Buffer impedance question

[armdnrdy]

I've read a few tutorials on impedance and to be honest I'm just as confused as before!  

I'm trying to figure out the impedance of this PNP buffer. What components are determining the impedance?

http://s1289.beta.photobucket.com/user/armdnrdy/media/Buffer_zps1bec30b0.jpg.html#/user/armdnrdy/media/Buffer_zps1bec30b0.jpg.html?&_suid=135836472091305619832366457627

[gritz]

My lazy answer is 47k in parallel with (10k + the impedance at the tranny's base*). So not much.

*if the tranny's HFE is reasonable then we can ignore the input impedance of the tranny itself and just take the 330k resistor into account. So the input impedance will be R101 in parallel with (R102 + R103), which is:

1/((1/47)+(1/340)) = 1/(.0213 + .0029) = 41K

*The input impedance of the transistor itself in this configuration (emitter follower) is the resistance at the emitter * HFE, so it's probably big enough to ignore. It's an emitter follower because the voltage at the collector will be held steady by C102, which in conjunction with R104 is presumably dropping a high supply voltage. I'm not sure what the power supply voltage is supposed to be, but at 9V it will barely bias itself on.

[armdnrdy]

Thanks for the reply.

Well 41K doesn't seem to be a very high input inpedance does it?

This is from the buffer/JFET switching board that was addded to the later Roland Jet Phasers.

I wanted to reproduce the original input/output inpedance while making it true bypass. The original lets the signal travel all the way through the effect board!

That can't be good!

[gritz]

Ahh, the Jet Phaser guy - I knew I recognised your username! o you have a link to the schematic you're referring to? I'm quite intrigued now.

[armdnrdy]

YES! It is I!! The Jet Phaser Guy!  

I just found an interesting link for calculating transistor amplifiers.

http://en.transistoramp.de/common-collector-circuit/

Here's the link for the re-drawn Jet Phaser.

http://www.aronnelson.com/gallery/main.php/v/diyuser/Roland+AP-7+Jet+Phaser.jpg.html

[armdnrdy]

If you were refering to the full gate board schematic it is included in the Jet Phaser PDF in the gallery.

[R.G.]

It's worse than that. I'll do the lazy thing and approximate heavily too.

The transistor is set up as an inverting feedback amplifier, similar to the inverting-gain opamp stage. This setup causes the inverting input to be driven to a "virtual ground", so the impedance is nearly zero. So the input impedance at the base to emitter being high doesn't help much.

My first estimate for the input impedance is 47K in parallel with something around 10K. It's not quite that bad, but I'd start there.

I've read a few tutorials on impedance and to be honest I'm just as confused as before!  

What part's confusing?

There's a lot of elaboration, but "impedance" is the generalized version of "resistance" to include inductors and capacitors. A capacitor lets high frequencies through easily, but low frequencies less easily. Inductors are the reverse - DC goes through easily, and higher frequencies get "impeded". The amount of "impede-ance" they offer is expressed the same way as resistance, that being how many volts does it take to force one ampere through the part. With inductors and caps, this is different at every frequency.

Once you have that down, what gets tricky is how individual parts get combined in series/parallel combinations.

[gritz]

It's worse than that. I'll do the lazy thing and approximate heavily too.

The transistor is set up as an inverting feedback amplifier, similar to the inverting-gain opamp stage. This setup causes the inverting input to be driven to a "virtual ground", so the impedance is nearly zero. So the input impedance at the base to emitter being high doesn't help much.

My first estimate for the input impedance is 47K in parallel with something around 10K. It's not quite that bad, but I'd start there.

I thought that at first, but if you look closely you'll see that the transistor's collector is decoupled to ground by C102, so there's no signal feedback to base.

[armdnrdy]

Thanks for the replies,

What I find confusing is I don't really understand what each component is doing.

I would think that C103 (the 100K pull down) would have an effect on impedance.

[gritz]

If you were refering to the full gate board schematic it is included in the Jet Phaser PDF in the gallery.

I'm almost embarrassed to admit how long it took me to find the "download document" link on that page. Small screen + bad eyesight = made of fail!

[armdnrdy]

If you were refering to the full gate board schematic it is included in the Jet Phaser PDF in the gallery.

I'm almost embarrassed to admit how long it took me to find the "download document" link on that page. Small screen + bad eyesight = made of fail!

I don't think it's just you!

I had a hard time figuring it out as well!

[gritz]

Thanks for the replies,

What I find confusing is I don't really understand what each component is doing.

I would think that C103 (the 100K pull down) would have an effect on impedance.

C103 is just the coupling capacitor to the signal input of the main board. The input impedance of the main board (which the emitter of Q101 will see in parallel with R106's 100k) is ~rather larger~ than 10k, so it's not making Q101 drag it's ass (to use a technical term in lieu of sucky equations).

Remember that back in the day it wasn't unusual to see low input impedance guitar fx (I'm thinking fuzzes and wahs amongst other things). There's also the remote possibility that Roland's service notes have a mistake in 'em. It's unlikely, as the 47k input resistor is referred to on the schematic and the pcb layout, but it's possible.

[gritz]

duplicated post...

[PRR]

> I would think that C103 (the 100K pull down) would have an effect on impedance.

Also whatever thing you connect to the output jack.

But all that junk is multiplied by hFE.

Assuming it turns-on at all (it's marginal), hFe is likely over 50, so 10K||100K||10K total loading at Emitter is 238K at Base. Comparable to the 330K, and way over-whelmed by the 47K.

I get a number like 39K altogether, though "41K" and "10% under 47K" are entirely equivalent for rock-n-roll.

This is lower than I like to see. But not absurdly so.

> impedance and to be honest I'm just as confused as before

Start with the easy stuff. The power line from the street to my house is 1 ohms. I plug in a 10 ohm lamp. How much of the 240V at the street actually gets to my lamp? How about if I find a 2 ohm lamp? A half-ohm lamp? A 100 ohm lamp?

If you can do these answers, then you can see why we *tend* to like input impedances considerably higher than output impedances. (We get more "light".)

It is mildly confusing when the impedance is not constant. A naked gitar pickup is 5K for bass to midrange and >100K at the treble resonance. You don't really need the whole curve. Figure the "drop" when either 5K or 200K drives a 40K load. It's substantial. Then when driving a 1,000K load (hardly any drop either way). The drop is greatest at guitar's treble resonance, the high-strings overtones.

However most guitars have volume pots and if you factor that, if the pot is not full-up the amplifier (or buffer) input has much less effect.

[R.G.]

I thought that at first, but if you look closely you'll see that the transistor's collector is decoupled to ground by C102, so there's no signal feedback to base.

Doh~!

Yes, you're right. Never mind what I said...   

[armdnrdy]

Thanks for the replies,

What I find confusing is I don't really understand what each component is doing.

I would think that C103 (the 100K pull down) would have an effect on impedance.

C103 is just the coupling capacitor to the signal input of the main board.

I meant R106 (100K pulldown) oops. My eyes are going as well!

[gritz]

I thought that at first, but if you look closely you'll see that the transistor's collector is decoupled to ground by C102, so there's no signal feedback to base.

Doh~!

Yes, you're right. Never mind what I said...  

I double, then triple checked after you posted, because I thought I'd got it wrong. I usually do!  

@ armdnrdy: I quickly breadboarded the buffer in your original post and got a voltage of 0.46V across the emitter resistor, so it does bias on - just. but there's not a huge wiggle room for signal though. Because the Jet has two batteries in series to power it I thought that maybe the buffer was powered by 18V, but the schemo suggests that the signal ground is connected to the centre tap of the batteries (as it should be, 'cos that's the way the mainboard's connected), so the supply to that buffer stage is -9V. Strange that the tranny is biassed so low.

Regarding R106 - it's ten times larger than the emitter resistor, so while it has an effect it's not a gut-bustingly significant one. If you think of C103 as providing a low impedance path to audio then it puts R106 in parallel with R105 - reducing the effective emitter resistance to 9.1K:

1/R = 1/10k + 1/100k = 11/100. 100/11 = 9k1.

So a slight reduction (and the input impedance of the main board will reduce it a bit further and so slightly increase the base current that the transistor needs to follow the input voltage), but it's no great shakes compared to the 47k at the input.

Hope this makes sense a bit.  

[armdnrdy]

Something that I should have mentioned earlier:

The 100K input pulldown on the main phaser schematic is not installed on the actual phaser.

That 100K was implemented on the first version without the gate board. I have good gut shots of both versions that I collected from various places on the net. The input, output and a 220K pull down resistors shown on the drawing were mounted on the trace side. The later gate board version is missing those three resistors.

[armdnrdy]

I've also verified the component values of the gate board from photos.

So, what you are saying is... this buffer design is just about useless?

[gritz]

I've also verified the component values of the gate board from photos.

So, what you are saying is... this buffer design is just about useless?

No, not at all. It is what it is*. If it's correct then it must be part of the sound. Even if it clips a bit.

Consider a 20-year-old electronics student that's  never seen a guitar and has only been schooled in "the right way to do stuff". If (s)he examined the circuit of a two transistor fuzz and was then told that it was going to be fed by a mahoosive coil of really skinny wire bolted to a guitar via twenty feet of cable / capacitor then our student would doubtless protest that it was all wrong, the input impedance of the fuzz was too low and that we're all crazy for even entertaining this kind of nonsense. But we know different. It works, even though it is technically "sub-optimal".  

* That buffer is a bit odd though. As you've checked those resistor values with actual photos then there's not much room for doubt. It would make more sense if the dropper resistor R104 was 1k, not 10k and if R101 was say 470k rather than 47k. Then there would be room for more signal "swing" as well as a bit less loading on whatever preceded it in the chain. But they made it that way for a reason...

I see they don't quote an input impedance in the spec.