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Custom pre-driver amplifier circuit.
Custom pre-driver amplifier circuit.
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Old 19th June 2018, 03:19 PM   #1
edbarx is offline edbarx  Malta
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Default Custom pre-driver amplifier circuit.

Hi everyone. I am a newbie on this forum. I am posting here to receive some feedback as to how my custom pre-driver circuit can be improved. This circuit works but it suffers from hum although this goes as soon as there is an input signal is present. Connecting a mobile phone as a sound source does not produce a mains hum. However, certain tracks play with significant noise.

I think, the solution lies in making a more complex input stage. I have seen some circuits that even employ positive feedback through a 10 Ohm resistor that is connected to ground and then to both the non-inverting and inverting inputs of the input differential pair but this is done indirectly.
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Old 19th June 2018, 03:32 PM   #2
rayma is offline rayma  United States
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Quote:
Originally Posted by edbarx View Post
I have seen some circuits that even employ positive feedback through a 10 Ohm resistor
that is connected to ground and then to both the non-inverting and inverting inputs of the input
differential pair but this is done indirectly.
The 10R isn't for positive feedback, it just breaks the ground loop that causes the hum.
That's just what you need. It goes from the input RCA ground shell (with no other connections to it)
to the audio circuit common. Avoiding Ground Loop Hum

Last edited by rayma; 19th June 2018 at 03:44 PM.
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Old 19th June 2018, 03:42 PM   #3
JonSnell Electronic is offline JonSnell Electronic  United Kingdom
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Re configure your bias network. Put the preset (that may go open circuit) in the BE side of the transistor. If it fails the bias will go low and not allow the output stage to destroy itself.
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Old 19th June 2018, 04:27 PM   #4
john_ellis is offline john_ellis  United Kingdom
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Your hum might be oscillation. Your input resistor should be 10k (right first time) to match the feedback resistor.
Add a 100pF capacitor across it to keep impedance low at high frequencies.
Possibly include a470 ohm to 1k resistor in series with the input for the same reason, this creates an RC filter with even more HF suppression.
Your 10pF capacitor to suppress oscillation is not large enough without emitter degeneration in the input stages. Either add 100 ohms in series with each input transistor emitter or make the capacitor 100pF for starters.
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Old 19th June 2018, 09:14 PM   #5
R Dijk is offline R Dijk  Norway
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Quote:
Originally Posted by john_ellis View Post
Your hum might be oscillation. Your input resistor should be 10k (right first time) to match the feedback resistor.
Add a 100pF capacitor across it to keep impedance low at high frequencies.
Possibly include a470 ohm to 1k resistor in series with the input for the same reason, this creates an RC filter with even more HF suppression.
Your 10pF capacitor to suppress oscillation is not large enough without emitter degeneration in the input stages. Either add 100 ohms in series with each input transistor emitter or make the capacitor 100pF for starters.
Keep in mind very high gain here: 10K/100 +1 =near 100 ... Current gain in amp stage is aproximately 10. At the same time current gain in vas stage is near 10 to 11. That i consider on the low side.

Last edited by R Dijk; 19th June 2018 at 09:21 PM.
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Old 19th June 2018, 11:17 PM   #6
john_ellis is offline john_ellis  United Kingdom
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The local feedback from the 90 ohm resistor helps to keep the frequency response high in the VAS, so there is still a lot of gain at HF. Needs a bigger compensation capacitor.
Not sure I follow the diagram correctly.
Are the resistors really 90 ohms and 69 ohms in the VAS and CCS - NPV's are 91 and 68.
Is the current source in the input stage really 480 ohms should this be 4.7k?
Why use a 10k across the base-emitter of the VAS?
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Old 20th June 2018, 02:39 PM   #7
edbarx is offline edbarx  Malta
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Excuse me for not explaining how my circuit works. The following is an explanation. The differential pair are supplied current through a chain connected to the negative side of the split power supply. Current first goes through two series connected diodes which produce a pseudo-constant voltage drop of around 07V x 2 = 1.4V. This is then used to directly drive the base of the final stage's constant current source. Moving upwards along the same chain, current first flows through a 480 Ohm resistor and then it branches into a 5.1V Zener diode that has its other end connected to ground and a resistor that supplies current to the differential pair. Since the Zener diode is essentially in parallel with the Base-Emitter junction of the input transistor and the 2.2K resistor, the current in the current source should be (5.1 - 0.7)/2200. The voltage across the input resistor that is connected to ground is in the range of a few micro-amperes, so its voltage drop can be ignored for calculation purposes. The output from the differential pair is developed across a bias diode that provides a guaranteed drop of 0.7 volts and a 1K resistor in series. The next stage is the pre-driver stage. The output from the differential pair is connected directly to a PNP transistor with it an Emitter resistor of 90 Ohms. Since we have a 0.7 volt voltage drop across the Base-Emitter junction and the diode at the output from the differential pair, the voltages that vary with a signal are across resistors, the 1K and 90 Ohm resistors. This means, the current gain is defined solely by these resistors which is around 1000/90. The 10K resistor connected across the Base-Emitter junction is to guarantee that the transistor's capacitance discharges when dV/dt is very large. In such a situation, the transistor is alsmost a switching transistor. Needless to state, I added this resistor after I found it greatly improved performance. Without it, there was distortion that could be noticed. Regarding the 10pF capacitor at the output instead of the usual 47pF - 300pF capacitors, I tried to use the standard range of capacitance but the circuit misbehaved. The output from the PNP transistor is connected to a constant current source with a preset current of around 0.7/69. This arrangement is common, so there is no need to re-explain it here. The dangling middle transistor is to provide an accurate bias for the driver stage.
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Old 20th June 2018, 08:42 PM   #8
R Dijk is offline R Dijk  Norway
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Quote:
Originally Posted by edbarx View Post
The 10K resistor connected across the Base-Emitter junction is to guarantee that the transistor's capacitance discharges when dV/dt is very large. In such a situation, the transistor is alsmost a switching transistor. Needless to state, I added this resistor after I found it greatly improved performance. Without it, there was distortion that could be noticed. Regarding the 10pF capacitor at the output instead of the usual 47pF - 300pF capacitors, I tried to use the standard range of capacitance but the circuit misbehaved. The output from the PNP transistor is connected to a constant current source with a preset current of around 0.7/69. This arrangement is common, so there is no need to re-explain it here. The dangling middle transistor is to provide an accurate bias for the driver stage.
Well if the 1K resistor could not do it togheter with the 90 Ohm in emitter. the 10K adjust bias and affect switching time by less than 10%
But great work Ua741
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Old 21st June 2018, 09:05 AM   #9
edbarx is offline edbarx  Malta
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I did further testing of the pre-amplifier circuit and found at high volumes there is distortion. How can I reduce distortion?
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Old 21st June 2018, 04:36 PM   #10
john_ellis is offline john_ellis  United Kingdom
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I was questioning the value of the 480 ohm resistor. The current flowing in it will be
(18-1.4-5.1)/480=24mA which seems high.
The tail current in the LTP is approx. 2mA. 5mA would be enough for the 5.1V zener diode, totalling 7mA which would only need a 1.5k resistor.
The distortion in such a circuit depends on the differential base-to-base voltage in the LTP.
Your 2.2k resistor is not a long enough "long tail" to keep the distortion low: there will be significant modulation of the tail current due to voltage modulation across this resistor.
You should use a proper CCS.
So my question is what loading have you got on the output to require the signal currents in the LTP to deviate from the quiescent values so much that there is high distortion.
If it is connected to an output stage it would be useful to see how/what.
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