hi and low inputs. what for?

[allsavy]

I'm building an 18 watt TBM clone (Ritchie of 18 watt.com) and although that particular design does not have high and low inputs on the clean channel I might consider installing them if at all useful. What exactly are they meant for? High and low impedance input? Or high and low signal level? Would I plug a lespaul into high and a strat into low? Active into high and passive into low? Eletric Guitar into high and microphone into low? It is the resistor across the input that defines the input impedance (usually 1meg) right?
what would one have to do to have the ability to use one of the inputs as a mic in?

[tubeswell]

It just gives you a different impedance option. In Fender circuits the Lo gain one cuts a bit of treble and provides a softer tonal option. If your pedals are giving you grief it can come in handy

[tbryanh]

A high impedance single-ende mic can probably be connected directly to the amp.

For a balanced, low impedance mic, a transformer between the mic and amp that converts the signal from a balanced low impedance to a single-ended high impedance would probably be best.

[txstrat]

In most Fender circuits: On the low input the guitars signal is fed to one 68k resistor with a 1 meg ground/grid resistor. On the high input a second 68k resistor is added in parallel to the first. That halves the whole resistance to 34k and therefore the signal going to the grid is stronger and drives the first stage a little more.

[Bruce / Mission Amps]

I think it's so two electric instruments or a HI-Z mic and one electric instrument can be plugged in at the same time.
However, as TXSTRAT mentions, that is the effect of it all if not used that way.
I doubt very seriously the jack combination was ever intended to be used for different impedance instruments, to lose a little input drive signal, tone anything down or roll off any highs.

[cminor9]

On the high input a second 68k resistor is added in parallel to the first. That halves the whole resistance to 34k

Wait, wouldn't running two 68k ohm resistors in parallel give the average resistance across the two, 68k?

[tubeswell]

Well I thought the two different channels on the deluxe etc was so you could have two different instruments of instrument mic at the same time. But I always thought the hi-lo inputs altered the input impedance, so you could have more choice over gain. (But I've been wrong before - so bite me.)

[chipaudette]

In my opinion, on a Fender, the high and low inputs are never to be used at the same time...the coupling between the two inputs is just two high. For instance, you change the volume on one guitar and it can affect the volume on the other. Bad.

Now, many old fenders have two SETS of high and low inputs. This is the so-called "Normal" channel and "Vibrato" channel (Deluxe Reverb, Twin, Vibrolux, Bandmaster, etc). These amps were definitely made to have two instruments plugged in...one plugged into the "normal" and the other plugged into the "vibrato". Because each channel had its own pre-amp complete with volume control and tone stack, there was no coupling between the two sets of inputs. Nice. The old 4-jack Marshalls had two pre-amps as well. You could plug two instruments into those if you really wanted to.

So, we get back to "high" vs "low" inputs. It's my opinion that Leo Fender...especially in the blackface years...was trying to give as clean a sound as possible. He was careful with his signal levels throughout the amp so that he could avoid distortion as much as possible. Unfortunately for him, there were a range of instruments that one might plug into a Fender amp. These instruments all had different signal levels. A single coil guitar, for example, puts out a lot lower signal level than a humbucker. Therefore, he included a "high" and a "low" input with different effective sensitivities.

The "high" input has a higher input impedance, which keeps the signals from a high impedance guitar pickup from being loaded down...ie, the signal from the pickup stays at full strength. It was basically made for Fender single coils. Humbuckers, however, put out a lot more voltage. It can cause distortion in the preamp before you distort the power amp. Leo didn't want this kind of thing to happen. So, he included the "low" inputs. They have much lower input impedance, which drags down the signal level put out by high impedance sources like guitar pickups. Since the signal level is now lower, it reduces the chance of your humbuckers distorting the amp too early.

So, while it is not quite correct to say it this way...the "high" input feels like it has higher gain while the "low" input feels like it has lower gain. I believe that Leo thought that you should plug instruments with low signal levels (single coils) into the "high" input while you should plug instruments with high signal levels (humbuckers) into the "low" input. This would keep the signal levels in the amp consistent and would minimize the chance of distortion.

Of course, most people like some amp distortion...especially players of humbuckers. Therefore, most people just plug into the "high" input and rock.

Chip

[martin manning]

The typical Fender input network with the two 68K resistors and one 1-meg resistor (wired to shorting jacks) has, for the #1 input, a 34K grid stopper (68K || 68K) with the 1-meg to ground before it. The 1-meg resistor effectively determines the input impedance here, so for practical purposes it is 1-meg.

Using the #2 input, the two 68K resistors are a voltage divider with the output taken from the junction of the two. This attenuates the input signal by half (3 dB), so the stage gain will be halved. The effective input impedance is now determined by the two 68K’s in series, so it’s now 136K, almost 1/10 that of the #1.

In summary, the #2 input will reduce the input stage gain by half, and show ~1/10 the input impedance of the #1, which will load down the source as previously stated.

MPM

[Enzo]

It cuts the voltage in half, so it is a 6db drop. Half power is 3db.

When your guitar is plugged into the jack, the pickups are in parallel with that 1 meg resistor, so the input impedance is not 1 meg, it is the impedance of your pickups. Effectively that 1 meg disappears. It is important when some effect with an output cap is connected there though. That 1 meg provides a DC path to ground for that grid.

Basically there are high and low sensitivity jacks. If your signal is too hot, plug into the lower sensitivity jack 2.

[bnwitt]

What Bruce and Enzo said. 34k ohms resistance for your signal in #1 and 68k ohms resistance in #2

See the two attached schems showing the signal path for each jack. Also I tend to buy Bruces explaination of why Leo did it. Back then there were no public address systems and a lot of players plugged in there guitar and a microphone to their amps.

[bnwitt]

Wait, wouldn't running two 68k ohm resistors in parallel give the average resistance across the two, 68k?

Cminor9, think of the two resistors as pipes, both 1 inch in diameter with a fixed water pressure feeding through. With one pipe you have X resistance, with two pipes the same diameter and length you have 1/2 X resistance. this is because you now have double the conduit or pathway through which the current can flow. Two 68k resistors in parallel give half their individual resistance of 34k as you are doubling the conduit or pathway through which the current can flow.

[martin manning]

It cuts the voltage in half, so it is a 6db drop. Half power is 3db.

When your guitar is plugged into the jack, the pickups are in parallel with that 1 meg resistor, so the input impedance is not 1 meg, it is the impedance of your pickups. Effectively that 1 meg disappears. It is important when some effect with an output cap is connected there though. That 1 meg provides a DC path to ground for that grid.

Basically there are high and low sensitivity jacks. If your signal is too hot, plug into the lower sensitivity jack 2.

Yes, my mistake; half the gain is –6 dB

Re impedance, isn’t it customary to state the input impedance of a device as that which would be seen by any driver looking into the input?

MPM

[martin manning]

What Bruce and Enzo said. 34k ohms resistance for your signal in #1 and 68k ohms resistance in #2

Do you mean with respect to the input impedance? It's the resistance to ground that matters. The series resistance to the grid doesn't contribute to this or attenuate the signal until the frequency is high enough to see a path to B+ (AC ground). That won't occur until 20-40 kHz, depending upon which input is used.

MPM

[bnwitt]

The series resistance to the grid doesn't contribute to this or attenuate the signal

Of course it does Martin. I don't know what theoretical realm you're in, but the series resistance between the guitar signal and the grid definitely attenuates the signal level at the grid. Try increasing the 68k's to 1 meg and see what happens to your guitar signal level at the grid.

[martin manning]

Of course it does Martin. I don't know what theoretical realm you're in, but the series resistance between the guitar signal and the grid definitely attenuates the signal level at the grid. Try increasing the 68k's to 1 meg and see what happens to your guitar signal level at the grid.

I think my theory is correct-

For input #1, you have the 1M to ground, and after that, 34K in series to the grid. There is virtually no current flow through the 34K, so whatever voltage is at the top of the 1M will show up at the grid. The series resistance after the 1M doesn't matter.

For input #2, you have a voltage divider made up of two 68K resistors (the 1M is shorted). That cuts the signal at the grid in half, but any two equal-valued resistors would do the same, as all the current is flowing to ground. The impedance seen by the guitar is the two 68K's in series, or 136K.

MPM

[bnwitt]

Whatever Martin. You believe what you want to believe but I suggest you do some hands on testing to verify your theory.

[cminor9]

Cminor9, think of the two resistors as pipes....

Ok, now that you explain it that way, it makes sense. Thank you!

[bnwitt]

Ok Martin, I slept on it and I think I understand where you are coming from on this. I get the resistance to ground for signal loss point. I'm not sure the statement that any two same value resistors would perform the same however. I'll have to slap in some 1 meg resistors in place of the standard 68k values and test that theory out. Somehow I think I'll find a huge reduction in signal on the grid but maybe you're correct.

[GibsonLover]

Back to the original question of "what for?" remember that in the good ol' days PA systems were slow to develop. Most small combo groups did not have PA's, and what was available looked like another guitar amp anyway. Your guitar amp was your PA. Mic in one and guitar in the other. The hi-low option is for the user to to suit himself based on his equipment and the guidance of information already posted.

[tubeswell]

or for mic'-ing a harp?

[martin manning]

Ok Martin, I slept on it and I think I understand where you are coming from on this. I get the resistance to ground for signal loss point. I'm not sure the statement that any two same value resistors would perform the same however. I'll have to slap in some 1 meg resistors in place of the standard 68k values and test that theory out. Somehow I think I'll find a huge reduction in signal on the grid but maybe you're correct.

Okay, cool, I'm interested to hear what you find out, since I've never tried it myself. For the input #2 case, the two 68K resistors will cut signal to the grid in half, and show input imedance of 136K. If you put in two 1M, then you still have half the signal at the grid as you have at the tip of the cable, but now you have 2M input impedance. You won't load down the source as much, and that in itself will affect the signal level at the grid. In that respect the 1M's won't perform exactly the same as the 68K's. Also, the high frequency break point will move down to 1.5 or 3 kHz. Going from 68K to 1M is a huge change, far from the 2x difference between 34K and 68K as seen in the original #1 vs #2 inputs.

MPM

[tubeswell]

Well the way I've had it explained to me by the pros, a voltage divider works on the ratio of the two resistors. The actual size of the resistors affects the amount of current getting through (and there is virtually no current on a grid anyway - unless its forward grid current - but that's another story), but it is the resistors' ratio that determines the voltage, e.g.;

The 1M is the bottom leg of the divider, so the voltage across it will be its percent of the whole resistance times the voltage. If you used a 100k ground leg resistor and 3k4 series resistor, the same voltage division would occur, except now there would be only 103k4 in total and therefore more of a load on the input device (because more of the already insignificant current would be getting to ground, so maybe you would opt for a 470k ground leg resistor and 16k series resistor?) As long as the overall resistance is high enough to prevent loading, use any resistor values you want as long as the ratio between them yields the voltage you want.

JM2CW