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.
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.
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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.
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.
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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Alain.
I did use Vbe multiplier in my design. I also added emitter resistors to output transistors.
houldn't they be the same?
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.
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.
If you used an ammeter to take those measurements, then pick it up, take it outside and throw it as far as you can manage.
Measure the voltage across the resistors.
Keep your suggestion of 820r for R6. and keep the output emitter resistors.
Add emitter resistors to Q1 & Q2. Try 100r to 150r. I would also add 10r to Q3 emitter.
Replace R3 with a 20k VR.
Use VR3 to balance the emitter currents of the LTP.
These emitter resistors will reduce the open loop gain. Since the higher LTP current will increase the open loop gain, we need to bring back some stability.
But, then we don't have a Sinclair any more.
Add emitter resistors to Q1 & Q2. Try 100r to 150r. I would also add 10r to Q3 emitter.
Replace R3 with a 20k VR.
Use VR3 to balance the emitter currents of the LTP.
These emitter resistors will reduce the open loop gain. Since the higher LTP current will increase the open loop gain, we need to bring back some stability.
But, then we don't have a Sinclair any more.
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