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Old 15th April 2012, 04:57 AM   #21
JGAN is offline JGAN  United States
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Ok so I read this page and I think that building a second circuit for just this project is more trouble than it's worth. I will just have to try a few resistors I guess.

Thanks again all!
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Old 15th April 2012, 09:13 AM   #22
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Quote:
Originally Posted by Simon B View Post
There is something of a "how long is a piece of string" to all this.

I am surprised at the low values being suggested here for Rs - my schematic was for a source follower, with the assumption that it would be used to drive a hi-z instrument amp input - I wonder if there might perhaps be bit of confusion about exactly what type of circuit needs what value of Rs? What purpose is served by using a value as low as 1k or 2k here? The higher the standing current the shorter the battery life.
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The point of a source follower is to give a low output impedance, and consumption is pretty low even with a 1K source resistor (4mA or so).

I don't see how a 47K could bias anywhere near half point?.

Incidentally. all the ones I've ever built used 2N3819 FET's.
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Old 15th April 2012, 09:30 PM   #23
Simon B is offline Simon B  United Kingdom
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JGAN, the page on piezos that you link to shouldn't worry you too much, so long as you have a high impedance preamp for your pickup to feed into, you should get a fairly flat frequency response. Please, don't let all this put you off actually building a preamp so you can get on with playing your cello...

Quote:
Originally Posted by Nigel Goodwin View Post
The point of a source follower is to give a low output impedance, and consumption is pretty low even with a 1K source resistor (4mA or so)..

I'd say that the point is to give a sufficiently low output impedance to drive the load it's to be connected to. There doesn't seem much point in shortening battery life by trying to take it way lower than that. For gigging musicians, a bug preamp that doesn't kill the battery if accidentally left on between gigs is quite useful - 1000 hour battery life really is better than 100. Naturally though JGAN has to undermine me here by wanting to burn his battery like there's no tomorrow!

Source follower output impedance, Zout, is given by:

Zout = Rs // (1/gm)

Vishay quote values for gm for their 2N3819 as 2mS min, 5.5mS typical. With even minimum gm of 2mS, Zout of the follower is Rs in parallel with 500ohms, so very little difference in Zout between Rs = 47k and 2k2 - it's the gm term that dominates.

What's left? Headroom, ie quiescent Vs. Whilst getting it half way between the supply rails does give best headroom for the frenzied attack that JGAN is no doubt planning, for this circuit the operating point is very much device dependent - I was indulging in wishful thinking suggestng he choose Rs to get it to 4.5v.

Both Fairchild and Vishay quote the following for their 2N3819s:

With Vds = 15 V, for Id = 200 uA,
Vgs = –0.5v min, –2.5v typical, –7.5v max

ie, highly variable.... The values for Vs that I got with a 9v supply feeding a source follower were:

Fet sample 1:
Vs = 1.21v with Rs = 2k2 (Id = 550uA)
Vs = 1.59v with Rs = 47k (Id = 33uA)

Fet sample 2:
Vs = 1.33v with Rs = 2k2 (Id = 605uA)
Vs = 1.73v with Rs = 47k (Id = 37uA)

Using Rs = 47k with my fets, which are towards the 'small' end of the range for required -ve Vgs, gets the operating point a little closer to our ideal Vs = 4.5v than using Rs = 2k2. For a different batch, at the other end of the scale, Vs = 47k would give a bit less headroom.

Only with the use of fets from batches falling in the middle of the range of possible Vgs-for-given-Ids will the best possible headroom be achievable. My suppliers tell me that Nigel Goodwin gets all those and I can't have any.......

It is worth checking what Vgs is in a source follower if headroom is tight, ie occasional moderate clipping is occurring. If Vgs is lower than 4.5v, increase Rs. If Vgs is higher than 4.5v, lower Rs. For a given fet, with source resistor biasing, we can't make that much difference to the quiescent Vs compared to the manufacturing spread.



Last edited by Simon B; 15th April 2012 at 09:37 PM.
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Old 16th April 2012, 02:32 PM   #24
Simon B is offline Simon B  United Kingdom
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I should add that the equation I give above for Zout of a source follower

Zout = Rs // (1/gm)

derives from the small signal model.

When driving a capacitive load, eg a long cable, then the only path available for discharge of that capacitance on negative going signal slopes is the source resistor Rs.

If Rs is too high in value for the capacitance being driven, then the available slewing rate will be too low and distortion ensues - there are applications for which lower values of Rs will be necessary.



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Old 17th April 2012, 01:49 AM   #25
JGAN is offline JGAN  United States
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Thanks for the replies Simon,
I placed an order for parts, and they should arrive in 1-2 weeks. In the mean time, I was going back over this thread, and you mentioned how R4 was a bleed resistor for C1. Did you mean C2? Because iirc, bleeder resistors are in parallel with their corresponding cap.
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Old 17th April 2012, 04:13 AM   #26
Simon B is offline Simon B  United Kingdom
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Hi JGAN,

No, I did mean C1. It keeps the output side plate of C1 free of dc voltage, ie at ground potential. With C1 an electrolytic, this is necessary, as a high impedance, dc coupled input that it might be connected to would not be happy otherwise.

"Bleed" as in "stops you getting bleedin' problems!"

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Old 17th April 2012, 08:54 AM   #27
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Quote:
Originally Posted by Simon B View Post
Hi JGAN,

No, I did mean C1. It keeps the output side plate of C1 free of dc voltage, ie at ground potential. With C1 an electrolytic, this is necessary, as a high impedance, dc coupled input that it might be connected to would not be happy otherwise.



Any correctly designed input wouldn't be DC coupled anyway

Quote:

"Bleed" as in "stops you getting bleedin' problems!"
To be fair, I don't really know what you'd call that resistor either?.

It might be a better idea to use a non-electrolytic?, a 0.1uF or 0.22uF would be fine (reactance at 20Hz roughly 80K and 40K) if you're feeding guitar inputs. Obviously if you're feeding low impedance inputs, then an electrolytic would be preferable.
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Old 17th April 2012, 01:38 PM   #28
Simon B is offline Simon B  United Kingdom
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Well I did think about saying something like "keeps the output side plate of that cap bled down to 0v" but the "bleedin problems" seemed a bit more fun, so I went with it!

Yes, dc amp input coupling isn't clever, but it is common, especially on valve amps, a fender twin for instance and plenty of others. Myself, I like a bit more between B+ and the instrument lead than the distance between the grid and plate of a small bottle with red hot bits in it .......

Also the bleedin' resistor helps stop dramatic cracks coming through when plugging in. Which is bleedin' nice.
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