high output impedance

[fuzzy645]

A few questions about output impedance:

1. What would be considered within range of normal output impedance for an overdrive pedal?  Would it be in the 5K - 20K range?, or is that too high?

2. What would be the symptom/problem one would experience if a pedal had too high an output impedance?

Since we are on the topic of impedance, I might as well as about input impedance too:

3.  I have noticed on many pedal schematics they put a 1M resistor to ground right on the input.  This seems to be standard on amps too.  My question is this:  Does the act of simply doing this set the input impedance to be high?

Thanks!

[amptramp]

A few questions about output impedance:

1. What would be considered within range of normal output impedance for an overdrive pedal?  Would it be in the 5K - 20K range?, or is that too high?

There are many within that range, but some use an op amp to drive the output and these may have a lower output impedance than 5K.  The op amp itself has a very low output impedance, on the order of tens of ohms or less, but most designers add series resistance just to ensure that if someone does something stupid (like plug two outputs into each other) that there is something to limit the current and also isolate the amplifier from the capacitance of the output cable, which could cause instability if the resistance was not there.

2. What would be the symptom/problem one would experience if a pedal had too high an output impedance?

The impedance of the output and the cable capacitance act as a low pass filter so the hgh frequencies would be muted.

Since we are on the topic of impedance, I might as well as about input impedance too:

3.  I have noticed on many pedal schematics they put a 1M resistor to ground right on the input.  This seems to be standard on amps too.  My question is this:  Does the act of simply doing this set the input impedance to be high?

Thanks!

The input impedance is set by the parallel combination of input resistance networks and active devices.  The 1M resistor is often used to ensure coupling capacitors have a return to ground so that you don't get a sudden switch pop when you connect the input to the source.  If this is not done, leakage across the capacitor would bring the input end up to the voltage at the other end (often half the battery voltage or 4.5 volts in many circuits).  When the device was switched from bypass to in circuit, you would get an enormous switch pop caused by the sudden grounding of a capacitor, sending a -4.5 volt pulse into the circuitry.  The 1M resistor eliminates most of the voltage caused by leakage current.

[earthtonesaudio]

1. Normal for an "overdrive" can be anywhere in a large range.  Some pedals have op-amp outputs with output impedance under 100 ohms.  Other pedals like the Fuzz Face may have a maximum output Z of over 250k ohms.  "Too high" or "too low" depends on the application.  5k might be considered average, 20k just slightly higher than average in the realm of guitar effects.

2. A high output impedance is more sensitive to the load placed on it.  That includes the cable leading to, and input impedance of, the next effect.  The cable capacitance can become a treble-cut filter for an effect with a high output impedance.  A high input impedance after a low output impedance maximizes voltage transfer.  Generally for audio gear this is desired and helps get the most signal and the least noise.  A low input impedance after a high output impedance maximizes current transfer, but most audio gear is not set up for this.  As a result you drop a lot of level and have to boost the gain to make up for this loss, and noise performance suffers.

3. To put it simply, no.  A 1M resistor to ground presents an impedance of 1M... but it's in parallel with the other impedances in the circuit.  So if you have a 1M in parallel with 100M the resulting equivalent impedance will be slightly less than 1M.  Conversely if you put 1M in parallel with 10k the 1M does basically nothing, and the input impedance becomes slightly less than 10k.

[edit] Ron typed faster...

[artifus]

http://www.experimentalistsanonymous.com/ve3wwg/impedance_matching

[fuzzy645]

Thanks.  Your answers have been helpful.

[fuzzy645]

Another question...

Does it make sense that the actual PCB would demonstrate a low output impedance (in the 1K range), but when connecting a 100K volume and 100K Tone pot on the output, the entire circuit would then demonstrate output impedance in the 40K - 50K range?   

Some circuit background:    the first thing the PCB output wire goes to is the tone pot (I assume it would be correct to say this 100K tone pot is in series with the PCB) so resulting resistance at that point in time I'm guessing would be 101K.  Next in the chain comes the volume pot, which I believe would be considered to be in parallel with the aforementioned tone pot, so we essentially have a pair of 100K pots in parallel on the output thereforeresulting output impedance (more or less) is in the 50K ballpark.   Assuming that is a BAD thing (perhaps I'm assuming wrong), what would be the remedy (if one is needed)?

1.  swap out the 100K pots for 50K pots.  That would bring output impedance down to 25K.  Would tone suffer??

2.  swap out the 100K pots for 25K pots.  That would bring the output impedance down to approx. 12.5K.  Wouild tone suffer?

Thanks!

[earthtonesaudio]

It makes perfect sense that adding a voltage divider volume control would raise your maximum output Z.

I say "maximum" because when the volume control is up all the way, you are essentially connecting the output terminal to the 1k output impedance of the circuit, and the output Z is about 1k.  But as you start to turn down the resistance in series with the circuit output increases, which raises the circuit's output impedance. 
Simultaneously, the resistance between the wiper and ground is decreasing, so you end up with a maximum output impedance at about 50% of the volume pot.  As you decrease the volume further, the resistance from wiper to ground decreases to zero eventually.  Technically the output impedance of the pedal is now zero ohms, but the output impedance of the circuit is 101k.  Not that it matters, though, since the volume is at zero.

Note this is only referring to the volume pot.  It's pretty standard to wire a volume pot as a voltage divider (signal in one lug, opposite lug grounded, and output taken from wiper), but there are many different tone controls and I didn't want to assume you were using a particular one.

It's difficult to say if tone would "suffer" for a couple reasons.  Mainly that's a subjective question, but also if you simply halve the value of the tone pot then you're changing the frequencies affected by the tone control.  So to maintain the same response at a new (lower) output impedance you would also have to change the value of the capacitor(s) involved.  This is also true for the volume control but generally you can just make the output coupling caps big enough to pass all frequencies of interest.

[therecordingart]

Everything has been answered so I'll I have to offer is this fun calculator to play with. This will give you the amount of signal loss to expect between different load and source impedances.

http://shure.custhelp.com/app/answers/detail/a_id/224/kw/load%20loss/session/L3RpbWUvMTMyMTM3MjA5MS9zaWQvY2RSTGFiSms%3D

[fuzzy645]

I say "maximum" because when the volume control is up all the way, you are essentially connecting the output terminal to the 1k output impedance of the circuit, and the output Z is about 1k. 

Ahhh.  You see when I measured the output impedance, I am getting in the 40K range when all controls are maxed on 10.  So you are saying the volume pot on 10 is a "non-issue" - but perhaps the culprit might then be the tone pot/cap combo. 

I will see what happens when I chop down the tone pot value to 50K.  The vol/tone stock is wired as depicted below with  tone functioning as a treble boost. C1 is a 2200 pf cap, and both R1 and R2 are currently at 100K.

[earthtonesaudio]

Okay, with that tone/volume arrangement the tone pot resistance is only adding to the output resistance because 100% of the pot is in series with the output.  For the volume pot, the upper half is in series and the lower half is in parallel with the output.

The capacitor C1 is in parallel with R1 and the tone control is formed by this parallel combination and the shunt resistance of the volume pot.  So the frequencies affected by the tone pot depend on both resistors, not just the value of the tone pot.  If you want to maintain the same frequency response you have to scale both resistors by some factor, and scale the capacitor by the inverse of this factor.

i.e. if you halve R1, you must also halve VOL but double C1.

[fuzzy645]

Okay, with that tone/volume arrangement the tone pot resistance is only adding to the output resistance because 100% of the pot is in series with the output.  For the volume pot, the upper half is in series and the lower half is in parallel with the output.

The capacitor C1 is in parallel with R1 and the tone control is formed by this parallel combination and the shunt resistance of the volume pot.  So the frequencies affected by the tone pot depend on both resistors, not just the value of the tone pot.  If you want to maintain the same frequency response you have to scale both resistors by some factor, and scale the capacitor by the inverse of this factor.

i.e. if you halve R1, you must also halve VOL but double C1.

Cool.. So if currently using 100K vol/tone with 2200 pf cap, I will try 50K vol/tone with approx 4400 pf cap.  I suppose it would be worth even seeing what happens with 25K vol/tone and approx. 8800 pf cap.

Thanks!

[fuzzy645]

Internal "Ah-ha" moment  - I was just thinking, I bet this is the very reason why designers build output buffers into these types of circuits.

http://www.muzique.com/lab/buffers.htm

[amptramp]

Internal "Ah-ha" moment  - I was just thinking, I bet this is the very reason why designers build output buffers into these types of circuits.

http://www.muzique.com/lab/buffers.htm

With the large variety of pedals available and the vast difference in input and output impedances, it pays to use a buffer to isolate controls like tone and volume.  The buffer is not to boost signal or establish a low output impedance, it is there to isolate the controls from external influences.