B1 Buffer Preamp

Jump to Latest
#1,990
I'm sorry if this has ben covered before, but I just wanted to clarify. In the construction notes, Nelson states:

The best performance generally comes from matching the Idss of Q100
and Q101, and also Q200 and Q201. The Idss is simply the current that
flows through the JFET when the Gate and Source are grounded and +10
volts or so is applied to the Drain. Often when you buy JFETs you can get
them in Idss grades. I use GR or BL grades for this project.
Click to expand...

Is that right? I would have thought that matching Q100 with Q200 and then Q101 with Q201 would provide better channel matching. Or am I missing something? Probably...but I thought I'd just check. I ask because I have 2x matching pairs, rather than a matched quad. They are all fairly close, but clearly 2x pairs.

Many thanks
Lucas
 
Last edited:
#1,991
I'm sorry if this has ben covered before, but I just wanted to clarify. In the construction notes, Nelson states:

The best performance generally comes from matching the Idss of Q100
and Q101, and also Q200 and Q201. The Idss is simply the current that
flows through the JFET when the Gate and Source are grounded and +10
volts or so is applied to the Drain. Often when you buy JFETs you can get
them in Idss grades. I use GR or BL grades for this project.
Click to expand...

Is that right? I would have thought that matching Q100 with Q200 and then Q101 with Q201 would provide better channel matching. Or am I missing something? Probably...but I thought I'd just check.

Many thanks
Lucas
 
#1,993
Q100 and Q200 are on one channel so if the combination of these two gives you a power of Y out of the B1 and the combination of Q101 and Q201 give you a power of Y+1 out of the other channel... then you have two channels that are not balanced perfectly.
If out of one channel you have Q100 giving X and Q200 giving Y and out of the other channel you have Q101 also giving X and Q201 also giving Y then both channels give you X+Y and they are balanced.
Uriah
 
#1,994
udailey said:
Q100 and Q200 are on one channel so if the combination of these two gives you a power of Y out of the B1 and the combination of Q101 and Q201 give you a power of Y+1 out of the other channel... then you have two channels that are not balanced perfectly.
If out of one channel you have Q100 giving X and Q200 giving Y and out of the other channel you have Q101 also giving X and Q201 also giving Y then both channels give you X+Y and they are balanced.
Uriah
Click to expand...

well said 🙂

However if one doesn't want to match them, they are for sale with aboard for 40$ matched from passdiy 🙂

J'
 
#1,995
udailey said:
Q100 and Q200 are on one channel so if the combination of these two gives you a power of Y out of the B1 and the combination of Q101 and Q201 give you a power of Y+1 out of the other channel... then you have two channels that are not balanced perfectly.
If out of one channel you have Q100 giving X and Q200 giving Y and out of the other channel you have Q101 also giving X and Q201 also giving Y then both channels give you X+Y and they are balanced.
Uriah
Click to expand...

Of course. Duh! Obvious.
 
#1,996
udailey said:
Q100 and Q200 are on one channel so if the combination of these two gives you a power of Y out of the B1 and the combination of Q101 and Q201 give you a power of Y+1 out of the other channel... then you have two channels that are not balanced perfectly.
If out of one channel you have Q100 giving X and Q200 giving Y and out of the other channel you have Q101 also giving X and Q201 also giving Y then both channels give you X+Y and they are balanced.
Uriah
Click to expand...

Except that according to Nelson's schematic Q100 and Q101 are on the right channel and Q200 and Q201 are on the left channel. The matching (of Q100 with Q101 and Q200 with Q201) ensures that the upper and lower JFETs in each channel are biased with very close Vgs and Vds.

Since these operate as constant current source loaded source followers with a gain of 1, it is not necessary to match devices between channels, though it won't hurt anything if you do.

Cheers,
 
#1,997
Nikolas Ojala said:
The B1 buffer is essentially a voltage follower.
A voltage follower is also an essential part of a Linkwitz-Riley filter.
See this: Active Filters

Has anyone tried the B1 voltage follower as a component in a Linkwitz-Riley filter?
Click to expand...

Rob is working on it at the moment http://www.diyaudio.com/forums/blogs/relder/142-b1-active-xos.html 🙂 And I've been thinking about it but didn't have enough knowledge to realise that the B1 was suitable, as I thought you needed gain to do an active filter.

Tony.
 
#1,998
Jacques Merde said:
Except that according to Nelson's schematic Q100 and Q101 are on the right channel and Q200 and Q201 are on the left channel. The matching (of Q100 with Q101 and Q200 with Q201) ensures that the upper and lower JFETs in each channel are biased with very close Vgs and Vds.

Since these operate as constant current source loaded source followers with a gain of 1, it is not necessary to match devices between channels, though it won't hurt anything if you do.

Cheers,
Click to expand...

Yeah, that's what I was getting at. So I am right then, according to the schematic Qxxx references.
 
#2,000
udailey said:
I have read a few people saying that there is no reason to match the devices in the B1.
Uriah
Click to expand...

True, it is not necessary to match Vgs or transconductance. Still, I would match (or select, if you will) for Idss as Nelson suggested. In the Building a symmetrical psu B1 buffer thread good results are reported when the device with the slightly higher Idss is used in the lower position. See post #487 in that thread, and the various follow ups.
 
You must log in or register to reply here.
Top Bottom