Time for a display of the vastness of my ignorance

[skyboltone]

What's the difference between the two inputs on many amps? One will have say 68K to the grid of V1A and the other will have that and 1 meg to ground too.

I'm fixin to have just the one input.

Thanks in advance
Dan

[bnwitt]

Here is an explanation from Dan Torres:

Two input jacks – found on Fenders and Marshalls (the Marshalls in question have four jacks.)

Why is one high gain and the other low? And how does the whole thing work, what are these resistors doing?

Overall the two high and low gain jacks are a very good trick using switching jacks and as few parts as possible.

Jack #1, the high gain jack has a 1 meg (1,000,000 ohms) resistor wired across the jack from the hot tab to the ground tab, and the grounded lead of the resistors is bent back and soldered to the middle tab – the “switch.” The switch is grounded on jack #1.

These switching jacks have an extra “leaf” that is touching the hot, or “tip” connection of the jack – but only when nothing is plugged in – called a “normally closed” jack. This can be made to do all kinds of things.

Trick #1. In the case of the #1 jack, when you do not have your guitar plugged into the amp (either jack) the “switch” is closed – since it is grounded, the amp is turned off for low noise, and no chance of runaway.

The 1 meg resistor is the “grid load” resistor for the first tube stage of the amplifier. All tubes have to have a grid load resistance or they will not operate.

Instantly many of you are thinking, “what about the guitar pot, or the pickups, aren’t they a grid load?” Yes, they would be – but – if the guitar had active electronics, or had a capacitor in series with the output signal from the pickups, there may be no direct resistance from the guitar. It is better to be safe and have the resistor there.

Ok, more on the input jacks. The second jack has it’s switch soldered to the first jacks hot lead. Each jack has a 68k (usually) resistor from its hot lead, and they join together to make one lead to the tube stage.

These “series resistors” are sort of input buffers – to keep the amp from getting high level spikes from the instruments. Not totally necessary for the vintage guitars these amps were made for, but a real good idea for modern guitars with high output pickups, and in the old days, accordions!

Here is trick #2. Jack #2 has it’s switch soldered to Jack #1’s hot connection, and a 68k resistor on it’s hot lead – when the switch is closed – nothing plugged into jack #2, it’s 68k resistor is placed directly in parallel with Jack #1’s 68k resistor, for half the value, 34k. Less series resistance.

Trick #3. If you plug into jack #2 only, check out what happens. We already learned that if nothing is plugged into jack #1 the hot is shorted to ground via the switch – so now Jack #1 represents a direct ground connection.

The signal from jack #2 would pass through the 68k resistor, and at that point see the 68k resistor from Jack #1 with its other end grounded!! So the signal has 68k in series (more resistance, a lower signal level gets to the amp) and a 68k grid load resistor – much lower resistance than the 1 meg. A lower value of resistance to ground reduces the signal to the tube (again.) Check out the schematics to see what is going on.

[skyboltone]

Thanks Ben:
I had all the switching business figured out. What I didn't know is that the higher the grid load the higher the gain. It makes sense when I think about it, because with a lower load resistor, more input voltage gets shunted off and doesn't go to the tube. I'll be putting in the 1M load and 68K series to start. The ring and the switch leaf will be grounded.
thanks

[PRR]

> the vastness of my ignorance

I think 99% of players also miss this point.

Hard to see the way Leo jammed it together.

Ignore the 1Meg. You hardly ever need it. But when you do need it, you DO. So put it in and forget about it.

See picture.

Plug in input "1", it pretty much just goes to the tube. The 34K has no effect on guitar, it dents any radio signals that a long cord may catch. You don't even need the 34K, until the day you DO: get a gig with a parking lot of CB radios, or under a big AM transmitter.

Plug in input "2". This is the clever bit. The signal is cut in half, with no user action except picking a jack. This appeared in the late 1950s. 1930-1950 it was all about getting enough signal to overwhelm tube hiss. But then came Alnico and then double-wound HumBuckers. Signal levels got hot enough to distort the first grid, before any volume control. (Not all guitars had volume controls yet.) So if you are breaking up despite what you set knobs at, try input "2".

The third condition is "1+2". This will mix two signals equally. Used to be we had more players than amps. Doubling-up was common. Now we have many more amps than players, so the original function is pointless; there's other things people do with the second jack.

In input "1" the load on the guitar is 1Meg (not shown).

In input "2" the guitar is loaded with 136K (and 1Meg). The guitar pickup is 5K at low frequencies but rings-up above 100K at the top of the guitar's strong overtones. So 136K isn't a heavy load, but not negligible, but you only go to "2" when your axe is distorting in input "1", so a bit of loading is OK.

With both inputs in use, things is strange. Each pickup feels the other pickup though 136K (and 1Meg hanging on the side). Maybe not ideal, but then two instruments on one amp is not ideal. If it gets you through a bad night, WTH.

> I'm fixin to have just the one input.

Your axe has a volume control? You don't do booster pedals (or accept/enjoy the overload they may cause)? Then fine. The two inputs is a hack for too-hot uncontrolled pickups, and a fallback when two players must share an amp.

[benoit]

>
In input "2" the guitar is loaded with 136K (and 1Meg). The guitar pickup is 5K at low frequencies but rings-up above 100K at the top of the guitar's strong overtones. So 136K isn't a heavy load, but not negligible, but you only go to "2" when your axe is distorting in input "1", so a bit of loading is OK.

2 things. 1, can you clarify the above a little? THe guitar pickup is 5k at low freqs means has a resistance of that to those frequenceies? And what do you mean by "rings up above 100k?" Sounds like frequency but doesn't really make sense to me. SOrry if my questions don't make sense, I'm just looking for a little elaboration on what I quoted.

Ben

[Darkbluemurder]

Thanks Ben:
I had all the switching business figured out. What I didn't know is that the higher the grid load the higher the gain. It makes sense when I think about it, because with a lower load resistor, more input voltage gets shunted off and doesn't go to the tube. I'll be putting in the 1M load and 68K series to start. The ring and the switch leaf will be grounded.
thanks

If you also play an electric acoustic with a non-preamped piezo through the amp you may want to choose a 3M or 5M resistor instead of the 1M. If not the 1M is good.

FWIW I never used the #2 input on any amp, and I get all my overdriven sounds from pedals. But I do not use them to boost and push the amp into overdrive.

[PRR]

> what do you mean by "rings up above 100k?"

Above 100K ohms, usually around 3KHz.

Why do you care? Do you know what to do with an impedance? Many folks, even trained engineers who should know better, don't. "Impedance Matching" has done more harm than good.

> The guitar pickup is 5k at low freqs means has a resistance of that to those frequencies?

Right. Use your DMM (DC) Ohms function. It forces a fixed current through the resistance, measures the voltage that happens, displays voltage divided by current, which is Ohms.

You can do the same with AC/Audio test signal. And at various frequencies. Though good impedance meters are rare. Partly because most technicians don't have much real use for impedance.

As you come up from sub-1Hz, you mostly "see" the copper resistance of a long piece of thin wire. About 5K, though a lot of variation depending on pickup purpose.

Around 100Hz the impedance starts to rise. Makes sense. The winding is inductive. Inductive impedance rises with frequency. We know from other techniques that a guitar pickup's inductance is around 5H (again, wide variation depending on purpose). 5H at 10Hz is 300 ohms, hardly noticed in series with the 5K resistance. 5H at 100Hz is 3K ohms, easily noticed in series with the 5K resistance. 5H at 1KHz is 30K ohms, now the 5K resistance hardly matters.

If that's all there were, the impedance would rise to 300K at 10KHz, 3Meg at 100KHz, etc.

But there is capacitance everywhere. Especially many turns on top of each other, or long cables. Say 500pFd. Capacitive impedance falls with frequency. 500pFd at 10Hz is 34Meg Ohms, at 1KHz is 340K Ohms, at 10KHz is 34K Ohms.

So impedance will rise from 5K below 100Hz, to 30K at 1KHz, but by 10KHz it will be 34KHz and at 100KHz down to 3.4K ohms. So you can plot a wide range of frequency versus impedance with simple thinking.

At about 3KHz, both the inductance and the capacitance have 100K impedance individually. Something interesting happens. If there is no resistance anywhere, the impedance goes to infinity. This is how you tune one radio station out of the hundreds of signals on the air.

But we always have resistance. The 5K copper loss limits the ideal impedance to about 2Meg. Real pickups show much lower impedances at resonance. The coil core is very lossy. A good model of core-loss is complicated and messy. I threw a 500K resistor across the winding to show the idea, but that's oversimplified. In general the resonant peak will be above 50K ohms but never as high as 2Meg, and will fall somewhere between 1KHz and 10KHz, often 2KHz-6KHz.

The schematic shows the idealized simplified pickup equivalent. "V1" is the magnetic wiggle at the pole, the "input". For impedance analysis, this is irrelevant.

To get impedance, I injected a 1A AC test-current, swept from 80Hz to 10KHz, and read the voltage developed across the pickup terminals. Yes, 1A is unrealistic... but it gives an easy answer (1V = 1ohm), and SPICE components are too stupid to melt when you have 100,000 Watts in a small coil. In real life, we might inject 1/1,000,000A (1uA) so the pickup don't melt, then multiply Volts by 1,000,000 to get to Ohms. The impedance comes out the same.

[KT66]

Skybolt my ingorance is much more vast than yours. I found this thread interesting and I have a question that is sort of related. I built a plexi and I wanted to internally jumper the channels. I looked at the signal path of using a patch cord on four inputs and it seemed to me that I could do it with less resistance in the signal path. So I diconnected 3 of the jacks, ran a piece of shielded cable over to V1 from the remaining jack (making sure it had the 1 meg grid load) and connected it to 2 68K resistors and connected each of those to the grids on V1. It works fine and I can blend the channels but I am curious as to how much this sounds different from the patch cord method. Incidentally, I made an amp once and forgot to put the 1 meg grid load, and it did work, allthough it was noisy. I assumed that the resistance in the guitar cord/ guitar wiring acted as the grid load. Sorry for being so long winded.