. The zero reference is smack dab in the middle of the two wave peaks. When the signal is below the zero reference it's a negative component. Since ambient audio sounds consist of vibrations in the air, there's a positive pressure wave followed by a negative pressure wave. Just as the AC sine wave predicts. A good amplifier magnifies both sides of the pressure wave and your audio speakers reproduce that pressure wave.
 

ElectricSpidey

Joined Dec 2, 2017
2,758
Tying all grounds together is not always straight forward.

For example when using a LM317 in voltage mode the sense divider ground should be located as close to the load as possible and not the source ground point.

So there can be some considerations.
 

Thread Starter

Hextejas

Joined Sep 29, 2017
187
Thank you Tony, that makes it clearer. I think.
Re the negative portion of the sound wave, and I have a hard time visualizing it. For my project, I have a 1k Hz tone as a test.
I input it on my little scope just for the heck of it can see the sine wave, plus I can hear it. And it has a negative component. How do I wrap all these observations together. A negative component, does the amp filter it out, should it, can I hear it, am I supposed to be able to ?
No wonder I never liked trumpets.
 

Tonyr1084

Joined Sep 24, 2015
7,853
In the case of your amp, yes, the negative component is muted. If you wanted to add that part in to recreate the full wave you'll need a dual supply. It's not being filtered out it's being clipped. Which means you're only hearing half the sound wave.

Trumpets? Well, I guess I've never liked the Harpsichord myself.
 

Thread Starter

Hextejas

Joined Sep 29, 2017
187
Very interesting, so I am imagining some circuit at the beginning that detects the negative somehow and makes it look acceptable to the rest of the circuit.
I wonder if that would make an appreciable difference to the listening experience.
Curious.
 

Thread Starter

Hextejas

Joined Sep 29, 2017
187
Tying all grounds together is not always straight forward.

For example when using a LM317 in voltage mode the sense divider ground should be located as close to the load as possible and not the source ground point.

So there can be some considerations.
Interesting Spidey and I will try and read more on it. Though I have no clue what you mean by using a LM317 in "voltage" mode. I thought that they were used as a voltage regulator and I don't know why the proximity matters.
 

Tonyr1084

Joined Sep 24, 2015
7,853
I wonder if that would make an appreciable difference to the listening experience.
Curious.
Yes, it would.

Have you ever heard SSB (Single Side Band) CB radio? It sounds like Star Wars where Red Leader is talking to Blue Leader. Or just try putting a diode in line on one of the speakers of your stereo. You'll notice the difference right away.
 

ElectricSpidey

Joined Dec 2, 2017
2,758
Interesting Spidey and I will try and read more on it. Though I have no clue what you mean by using a LM317 in "voltage" mode. I thought that they were used as a voltage regulator and I don't know why the proximity matters.
The typical linear regulator can be used in either voltage or current regulation mode.

The reason the ground leg of the sense divider is placed close to the load is to provide better regulation than placing it at say...the main filter cap, so the two places are technically "ground" but they are at a slightly different potentials.
 

WBahn

Joined Mar 31, 2012
29,978
The downward pointed triangles are ground points. Yes, they can be tied together. As for the AC component, yes, AC can go below ground. A 5 mV signal isn't going to hurt anything but the negative component will be cut off. So your amplifier will only amplify your signal when it is going positive. The capacitor (C1) will block the DC component from coming back into your AC source, so I don't see any problems with tying all the grounds together.
Why is the negative component going to be cutoff? It is AC coupled via C1. So the AC signal that appears on the left side of C1 will be replicated, to a reasonable approximation, on the right side of C1 just shifted upward by the DC voltage on C1, which will be around 1.4 V in this case.
 

WBahn

Joined Mar 31, 2012
29,978
As you can see from the title, I am mightily confused about the - (minus) side of things.
Let's look at this issue in general.

When you have a voltage source, such as a battery, ALL the negative terminal means is that the positive terminal of THAT source has a positive voltage difference relative to the negative terminal of THAT source. So the positive terminal of a 12 V battery is 12 V higher (more positive) than the negative terminal of that battery. There is NO intrinsic relation between the negative terminal of one source and the negative terminal of another source or between it and any other point in the schematic. Any relation between them is imposed by the schematic (i.e., how things are wired together).

A "ground" (usually better referred to as a "common") is simply a point in the circuit that we choose to refer the voltages at other points in the circuit to. We declare that our common has a voltage of 0 V. Since ALL voltages are potential differences, they always refer to the voltage at one point relative to the voltage at another. But having to always indicate which two points we are using is tedious and hampers communication. So we define a common reference point and then any time we talk about the voltage at Node A, it is understood that we are talking about the difference between the voltage at Node A and our common reference.

All common nodes in a schematic can be tied together. In the physical circuit, how this is done can have an impact on the circuit, but this is seldom modeled on simple schematics.

Tying points in the schematic together this way is known as "connection by name". Every node that has the same name given to it is electrically tied together as far as the simulation and PCB tools are concerned. It is very common to use connection by name for the common reference and the power supplies because this considerably declutters the schematic and it is very clear to the reader what net the connections are made to. Some people use connection by name for other signals in order to avoid routing wires. This practice is generally frowned upon because it can make reading the schematic, particularly debugging it, extremely difficult.
 

Tonyr1084

Joined Sep 24, 2015
7,853
the AC signal that appears on the left side of C1 will be replicated, to a reasonable approximation, on the right side of C1 just shifted upward by the DC voltage on C1, which will be around 1.4 V in this case.
I don't see any DC offset. Am I missing something?

Way I see it - when the signal swings negative then Q1 is shut down. No ? ? ? Teach me if I'm wrong. I'm open to admitting when I screw things up. I always thought you'd need a push - pull amplifier. I know you Can use a single sided supply, but don't you have to offset the input? Bias the transistor?
 
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WBahn

Joined Mar 31, 2012
29,978
I don't see any DC offset. Am I missing something?
Ground the AC source (set it to zero) and do a DC analysis.

Because of C1, no DC current can flow in R9 or R2, so they have no voltage across them.

What is the voltage on the left side of the capacitor? 0 V because the AC source is set to 0 V and there is no DC voltage across R9.

What is the voltage on the right side of the capacitor? 1.4 V because the current source biases them on (and I'm assuming they are not in saturation) so the base of Q2 is at about 1.4 V and there is no DC voltage across R2.

So the DC voltage across C1 is about 1.4 V (right side relative to the left side).
 

WBahn

Joined Mar 31, 2012
29,978
Well, maybe this is a thread I should do more reading with my mouth closed. Sorry if I misled anyone.
The discussion needed to help you will probably help the TS, too.

Way I see it - when the signal swings negative then Q1 is shut down. No ? ? ? Teach me if I'm wrong. I'm open to admitting when I screw things up. I always thought you'd need a push - pull amplifier. I know you Can use a single sided supply, but don't you have to offset the input? Bias the transistor?
Since Q1 has to have 1.4 V at its base in order to be "turned on", then a 50 mV positive signal is hardly going to accomplish that. So there MUST be some bias in order to put the transistor in a state such that a 50 mV positive signal can affect it.

In essence, what the input is doing is injecting or extracting current via C1. When the signal goes positive, it injects a little current into the node that has the base of Q1 and that gets added to the current coming down through R4 and into the base of Q1. When the input goes negative, it extracts a bit of current from that same node which robs Q1 of some of its base current.
 

Tonyr1084

Joined Sep 24, 2015
7,853
In essence, what the input is doing is injecting or extracting current via C1. When the signal goes positive, it injects a little current into the node that has the base of Q1 and that gets added to the current coming down through R4 and into the base of Q1. When the input goes negative, it extracts a bit of current from that same node which robs Q1 of some of its base current.
Sincerely, thank you.
 
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