Help me understand...

[adamct]

Let me start out by admitting that I am asking a dumb question. It is what it is. I'm trying to learn and thought I was making OK progress, yet I find myself confused by the simplest issue here.

I want to build a headphone dummy load, as described here: http://tangentsoft.net/audio/hp-dummy.html

But as I was considering the circuit, I found myself confused (this is the part where everyone probably starts either (a) throwing up or (b) laughing hysterically at my incompetence).

There is current coming in through the signal wires, goes through the switch, to the resistors and then to ground (the case), then back through the ground shield on the headphone cable (I know it isn't a DC signal, this is just how I think about the electrical path). What I can't figure out is: since the case is metal and the outside doesn't appear to be insulated, wouldn't you get some kind of shock if you touched it? Is the current/voltage too low to be noticeable? Or is it that skin is a worse conductor than metal, so the signal just goes through the case, even if you touch it? Or is there some reason why there actually isn't any electricity flowing through the case?

Once you've stopped laughing and/or feeling queasy, please help me out here...

Thanks!
Adam

 

[Doc B]

Yes, you got it. The potential, i.e. voltage of the ground and anything connected to it is 0 volts. This is an interesting subject. These days everything pretty much has a grounded chassis. But I just completed the restoration of a 1953 vintage Zenith clock radio in which the circuit is what is known as an ACDC radio. There is no power transformer, the tube heaters are wired in series and a rectifier connects right to the AC mains for B+. The scary part is that to keep the noise down the power switch is in the neutral leg of the AC mains rather than the hot leg. So when you switch the radio OFF, the chassis is live with 120VAC! Back in those days they relied upon the plastic case and knobs to keep you from getting zapped.

I was asked to wire an iPod input jack to the radio, but I won't do it. I bought an iTrip instead and will instruct the user how to use it to play the iPod thru the FM section of the radio.

 

[adamct]

Yes, you got it. The potential, i.e. voltage of the ground and anything connected to it is 0 volts.

Unfortunately, I didn't get it. 

What does that mean? Why is the voltage of anything connected to ground zero? And at what point does it become zero? Presumably there is some voltage on the incoming signal wire. Do you have voltage on the resistor leg that is connected to the switch, but not on the leg that is connected to the case?

Also, when thinking about what resistors to use, it makes sense to me to use 32 Ohms and 300 Ohms as the two most common headphone impedances (for my purposes, at least). But which would be more useful for the 3rd? 16 Ohms, 120 Ohms or 600 Ohms? 120 Ohms actually strikes me as a pretty unusual impedance, and I would have thought that it is more useful to test amps at extremes, so 16 or 600 would be more useful, but I don't know which (I would use both, except I can't seem to find a DP4T toggle switch and mouting a slide switch would be more effort than I care to put into this project).

As for the clock radio...yowza! 

Best regards,
Adam

 

[Doc B]

I am going to suggest looking up some sites about basic electronics. Voltage and current are not the same. Voltage is the measurement of potential energy, that is how much energy is available to be released, maybe into a tube plate, maybe into your hand...

Current is how many electrons flow when that potential energy is released.
The analogy usually given is to imagine a piece of wire as a stream. The height of the stream determines the potential energy, or, if you will, it's "charge" (i.e. a battery connected to the wire would be equivalent to raising the stream to a higher elevation). How much water moves past a certain point as the water moves downhill is the current ( in the wire it's electrons flowing rather than water). How fast the water goes is determined by the resistance of the rocks and stuff in the stream bed. Resistors in a circuit serve the same purpose

Your chassis is a pool that the stream empties into, the point of lowest potential. The still water there can't knock you off your feet but the water in the rapids, where the current created by the released potential energy is high, can.

AC is a bit different in that it alternates its flow direction. But the basic concept that there is a point of lowest potential still holds.

 

[adamct]

I understand voltage and current (although looking back, I should have worded my original post better). But I still don't really understand how to think about ground.

To continue with the common water analogy (assuming DC), I get that the water has high voltage at the top of the waterfall. And I get the resistance/rocks. I also understand the current/volume of water flowing down the waterfall. But once the water empties into the pool, it can't just pile up there forever, right? It eventually has to go somewhere or the water level will rise, and it can't rise indefinitely, can it?

Where does the water go? And when it goes, there will be current, right? And you can't have current with zero voltage, can you? I mean, if the water is flowing, then it has to have some energy, even if it is moving very, very, very slowly, doesn't it? Is the point just that the voltage is so minimal it is irrelevant, or is it actually zero?

If we switch over to AC and use the pipe analogy, instead of a waterfall, I think of it as a large pool of water (ground)connected by a pipe (hot/signal wire) to a pump (amp). The pump alternately pumps water through the pipe and into the pool and sucks it out. Over time, the net effect on the volume of water in the pool is zero. Within the pipe, when the pump is actually pushing or sucking (as opposed to the moment when it is switching over from one to the other), the pressure (voltage) is high and there is current.

So far (I think) so good. But I'm having trouble with the idea that there is no effect on the large pool of water (ground) at any given instant (as opposed to over time) when the pump is actively pushing or sucking. Even if the pump and the pipe are moving small amounts of water relative to the pool, there will be some effect on the pool, right? When the pump is sucking, for example, it has to pull some water from the pool into the tube, doesn't it? Won't that create some minimal amount of current and voltage in the pool?

Or do I just need to give up on this water analogy?

 

[Paul Birkeland]

Instead of thinking about water emptying into a pool, think about water coming out of your hose and emptying into the ocean.  Essentially, the water level can rise, but the new level to which it has risen is still considered "ground".

The movement at 0V/ground is lateral movement, not vertical.  In our model, the lateral movement does not change the potential, and has zero resistance (water moving in water).  Even though this is irritating to a physicist, it helps conceptually. 

The AC analogy is tougher, since the voltage pulses and current pulses don't quite happen at the same time and all of this can happen without a ground reference.

As far as the impact on the pool, the pipe would leave the pump, curve around for a bit, then empty back into the pool (think circuit=loop).  At the instant that the pump is sucking water from the pool, it is pushing the same amount of water through the pipe and back into the pool.  Other than some ripple from the water moving around, the net level of the pool never changes. 

 

[adamct]

OK, that is helpful in terms of understanding the theory. Now to go back to the example of the headphone dummy load...if voltage is the difference in energy potential between two points, then am I correct about the following?

- The voltage measured between the signal cable and an any point in the circuit up to the resistor (e.g., the input to the switch, the output of the switch, or the wire from the switch to the resistor) will be zero (because there is no difference between any of those points).

- The voltage measured between any point before the resistor, and any point after the resistor (the out leg, the wire to the case, or any part of the case itself) will be high (or not, whatever, but the point is that it will not be zero).

What then happens if I pick up the case of the dummy load with it connected to a headphone amp that is powered on? If I measure the voltage between my hand and the case, I assume voltage is zero? Why is that? Why do I have the same energy potential as the case if I am not grounded?

Is the answer just that the resistance of my skin is high, and the flow of current between my hand and the case is low, so Ohm's law tells us voltage is therefore also low? That kind of makes sense, but it still feels odd that my hand has the same energy potential as the case...

Best,
Adam

 

[Paul Birkeland]

- The voltage measured between the signal cable and an any point in the circuit up to the resistor (e.g., the input to the switch, the output of the switch, or the wire from the switch to the resistor) will be zero (because there is no difference between any of those points).

This is a little on the vague side.  There will be 0V DC, because the amplifier has no DC at the output, and the load resistors are not a source of voltage.

If you have AC signal (like a 60Hz tone), then you will have some AC voltage across whichever resistor is switched in, and this voltage will vary with the position of the volume control.

On the signal cable itself, you have two hots and a ground, so a lot of different results can be obtained based on what you are actually measuring.

- The voltage measured between any point before the resistor, and any point after the resistor (the out leg, the wire to the case, or any part of the case itself) will be high (or not, whatever, but the point is that it will not be zero).

The resistors go from signal to ground, it's not really an in or an out type of thing.

What then happens if I pick up the case of the dummy load with it connected to a headphone amp that is powered on? If I measure the voltage between my hand and the case, I assume voltage is zero? Why is that? Why do I have the same energy potential as the case if I am not grounded?

It's not really whether or not you are grounded, it's that you're not a source of voltage.  Go rub your feet on some shag carpet with socks on, then you can make yourself a voltage source

On the other hand, if the box happened to be plastic, or it wasn't connected to the ground wire of the headphone cable, nothing would happen. 

-PB

 

[adamct]

How does ground work in the Quickie? Is the amp grounded through the interconnects? If I were running it using interconnects connected to, say, an ipod (don't worry, I have no intention of doing that), would it be ungrounded? I've been thinking of ground as needing to be connected to the ground from a wall outlet, but that obviously doesn't apply to the battery-powered Quickie. Or is there something in the circuit that acts as ground?

Best regards,
Adam

 

[Paul Birkeland]

How does ground work in the Quickie? Is the amp grounded through the interconnects?

Yes.  When we get a user who has ground loop hum with the Quickie, we advise to use a clip lead to establish a connection between the ground in the Quickie and the chassis of a nearby piece of gear.

-PB