How to figure out input impedance of a buffered circuit

Mugshot · Post by **Mugshot** » 17 Oct 2011, 00:58

FiveseveN wrote:
considering the low quality of the transistors and caps
Are Asian electrons inferior to American electrons?

sig'd

kleuck · Post by **kleuck** » 17 Oct 2011, 06:49

By the way, and as a response to the topic :

Transistor alone input resistance = Hfe/(39*Ic) with Ic=collector current=current through the transistor,
What is the current trough the transistor ? Bias voltage is 4,5 volts if we assume a 9 volts battery, the direct voltage drop between base and emitter is about 0,6 volts with a silicon Si transistor, so voltage on the emitter is 4,5-0,6=3,9 volts.
With ohm's law, current is 3,9/10000 as the emitter resistor is a 10K one, so 0,39mA,
Therefore, Rt=200/(39*0,00039)=13 Kohms to 400/(39*0,00039)=27 Kohms
But as we have a total feedback between the resistor's emitter and the base, we must add to that the value of the resistance*Hfe.
Rt is therefore negligible : 10000*200=2 Mohms (worst case) to 10000*400=4 Mohms.
And the actual resistance is Rt' paralleled with the bias resistor : (470 K*2M)/(470 K+2M)=380 K (worst case) and in the best case 420 K.

LC filters cutoff : 1/(2*Pi*√(L*C))
So 2 H and 250 pf >> 1/(6,28*√(2*250*10^-12))=7,1 Khz,

JakeAC5253 · Post by **JakeAC5253** » 17 Oct 2011, 16:11

kleuck wrote:By the way, and as a response to the topic :

Transistor alone input resistance = Hfe/(39*Ic) with Ic=collector current=current through the transistor,
What is the current trough the transistor ? Bias voltage is 4,5 volts if we assume a 9 volts battery, the direct voltage drop between base and emitter is about 0,6 volts with a silicon Si transistor, so voltage on the emitter is 4,5-0,6=3,9 volts.
With ohm's law, current is 3,9/10000 as the emitter resistor is a 10K one, so 0,39mA,
Therefore, Rt=200/(39*0,00039)=13 Kohms to 400/(39*0,00039)=27 Kohms
But as we have a total feedback between the resistor's emitter and the base, we must add to that the value of the resistance*Hfe.
Rt is therefore negligible : 10000*200=2 Mohms (worst case) to 10000*400=4 Mohms.
And the actual resistance is Rt' paralleled with the bias resistor : (470 K*2M)/(470 K+2M)=380 K (worst case) and in the best case 420 K.

LC filters cutoff : 1/(2*Pi*√(L*C))
So 2 H and 250 pf >> 1/(6,28*√(2*250*10^-12))=7,1 Khz,

Thank you! Finally some field applicable theory. So would the Rsource, Rinputresistor, and Remitter all be calculated in parallel to find out the total impedance of the network?

kleuck · Post by **kleuck** » 17 Oct 2011, 17:23

I don't' understand, what network ? Impedance of the emitter is separated from the input's one (though it's multiplied on it)
Resistance of the emitter is another calculation, i can give it if you want, let me the time to find the formula (i don't know it by heart)

drbob1 · Post by **drbob1** » 18 Oct 2011, 01:04

Wow, I missed this catfight. I actually have done some experiments with both sides of this problem as well as read some fairly well thought out pieces using SPICE modeling for the interaction between a guitar, cable and plugs. And yes, I can tell you from experience that the cable, irrespective of plugs, does make a difference in a passive pickup situation. I'm sorry that I'm too lazy to do the calculations myself, but here's my understanding and experience...

1. Pickups, particularly single coils, form a resonant network with the cable and input impedance of the amp/device. Higher cable capacitance lowers the resonant peak in both frequency and gain. Higher input impedance raises the resonant peak in both. The effect is less marked although still there with humbuckers.

2. This resonant network effect is audible in the lengths of cables used for guitar. Running a 10' cable to a looper with 5 independent loops, then an output 10' cable (both decent quality, Neutrik ends and 25 pf/ft capacitance) into an amp gives a baseline tone. Running 5 8" pieces of Canare cable, soldered Switchcraft connectors, the signal is noticeably darker. Even 2 loops of Canare can do this. I have some "Quantum Oxygen Free" cable that sounded really cool when I bought it. It's big and hard and, when opened up has really thick insulators but poor copper coverage of the shield layer. This stuff, even one or two loops (again, soldered Switchcraft ends) sounds dark as heck. Stringing up the 5 loops with George L's cable, soldered or press fit ends, there's very little high end loss. BTW, if anyone thinks this is observer bias, remember, I bought the Canare kit believing it was BETTER than the George L's cable and was absolutely floored when I found that it dropped the high end out of my signal!

3. Drop ANY buffer at the beginning of the chain and the effect completely goes away. I can use any patch cable from the cheapest Hosa to the longest cable I've got (I tested it with 20' cables before the looper and in each loop for a total of 140' of cable, with an SHO as a buffer before and there was no treble loss). And yes, Mictester, you're right, a buffer in the guitar will insulate you from any effect of your signal chain on your guitar pickup's output. Unfortunately, playing that system is part of the art of guitar, you won't get the right response from a Germanium fuzz if you have a buffer before it (it won't clean up as nicely when you roll back on the volume) among other things. EMG built their rep on preamped pickups and a lot of folks feel that system is a bit sterile because of the decreased interaction between pickups and the rest of the signal chain.

4. I've got Boss pedals (buffered bypass), Ibanez (buffered), Klon, Cornish and a wall full of true bypass. Without a buffer before 5 TB pedals, there's some high end loss (probably as much from all those 1/4" plugs as the actual cables/pedals), with a buffer, ANY buffer, they sound fine. I don't have golden ears, I can't hear that the Klon's buffer or the Cornish buffer is better than the Boss buffers. I also can't hear a difference with a number of buffers in the chain. In theory, all those solid state devices should be a problem, building up noise like applying wax to your car too often, but I can't hear it. I can hear that the SHO, which is a very high impedance device, is very bright when engaged, but I'm suspecting that's a result of the other stuff in the pedal, not the input buffer.

rocklander · Post by **rocklander** » 18 Oct 2011, 01:14

so based on that, if I was (and I am) building up a board of all TB pedals, and wanted to include some clever stuff integrated into the board (like a rechargeable Li battery and a tb switch to tuner/kill) would it also be a good idea to include a buffer on the 'signal in' side of the in/out jacks? if so.. what cct? what's the simplest one to use? I really like the buffer in the black toast, but haven't a clue how to separate it from the black toast cct as an autonomous effect.

juanro · Post by **juanro** » 18 Oct 2011, 01:33

rocklander wrote:would it also be a good idea to include a buffer on the 'signal in' side of the in/out jacks?

"Signal in" or at least near it... if you give it the chance to insert the buffer in another place, you can (eventually) put your true bypass FF first, then buffer, then all the rest.

As for the buffer itself... I've used many (single FET, opamp) and all sounded more or less the same to me... even the "crappy" bipolar ones. Point me to the black toast schematic and I'll see if I can isolate the buffer for you.

Juanro

rocklander · Post by **rocklander** » 18 Oct 2011, 01:36

https://www.freestompboxes.org/viewtopic.php?f=13&t=2752

members/bajaman/Baja/Baja%20Black%20Toast/Schematic.png
(too large to embed

)

juanro · Post by **juanro** » 18 Oct 2011, 01:44

Keep the parts marked green and connect sw1b directly to out.
(That's pretty much the "single FET" I said before)

Regards,
Juanro

juanro · Post by **juanro** » 18 Oct 2011, 01:47

BTW, just to be a little on-topic: input impedance will be 1.7875 M

Juanro

rocklander · Post by **rocklander** » 18 Oct 2011, 01:53

thanks for that.. I tried doing the inverse a while ago (on paper here) but was told that I couldn't really do that cos ... um.. I duno why.. tried to segregate out the booster only but had it pointed out that it was not going to work like that ..

juanro · Post by **juanro** » 18 Oct 2011, 01:56

Mhm, don't see why not... the booster part is also a high input impedance circuit and should work just fine fed directly from the guitar. Perhaps Bajaman could chime in here?

Juanro

RnFR · Post by **RnFR** » 19 Oct 2011, 06:27

that should work fine!

I always put my buffer(tuner) on the end to drive the cable to the amp. am I missing something? I don't understand why u would need one in the beginning of the chain if it isn't going directly from the guitar. is it because of all those jacks?

rocklander · Post by **rocklander** » 19 Oct 2011, 06:34

kinda makes sense, but I would do something stoopid like forget to turn off some modulation (or other) pedal and mess the tuner up

my gig board has a 3pdt at the start which goes to my tuner (and then nowhere) or to my pedals.. so it's a kill as well.

merlinb · Post by **merlinb** » 19 Oct 2011, 10:00

kleuck wrote:You obviously don't know what is an RLC resonant circuit or how to calculate it's cutoff,

It looks like mictester used 25pF in his calculation, rather than 250pF. Genuine mistake? You're right though, he should know better than to dismiss cable capacitance when there are henrys of inductance in involved. Not that I believe he is an EE or anything.

stonefreed · Post by **stonefreed** » 19 Oct 2011, 14:57

I like the first part of the black toast very much but the boost doesn't sound very good imho. It's sounds much too harsh.

I started with the buffer only and then added the second part because the buffer isn't quite unity gain.

I'm going to leave it out again.

Alf