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Old 9th July 2007, 02:34 AM   #1
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Default Input stage biasing - BJT

Hi,

I've been trying to gather information on my way to building my first amp. I have included a schematic for the input stage in this post. I Don't quite understand the function of the Sziklai stage as I have it - I know it seeks to linearise the performance but the values I saw in a design seem to saturate one side of the leg more than the other ...

Should I be balancing the quiescent current in each leg of the diff amp?

That is, should the current in R108 match that in Q105?

I've noticed also in the simulation that although R109 and R110 are relatively high (they need to be above 22R to have any real effect in balancing the currents I believe) there is still quite an imbalance and I changed the values of R107 and R108 to correct this imbalance. I'm not sure that this is best practice. I've got a feeling that I should leave all values similar to one another down the legs of the diff amp ... Comments please.

The values I have in my circuit come from either some simulation or stuff I've gathered from a variety of sources. The transistor types are real - they have been purchased and will be used. The rails will have the same voltage as the power end but are from a separate transformer and are +-35V.

I'd appreciate any help in this issue as I can't seem to source anything more of great value.

Cheers,
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Old 9th July 2007, 07:38 AM   #2
beeger is offline beeger  United States
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Howdy; You have grossly oversized r105-106. Try 22-47 Ohms. The circuit looks pretty good other than that.

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Old 9th July 2007, 09:18 AM   #3
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Hi beeger,

Yes. I have seen values as low as 22R. I believe these R's control the gain - the higher the value the lower the gain.

I will proceed to try and simulate this soon.

Cheers,
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Old 9th July 2007, 01:52 PM   #4
AndrewT is offline AndrewT  Scotland
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Hi,
r105 & r106 can be set in the range 0r0 to 330r.
Your present values drop about 20V when the 3mA passes through them. I suspect that is almost shutting down the amplifier.
Try 100r for starters.

R107 allows about 0.27mA through Q103 leaving about 2.7mA to pass Q104. This will probably work but I would reduce R107 maybe as low as 1k0.
C101 is a bit (a lot) big, try 15uF//1uF film.
You could even omit the electro and try 2u2F or 4u7F film alone, but this will cut low bass slightly.
Similarly C107 is a little big but probably OK if you parallel it with a film cap.
You need a resistor in the collector of Q107.
You can improve PSRR by adding a cap across CE of Q101.
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Old 9th July 2007, 04:34 PM   #5
AMV8 is offline AMV8  United Kingdom
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Hi

I have used a similar circuit on a pre amp. As everyone else advises I have used 47ohm for r105 and r 106. But I do not know your line voltages and you will need to experiment. In my case I reduce the value on either r105 or r 106 and add a variable resister but as I do not know what your input stage drives this may not help.

|Don
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Old 9th July 2007, 05:16 PM   #6
anatech is offline anatech  Canada
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Hi Steve,
Have you made any attempt to match those parts yet? If you're still in sim mode, they will be if you specified the same device. In real life you will have to do this.

-Chris
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Old 9th July 2007, 07:17 PM   #7
CBS240 is offline CBS240  United States
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Default Re: Input stage biasing - BJT

Quote:
Originally posted by steve_mak
Hi,

I've been trying to gather information on my way to building my first amp. I have included a schematic for the input stage in this post. I Don't quite understand the function of the Sziklai stage as I have it - I know it seeks to linearise the performance but the values I saw in a design seem to saturate one side of the leg more than the other ...

Should I be balancing the quiescent current in each leg of the diff amp? That is, should the current in R108 match that in Q105?
Hi
Having balanced current in both legs of the diff is important since the current mirror presents an equal current load to each leg. Q104/105 help keep current change low in the diff transistors, Q103/106, so they conduct a more constant current, that is basically 0.6V/R107, and 0.6V/R108. It is essentially a common emitter driving an emitter follower, for both sides of the differential.

Quote:


I've noticed also in the simulation that although R109 and R110 are relatively high (they need to be above 22R to have any real effect in balancing the currents I believe) there is still quite an imbalance and I changed the values of R107 and R108 to correct this imbalance. I'm not sure that this is best practice. I've got a feeling that I should leave all values similar to one another down the legs of the diff amp ... Comments please.
R109 and 110 are degeneration resistors, as is R105 and 106. This is a form of feedback to better match the components. Every BJT has a resistive component from base to emitter known as 're'. Ideally you want to have two transistors in a diff or mirror that have equal re. Guessing for small signal BJT, re is around 20-40 Ohms.? In simulation world, this is easy as copy/paste. But the real world is not so simple. In order to match the re's of the transistors, small emitter resistors are added(in series with re) to decrease the percent difference of the overall (Re + re) in the transistors, effectively making them more equal. Gain is sacrificed for equality in this way to better establish equal currents. There are monolithic type transistor arrays that are better matched for this function, requiring little or no degeneration. For example...http://www.thatcorp.com/datashts/300data.pdf Unfourtunately they usually cost more.

Quote:

The values I have in my circuit come from either some simulation or stuff I've gathered from a variety of sources. The transistor types are real - they have been purchased and will be used. The rails will have the same voltage as the power end but are from a separate transformer and are +-35V.

I'd appreciate any help in this issue as I can't seem to source anything more of great value.

Cheers,
IMO, R107/108 might be a little bit large. For Q103/106, Iq=0.6V/2K2=273uA. I would think you might want to bias around 0.9mA, judging from the H parameters vs Ic graph from the datasheet. http://www.datasheetarchive.com/data...article=339498
Then R107/108 =0.6V/0.9mA=667 Ohms or 680 Ohms as an actual value.

IMHO, adding a cascode to this differential, making it a constant power diff might also improve linearity. Also, IMHO, the collectors of Q104/105 could bypass the degeneration resistors, R105/106 and connect together with the collector of Q102.

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Old 10th July 2007, 03:47 AM   #8
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Thanks guys,

Your replies are much appreciated. I've fiddled with a circuit simulation program overnight and I've learnt a bit more but the program fails after some time or too many circuit changes! (It had a current of 15mA down one leg of the diff amp ....) I'll try again later tonight.

Andrew, thanks for your thoughts. Your estimate of R107/R108 being down to around 1k is a thought I had after my work last night too. I don't know why though! - It just seemed right to have something like 25% of the quiescent flowing through the transistor Q103/Q106. The resistor in the current mirror is missing because I'm a little green and just assembling from bits'n'pices I find... I like the thinking of putting a cap in the current source I've not seen that in my hunting so far. I didn't know that bipolars were so bad either until I found a site this morning poo pooing them badly. I'll look into the film suggestions.

AVM8 - I am thinking the current sharing is best affected by the resistors R105/R106. My simulating tells me that changing these values makes more difference than does changing R109/R110. The higher R105/R106 the better the current sharing and the lower the gain I think. I also found that R109/R110 need to be large enough to bias the darlington that is attached to the 'Vforward' tag in my diagram. In fact, the value of these resistors (R109/R110) are pretty much dictated by the issue of biasing the Darlington I believe.

Anatech - I am hoping to God I don't have to do too much matching! (I know it wont work - the hoping bit ...) I really want to just slap a couple in there and be done!

CBS240 - Your ideas of the bias levels accord with Andrew also. I'm thinking of around 1k2 for R107/R108 but I might go lower. Thanks for the explanation of how it works ... please wait while I process the information!

It looks as though I should expect to cut up my lovely new pcb because I've stuffed up somewhat. Thanks again to you all for your thoughts.

Cheers,
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Old 10th July 2007, 07:22 AM   #9
AndrewT is offline AndrewT  Scotland
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Quote:
am hoping to God I don't have to do too much matching! (I know it wont work - the hoping bit ...) I really want to just slap a couple in there and be done!
Hi,
you should be using matched pairs for
Q103/106,
Q104/105,
Q108/109.
The last two pairs could be matched for Vbe only at their operating current.
The first pair should be matched for Vbe at operating current and a close match for hFE maybe <+5%. It would also help if this input pair could be high gain.

I would also recommend matching the pairs of resistors to better than 0.5%

Keep in mind that your current mirror is trying to sink equal currents into their collectors.
The input pairs need equal currents in their emitters. All dead easy until you pull some current out of the Vforward tapping point. Even a FET VAS will pull some alternating current in charging and discharging it's gate. A BJT for VAS is even worse.
If you remove the mirror and substitute a resistor as the load then you can fine tune the resistor value to get an accurate DC equality of LTP currents.
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Old 10th July 2007, 09:11 AM   #10
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Good morning Andrew! (Getting dark for me ...)

Yes I hear you. I have purchased 2SA970BL and 2SC2240BL from B&D Entrerprises in the US. I think the BL equates to devices with hfe in the upper range for the device. This should narrow the number of devices somewhat but I have resigned myself to sitting down like a good boy and doing the tests.

I have learnt a lot in the last 24 hours and I appreciate it. I'm sitting down now to draw a new schematic.

Cheers,
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