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Old 30th December 2011, 06:50 AM   #1
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Default DC offset adjustment add-on

Hi to all.
I have built a discrete transistor amp. Would like to add a dc offset trimpot. The attached circuit is almost similar to mine. At least the input ltp, Vas is similar. I have already matched the LTP transistors. Still an offset voltage of 0,5V.
Any ideas
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Old 30th December 2011, 07:55 AM   #2
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Well 0.5V offset is too high and this is sign for potential problem in your amp.
With this kind of value trim pot wont help you but the easiest way to add trimer is to replase the 2k2 resistor on LTP with 1.5k and connect in series with it 1k trimmer. This is the easiest but not the best way. But first check for problems like oscillation or faulty device.

Last edited by astankov; 30th December 2011 at 07:58 AM.
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Old 30th December 2011, 08:02 AM   #3
Mooly is online now Mooly  United Kingdom
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A circuit that is "almost similar"... we need to see exactly what yours is really.

Its a very basic circuit (as posted) but capable of being refined.
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Old 30th December 2011, 09:02 AM   #4
alayn91 is offline alayn91  France
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Hello,

How did you match LTP: Vbe or Hfe ??

It should be Hfe.

Regards.
Alain.
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Old 30th December 2011, 09:26 AM   #5
alayn91 is offline alayn91  France
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hello,

Another thing;
I(Vas) is about 6mA.
So, V(R5) is about 1.2V.
I'm afraid it's not enough to bias Q6, Q8 and Q7.

Why, don't you use a Vbe multiplier ?

And, how do measure Ibias in Q8 & Q9 ?

Regards.
Alain.
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Old 30th December 2011, 11:28 AM   #6
AndrewT is online now AndrewT  Scotland
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Quote:
Originally Posted by alayn91 View Post
How did you match LTP: Vbe or Hfe ??

It should be Hfe.
No. I disagree.
For good side to side matching of amplification, the two halves of the LTP must be matched for Vbe @ the current they will operate in the LTP.
Once you have found a few that are very close for Vbe, then measure hFE and try to get a pair that are within 10% hFE.
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Old 30th December 2011, 11:44 AM   #7
AndrewT is online now AndrewT  Scotland
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Diy,
measure the current through R3 and through R6.
The currents should be exactly in the ratio 2:1
The "1" value of current is the test value for the selection of your Vbe and hFE.

I know that if the VAS is operating correctly that the VAS Vbe is between 550mV and 700mV.
I'll guess at 600mV for this example.
The voltage across R6 2k2 is 600mV.
The current is 600/2200 = 0.2727mA
That is acceptable but unusually low. I'd prefer to see a lower value adopted for R6.

I would expect the current through R3 to be 2 * 0.272727 = 0.5454mA
The voltage drop across R3 must be 18 * 0.5454 = 9.818V
For this particular set up.
If Vr6 = 600mV then Vr3 must be 9.818 V. If not then the LTP is not balanced.

Ah!
There is current being injected from the VAS to the Q1 collector. I will assume it is much smaller than the R6 current. But that is a bad assumption.

Now to the voltage at the power supply rails.
Vr2 is ~100mV
Vbe of Q1 ~ 600mV as is Vbe of Q4 and Q5.
The total voltage drop from the signal ground at the bottom of R2 is approximately 100+600+9818+600+600 = 11.7V
The schematic shows -20V That's an error of ~8V and it's not because of that bad assumption made earlier.
There are some poor component value choices in the schematic.

And selecting two perfectly matched hFE devices will not help balance the LTP, not one iota.
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Last edited by AndrewT; 30th December 2011 at 11:47 AM.
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Old 30th December 2011, 01:09 PM   #8
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Quote:
Originally Posted by alayn91 View Post
hello,

Another thing;
I(Vas) is about 6mA.
So, V(R5) is about 1.2V.
I'm afraid it's not enough to bias Q6, Q8 and Q7.

Why, don't you use a Vbe multiplier ?

And, how do measure Ibias in Q8 & Q9 ?

Regards.
Alain.
Alain.
I did use Vbe multiplier in my design. I also added emitter resistors to output transistors.
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Old 30th December 2011, 02:03 PM   #9
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Ok so ive measured the current in the two collectors. I Q1=5,6ma, and I Q2= 16ma. Shouldn't they be the same?

Last edited by diy didi; 30th December 2011 at 02:15 PM.
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Old 30th December 2011, 03:57 PM   #10
alayn91 is offline alayn91  France
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Hello,

It's not possible !!!

The sum of IcQ1) & Ic(Q2) is about 20V - 1.2V(Vbe(Q5) + Vbe(Q6)) divided by R3, so 1mA.

It's sure The LTP is not balanced, because you have 0.5V on the base of Q2.
So, Q2 is almost saturated and Q1 is almost off.

As AndrewT said, you can try to decrease R6, for example 820R.

Regards.
Alain.
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