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Audio Mixer using transistor

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electronic mind

New Member
hi
this circuit is an audio mixer.... i can beleive that but can someone explain to me how??

i have an idea i hope its right ,,, and if im wrong plz correct me::::

The input node works as a current adder and then the total current is amplified by the transistor
 

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Optikon

New Member
electronic mind said:
hi
this circuit is an audio mixer.... i can beleive that but can someone explain to me how??

i have an idea i hope its right ,,, and if im wrong plz correct me::::

The input node works as a current adder and then the total current is amplified by the transistor
This shows an N-channel JFET. My understanding is this device is not current controlled but voltage controlled Vg-s and is symmetric from D-S.
Since no current flows into the gate, the current flowing in R2 is equal to that of R4. This current is I = (a*Vin1 - b*Vin2)/(R2+R4) where Vin1 and Vin2 are the two inputs respectivly and a & b are scale factors from the input divider network (assuming C coupler is good short).

This current produces a voltage at the gate via the R2 (or R4) drop. This voltage (work the math) will be a scaled sum (or difference) of the two inputs Vin1 & Vin2. Vg = aVin1 - I*R2.
So it does sum the voltages and I suspect that with a properly biased JFET (Doesnt look biased to me) this voltage will be amplified at the output via the source resistor. I'm no expert with mixer circuits, but this seems like a crude circuit.. Someone else on here can likely provide better insight.
 

audioguru

Well-Known Member
Most Helpful Member
R2 and R4 mix the inputs simply because they are connected together. You make a mono (L + R) signal from a stereo source the same, with a resistor from each channel then the resistors are connected together.

The entire circuit works with voltage, not current.

The transistor isn't an amplifier, it is a follower with a small loss.

With the volume controls at maximum, if the same signal appears on both inputs, the output is a little less than the inputs. If only one input has a signal, the output is a little less than half of the input since the resistors are a voltage divider.

The volume controls affect each other, unlike a "real" mixer where the volume controls are completely independent.
 

mastertech

New Member
with this type of input network (fet gate side of circuit) the resistrors not
the pots should be considerably larger than the pots themselves in value
in order to avoid signal loss at the setting of the next pot, as you increase
inputs you need to take in to account this shortcoming of this circuit
arrangement, basicly keep the pots as small as possible (say 10k-20k this will also effect the values of the coupling capacitors when ) and determine the
values oF Rs in relation to the input impendace of the amp(fet) in this
instance in Mohms
 

audioguru

Well-Known Member
Most Helpful Member
The author of this mixer project claims, "As many or as few channels as are required can be added to the mixer. Do this by just duplicating the input "sections" which are clearly shown on the schematic. One version of this mixer I saw had 25 inputs!"
Just think about how low-level its output would be.

On his web-forum, many people are complaining that the output level of this "mixer" that they built is verrrrry low.
 

electronic mind

New Member
Optikon said:
electronic mind said:
hi
this circuit is an audio mixer.... i can beleive that but can someone explain to me how??

i have an idea i hope its right ,,, and if im wrong plz correct me::::

The input node works as a current adder and then the total current is amplified by the transistor
This shows an N-channel JFET. My understanding is this device is not current controlled but voltage controlled Vg-s and is symmetric from D-S.
Since no current flows into the gate, the current flowing in R2 is equal to that of R4. This current is I = (a*Vin1 - b*Vin2)/(R2+R4) where Vin1 and Vin2 are the two inputs respectivly and a & b are scale factors from the input divider network (assuming C coupler is good short).

This current produces a voltage at the gate via the R2 (or R4) drop. This voltage (work the math) will be a scaled sum (or difference) of the two inputs Vin1 & Vin2. Vg = aVin1 - I*R2.
So it does sum the voltages and I suspect that with a properly biased JFET (Doesnt look biased to me) this voltage will be amplified at the output via the source resistor. I'm no expert with mixer circuits, but this seems like a crude circuit.. Someone else on here can likely provide better insight.
so the two voltage inputs r summed and the amplified
 

audioguru

Well-Known Member
Most Helpful Member
electronic mind said:
so the two voltage inputs r summed and the (then?) amplified
No, The JFET in this circuit is a follower and doesn't amplify. The more inputs that this mixer has, the lower its output level.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
audioguru said:
electronic mind said:
so the two voltage inputs r summed and the (then?) amplified
No, The JFET in this circuit is a follower and doesn't amplify. The more inputs that this mixer has, the lower its output level.
Yes, it's a REALLY CRAP 'design' - not recommended at all!. By far the easiest way to make a mixer is a simple opamp, as a virtual earth mixer.
 

audioguru

Well-Known Member
Most Helpful Member
Optikon said:
I suspect that with a properly biased JFET (Doesnt look biased to me).
It has (or might have) self-bias. I was trying to figure out its biasing, but a 2N3819 FET has such a wide range of unbiased current and cutoff voltage that I think many FETs won't even work in this REALLY CRAP 'design'.
 

mastertech

New Member
to make the dc bias independent of the resistors and pots,connect a bias
resistor from gate to ground and use another coupling capacitor at that point, and dont forget that this new bias resistor now becomes the input
impendance of the circuit
 

audioguru

Well-Known Member
Most Helpful Member
Hi Mastertech,
The circuit already has resistors from the gate to ground. Do you mean to bias the gate with a voltage divider?
If you bias the gate at a few volts and end-up with a 2N3819 FET that is close to its cutoff voltage limit of 8V, then this circuit won't work from a 9V supply.
 

mastertech

New Member
not with voltage divider, just from gate to ground, the reason for this
is there's a small amount or reverse current flowing from source to
gate and this develops a positive voltage on gate that might upset
the voltage bias level of the source, since the original circuit has
variable resistances gate to ground this will definately shift the bias at
source with changed input level settings
 

audioguru

Well-Known Member
Most Helpful Member
mastertech said:
not with voltage divider, just from gate to ground, the reason for this is there's a small amount or reverse current flowing from source to gate and this develops a positive voltage on gate that might upset the voltage bias level of the source, since the original circuit has variable resistances gate to ground this will definately shift the bias at
source with changed input level settings
Hi Mastertech,
This is a FET, it doesn't have input current. You are correct, the leakage current is small and is typically only 0.002nA, which would cause a bias voltage shift across two of the mixer's 10K variable resistors of only 0.01mV. Its maximum leakage current is 2nA which would cause a bias voltage shift of only 10mV. The leakage current is rated at 15V and will be much less at the few volts in the mixer project.
 
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