I know there are a host of descriptions of ways to lower the 405 input sensitivity. They are not all alike, some refer to 405-II's, and some are in the context of other modifications, and others are just different than each other. I have an ISS-9 405, I'm not doing anything other than I've re-capped, replaced transistors, and would like to lower the sensitivity without changing anything else. Some say use a 47k resistor and some say 4.7k. One uses multiple additional components, etc.
Is there a for sure way to do it? Is just the replacement of the C2 with 33uF and the replacement of R4 with 47K the correct solution?
Thanks😉 I have this amp working and don't want to see any more flames🙂
Is there a for sure way to do it? Is just the replacement of the C2 with 33uF and the replacement of R4 with 47K the correct solution?
Thanks😉 I have this amp working and don't want to see any more flames🙂
No. You need to both lower R6 and raise C4 in the desired ratio, and do the same to either R5 or C2, not both, this time lowering either the resistor or the capacitor (not raising it).
It makes no difference what version you have.
You can accomplish it either by replacing components or by adding the appropriate values in parallel, typically on the other side, to achieve the above.
There is no solution that can use 47K, unless you want some crazy sensitivity like 3.5V. Where did you read that?
It makes no difference what version you have.
You can accomplish it either by replacing components or by adding the appropriate values in parallel, typically on the other side, to achieve the above.
There is no solution that can use 47K, unless you want some crazy sensitivity like 3.5V. Where did you read that?
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You could simply split R1 into a potential divider in the ratio that you want to attenuate. No messing with the gain of IC1 or changing the input impedance. Personally I would replace R1 with a 250K potentiometer if there is a tidy way to mount it on the rear panel.
The least invasive solution is to put a series connection of a resistor and a capacitor (per channel) in the connecting cable and to leave the amplifier itself as is.
You probably have a special cable to drive the unusual DIN connector at the input. When you want to reduce the sensitivity by a factor x, for each channel, put a resistor of x - 1 times 22 kohm in series with an MKT capacitor of 680 nF/(x - 1) in the cable and you are done.
When you do the maths, you find that with this method, resistor R1 affects the low-frequency response, but all it does is cause somewhat faster roll-off below 1 Hz.
You probably have a special cable to drive the unusual DIN connector at the input. When you want to reduce the sensitivity by a factor x, for each channel, put a resistor of x - 1 times 22 kohm in series with an MKT capacitor of 680 nF/(x - 1) in the cable and you are done.
When you do the maths, you find that with this method, resistor R1 affects the low-frequency response, but all it does is cause somewhat faster roll-off below 1 Hz.
The 405 ISS9 has a midband gain of 35db (so 500mv rms input for 100 watt into 8 ohm).
What new sensitivity do you want?
1.5V seems to be the most common choice and makes sense I would imagine for a preamp that goes to 2V. I haven't found a preamp yet for this so I'll probably roll with that as a good place at least to start.
"The quick answer to your question is to replace R4 (1k6) to 4k7 and C2 (100uF) to 33uF. "
"Replace R1 220k in both channels with a 100k and 47k in series, 47K end to common/ ground and C1 to the joint between them, and top of pair to amp input terminal. If lower 'Z' input is preferable, use 22k for the smaller value and 47k for the larger."
"R6: 100K, R4: 6K8, C4: 150nF"
Some of the other recipes I've come across online.
"Replace R1 220k in both channels with a 100k and 47k in series, 47K end to common/ ground and C1 to the joint between them, and top of pair to amp input terminal. If lower 'Z' input is preferable, use 22k for the smaller value and 47k for the larger."
"R6: 100K, R4: 6K8, C4: 150nF"
Some of the other recipes I've come across online.
Well the first one is nonsense.
1. R4 is not 1K6 but 22K. There are no 1K6 resistors in a Quad 405. They were not readily available when it was designed.
2. 4K7 only makes sense if you want a sensitivity of > 2V.
3. You don't replace both R4 and C2. Otherwise you foul up the DC servo time constants well and truly, compromising bass performance.
4. Replacing R4 or C2 alone does nothing to the audio sensitivity, which is controlled by R6 and C4. It only adjusts the DC servo, which is also necessary but not sufficient.
The second one is another attenuating solution like the two already posted here. They all have the disadvantage of increased noise, and this one has the further disadvantage of intruding on the amplifier.
Your final recipe in quotes agrees with what I posted above for a ratio of 1/3.
1. R4 is not 1K6 but 22K. There are no 1K6 resistors in a Quad 405. They were not readily available when it was designed.
2. 4K7 only makes sense if you want a sensitivity of > 2V.
3. You don't replace both R4 and C2. Otherwise you foul up the DC servo time constants well and truly, compromising bass performance.
4. Replacing R4 or C2 alone does nothing to the audio sensitivity, which is controlled by R6 and C4. It only adjusts the DC servo, which is also necessary but not sufficient.
The second one is another attenuating solution like the two already posted here. They all have the disadvantage of increased noise, and this one has the further disadvantage of intruding on the amplifier.
Your final recipe in quotes agrees with what I posted above for a ratio of 1/3.
So for a sensitivity of 1.5 V, you would need 44 kohm and 340 nF between the signal source and the amplifier. Rounded to more practical values, either 47 kohm and 330 nF (preferably MKT), or 39 kohm and 390 nF (preferably MKT).
I neglected the current division between R1 and R3 in my earlier post. Accounting for that, the combination 39 kohm and 390 nF quite accurately reduces the sensitivity by three, while the other combination reduces it a bit more.
This approach leads to reduced noise at the amplifier output when you compare it to a QUAD-405 driven from a low-impedance source. The output noise is not reduced as much as it can be when you change the amplifier itself, but it is reduced.
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Looks good 👍
This is the original vs modified (47k and 330nF). One thing I would seriously look at is replacing that LM301 opamp which really is a bit grim by todays standards. The 3.3pF between pins 1 and 8 would need removing if you used something like a TL071 which Quad started using in later production.
In a nutshell: the equivalent input noise increases by a smaller factor than the gain decreases, hence the noise at the output decreases.
I will be happy to write a more extensive explanation with schematics and noise transformations, but I haven't got the time to do that right now.
I will be happy to write a more extensive explanation with schematics and noise transformations, but I haven't got the time to do that right now.
Given the fact that soldering components in a DIN or RCA cable is not that easy, I prever changing the components on the PCB. I think the noise reduction will be higher in that case, instead of the case of increasing the input resistance. I don't think the noise explanation is that difficult in the case of adding series resistance. A simplified explanation, only focussing on the voltage noise of the resistor: if the series input resistance is doubled, the noise voltage of the resistor is increased by the square root of two, so 1.4 times, the gain is halved, so the total noise output will be 0.7 times the original noise output. If we change the gain of the op-amp circuit with the same amount by increasing the feedback factor by reducing the feedback resistor R6, the noise output will be halved (0.5) compared with the original noise output. Joost Plugge.
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Regarding the noise, Joost already summed it up quite nicely, but didn't mention that the noise voltage of the op-amp also causes less noise at the output due to the reduced gain with the higher driving impedance. The following is my explanation with noise transformations, it is a bit more formal but boils down to the same.
First, I hope we can agree that the part to the right of R3 has these three properties:
1. It has a virtual ground input
2. It has a low-impedance output
3. It is not changed in any way by putting a resistor and a capacitor in the input cable
Therefore, everything to the right of R3 can be seen as a current-controlled voltage source with a transimpedance Z21 that just stays as is. It also generates some noise. The noise of a noisy two-port can be represented with an equivalent input noise voltage source and an equivalent input noise current source.
As a result, we end up with part A of this figure for a QUAD-405 driven from a low-impedance source and part B when you use my trick to reduce sensitivity.
To simplify things further, when we are interested in the output noise with no input signal, R3 and everything to the left of it can be replaced with just one impedance Z. This impedance is, of course, different for the two cases. It generates a thermal noise voltage with a density sqrt(4 k T Re(Z)).
As the transimpedance stays as is, the output noise is proportional to the current flowing into the virtual ground. Looking at the three noise sources in part C of the picture:
The noise current inoise just flows completely into the virtual ground, no matter what.
The noise voltage vnoise causes a current to flow into the virtual ground that is inversely proportional to the magnitude of Z. Hence, the higher the magnitude of the driving impedance Z, the less noise ends up at the loudspeaker due to the equivalent input noise voltage of everything to the right of R3. This basically just means that the noise gain drops. (There is no +1 term due to the place where I defined the noise voltage.)
The noise voltage of the real part of Z increases with the square root of the real part of Z, but the resulting current through the virtual ground is inversely proportional to the magnitude of Z. When Z is real, which it is over most of the audio band, the result is just an inverse proportionality to the square root of Z. Hence, this noise term also drops when the impedance is increased. (To be precise, when the impedance is not real, you can show that the current is proportional to the square root of the real part of 1/Z and therefore still drops with increasing impedance.)
First, I hope we can agree that the part to the right of R3 has these three properties:
1. It has a virtual ground input
2. It has a low-impedance output
3. It is not changed in any way by putting a resistor and a capacitor in the input cable
Therefore, everything to the right of R3 can be seen as a current-controlled voltage source with a transimpedance Z21 that just stays as is. It also generates some noise. The noise of a noisy two-port can be represented with an equivalent input noise voltage source and an equivalent input noise current source.
As a result, we end up with part A of this figure for a QUAD-405 driven from a low-impedance source and part B when you use my trick to reduce sensitivity.
To simplify things further, when we are interested in the output noise with no input signal, R3 and everything to the left of it can be replaced with just one impedance Z. This impedance is, of course, different for the two cases. It generates a thermal noise voltage with a density sqrt(4 k T Re(Z)).
As the transimpedance stays as is, the output noise is proportional to the current flowing into the virtual ground. Looking at the three noise sources in part C of the picture:
The noise current inoise just flows completely into the virtual ground, no matter what.
The noise voltage vnoise causes a current to flow into the virtual ground that is inversely proportional to the magnitude of Z. Hence, the higher the magnitude of the driving impedance Z, the less noise ends up at the loudspeaker due to the equivalent input noise voltage of everything to the right of R3. This basically just means that the noise gain drops. (There is no +1 term due to the place where I defined the noise voltage.)
The noise voltage of the real part of Z increases with the square root of the real part of Z, but the resulting current through the virtual ground is inversely proportional to the magnitude of Z. When Z is real, which it is over most of the audio band, the result is just an inverse proportionality to the square root of Z. Hence, this noise term also drops when the impedance is increased. (To be precise, when the impedance is not real, you can show that the current is proportional to the square root of the real part of 1/Z and therefore still drops with increasing impedance.)
So now I have two solutions based on responses here:
1) "R6: 100K, R4: 6K8, C4: 150nF"
2) "So for a sensitivity of 1.5 V, you would need 44 kohm and 340 nF between the signal source and the amplifier. Rounded to more practical values, either 47 kohm and 330 nF (preferably MKT), or 39 kohm and 390 nF" (although not explicit which R and C are being referenced)
Are these both equally good solutions?
1) "R6: 100K, R4: 6K8, C4: 150nF"
2) "So for a sensitivity of 1.5 V, you would need 44 kohm and 340 nF between the signal source and the amplifier. Rounded to more practical values, either 47 kohm and 330 nF (preferably MKT), or 39 kohm and 390 nF" (although not explicit which R and C are being referenced)
Are these both equally good solutions?
1) worsens the noise gain around the op-amp by reducing R4. Reducing R4 also increases the noise current that R4 injects into the virtual ground.
2) refers to extra components to be connected between the preamplifier and the main amplifier, so you could keep the amplifier itself as original as possible if you like. This was explained in the post quoted at the top of post #11.
As you have decided to go for a FET op-amp, I have some more alternatives you might like.
2) refers to extra components to be connected between the preamplifier and the main amplifier, so you could keep the amplifier itself as original as possible if you like. This was explained in the post quoted at the top of post #11.
As you have decided to go for a FET op-amp, I have some more alternatives you might like.
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With a FET op-amp and when you don't want to keep it original, you can increase the impedance of the DC feedback. It reduces the noise gain/increases the feedback around the op-amp, thereby reducing its noise contribution and its distortion, reduces the noise current R4 injects into the virtual ground, and it allows you to use a simple film capacitor for C2 instead of a tantalum electrolytic.
R6: 100 kohm
C4: 150 nF
R4: 47 kohm
R5: 39 kohm
C2: 2.2 uF
C1 and R3 as is (680 nF and 22 kohm)
R6: 100 kohm
C4: 150 nF
R4: 47 kohm
R5: 39 kohm
C2: 2.2 uF
C1 and R3 as is (680 nF and 22 kohm)