Ok, I removed C9 as well to see the difference and after some testing I can hear that there IS a difference with and without C9.
Removing it dropped the gain so low that I had to make drain R9 and R11 1.2k each to bring the gain to normal level. Bias trimmer is at about 10k value. This way I think is sounds the cleanest as it can be and the last bit of "compression like" sound is gone. I wasn't sure if there was a compression before removing the cap but now I'm sure the sound is better without it when I compared the two. So that's good.
But now the current consumption is 2.23mA. I guess it's cause of the low R9 and R11 value. So cause of the even cleaner sound without both C7 and C9 I'd like to keep it this way but of course bring the current down and maintain decent output level. With the values now gain is about 1 which is ok. What are the options of doing this? Bias trim doesn't do much with drain resistors being 5k or so.
Removing it dropped the gain so low that I had to make drain R9 and R11 1.2k each to bring the gain to normal level. Bias trimmer is at about 10k value. This way I think is sounds the cleanest as it can be and the last bit of "compression like" sound is gone. I wasn't sure if there was a compression before removing the cap but now I'm sure the sound is better without it when I compared the two. So that's good.
But now the current consumption is 2.23mA. I guess it's cause of the low R9 and R11 value. So cause of the even cleaner sound without both C7 and C9 I'd like to keep it this way but of course bring the current down and maintain decent output level. With the values now gain is about 1 which is ok. What are the options of doing this? Bias trim doesn't do much with drain resistors being 5k or so.
Just decided to experiment with what I know and I realized it's not the capacitor as a capacitor itself making this going crazy, just the overall "resistance" on the drain. So keeping both a capacitor and a resistor (C9 and R11) can work if the R is between 10k and 20k.
So adjusting the bias, R9 without a cap and R11 with the cap can give very clean output with current draw between 0.5 and 0.8mA which is not bad at all I think. A lot of fiddling around to get the right settings but can be done this way without having the current sky rocket whilst keeping the clean sound. It looks like it's everything into hitting the sweet spot.
I guess this will be one solution that works nice, according to me a really good outcome. I'd like to hear your thoughts on this solution or is there something better..? Thanks
So adjusting the bias, R9 without a cap and R11 with the cap can give very clean output with current draw between 0.5 and 0.8mA which is not bad at all I think. A lot of fiddling around to get the right settings but can be done this way without having the current sky rocket whilst keeping the clean sound. It looks like it's everything into hitting the sweet spot.
I guess this will be one solution that works nice, according to me a really good outcome. I'd like to hear your thoughts on this solution or is there something better..? Thanks
Put the Drain resistors back to 10K. Make gain adjustments at the Source resistor. If you had approx the desired output level with just one of the 50uF Source bypass caps fitted, that gives us a clue to some new values. Without getting bogged down with the arithmetic (other members are far better than I for that), I would suggest trying 1k to 1k5 for the first stage; maybe 1k8 to 3k3 for the 2nd. Then it may require some fiddling to get the bias to where the clipping is symmetrical. You could use the same value for both, but that would have a higher noise level.
You've provided a flash of insight for me: I've never been in the personal space of an Aguilar bass (touched, played, or taken apart), but if I had any money I'd bet they're winding very low output pickups, with very low parasitic capacitance. That would explain needing all that gain. Aren't they also known for their ability to play extremely bright? Low parasitic capacitance!
Cheers
You've provided a flash of insight for me: I've never been in the personal space of an Aguilar bass (touched, played, or taken apart), but if I had any money I'd bet they're winding very low output pickups, with very low parasitic capacitance. That would explain needing all that gain. Aren't they also known for their ability to play extremely bright? Low parasitic capacitance!
Cheers
OK, i'll try this. I'll put 10k trimmers on source with 10k on drain and see how's gonna work.
Two questions though:
1. What is symmetrical clipping and how do I know I do this right?, and
2. What would the difference be when I try this and get the same sound as now, asking cause want to learn some stuff here and why is this a better solution if it would sound the same [emoji4]
I wouldn't know about Aguilar pickups, but the Aguilar preamps go especially well with the LP's which are indeed very low output pickups and sound outstanding. Now I got a Delano HE MM which is very close to them with somewhat higher output I guess, can't really tell as the equipment I'm using now is all different to the one I used beforeYou've provided a flash of insight for me: I've never been in the personal space of an Aguilar bass (touched, played, or taken apart), but if I had any money I'd bet they're winding very low output pickups, with very low parasitic capacitance. That would explain needing all that gain. Aren't they also known for their ability to play extremely bright? Low parasitic capacitance!
Let's set aside the possibility of asymmetric clipping for two reasons: 1) You have 18V supply, and 2) You won't know, without a 'scope.
2. It won't sound exactly the same .. it should sound better. It will be more linear and use less power.
What you had with both 50uF Source bypass capacitors fitted, is what the RF design guys call Maximum Available Gain, and it has several drawbacks including poor linearity, and higher sensitivity to device characteristics. Removing them provides Source Degeneration, which greatly improves linearity.
In this case, with 5k Source resistors and 10k loading the Drain, each stage is only capable of 6 dB(V) of gain. If it turns out that you only need 12 dB total, that's OK. But the 5k Source resistors eat up a lot of dynamic range -- especially for a battery powered device. It may be entirely suitable for FETs of that gain range; but probably not for the ones you have.
Still need some DC voltages. It's no guarantee of freedom from clipping, but verifying that the quiescent Drain voltages are in the range of 40 - 60% of B+ (or mid-point between Source and B+, if the former is significantly positive). Those of us spectators out here still have no idea of the unique properties of your particular transistors.
Cheers
2. It won't sound exactly the same .. it should sound better. It will be more linear and use less power.
What you had with both 50uF Source bypass capacitors fitted, is what the RF design guys call Maximum Available Gain, and it has several drawbacks including poor linearity, and higher sensitivity to device characteristics. Removing them provides Source Degeneration, which greatly improves linearity.
In this case, with 5k Source resistors and 10k loading the Drain, each stage is only capable of 6 dB(V) of gain. If it turns out that you only need 12 dB total, that's OK. But the 5k Source resistors eat up a lot of dynamic range -- especially for a battery powered device. It may be entirely suitable for FETs of that gain range; but probably not for the ones you have.
Still need some DC voltages. It's no guarantee of freedom from clipping, but verifying that the quiescent Drain voltages are in the range of 40 - 60% of B+ (or mid-point between Source and B+, if the former is significantly positive). Those of us spectators out here still have no idea of the unique properties of your particular transistors.
Cheers
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Welcome to the world of designing
50% idea drawn on the back of a napkin, plus 50% theoretical design with datasheets, Math, calculator, simulation, plus 50% playing, testing, tweaking, repeating.
But ... but ... that does not add up!!!
Who says it does?
... but that´s the path.You need all 3 legs for the table to hold stable.
Uuuuhhh, thankls!! Don´t overdo it!
Ever been between a rock and a hard place?
I always needed to measure something for which I do not have the needed instrument (whatever: magnetic flux, temperature, match transistors or tubes, SPL, you-name-it) and had to kludge something.
Always some minimalistic kludge saves my bacon.
In this case I had to improvise a Scope when on Tour with some large (Argentine) band and had to work on a Hotel room table using whatever was available at the corner shop.
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This could work, but I would need to skip adding additional components as I got the PCB made already. So I can either swap some components and values or take out, adding would be a pain :/
But thanks for the thought, I'm sure it will bring some result
I know liner and logarithmic, is this linearity you're talking about meant in the same way, like for the sound or what? How does that work?
OK, this is what I have now in the circuit:
Bias - 10k
R5 and R7 - 10k
R9 - 4k7 (C7 out)
R11 - 10k (C9 in)
With this I'm getting pretty clean sound with probably 18dB boost, it's definitely more than 12, and 0.59mA current draw. I compared this circuit with the opamp one that I made (from the other thread) and Ilto my ear they sound same, can't hear any difference in terms of sound quality, clipping or anything. They sound pretty much identical just with their own EQ curves. In this regard the opamp one is a little more transparent as it uses Baxandall tone control.
Still need some DC voltages. It's no guarantee of freedom from clipping, but verifying that the quiescent Drain voltages are in the range of 40 - 60% of B+ (or mid-point between Source and B+, if the former is significantly positive). Those of us spectators out here still have no idea of the unique properties of your particular transistors.
Oh I don't know the "unique" properties on the transistors either... B+ to source (pin 2) is 16.7V on both transistors, B+ to gate (pin 3) is 8.5V on one and 11V on the other. Am I measuring this correctly? Gate to B+ seems to be around half point of power supply instead of the source... Not sure if this is good.
So yeah I'm not sure where I'm at I just know it sounds pretty clean and more or less the same compared to the opamp one unless I'm missing something, I can record an audio sample if needed...
Voltages are generally taken to be 'to ground', so for future measurements just clip your black probe on it and leave it there. I'm gonna need a little more time to do the arithmetic on your current measurements and study how things are. In the meantime:
- The 'gain structure' is backwards. I'm not keen on the term, but a subset of it applies here -- earlier stage(s) should have more gain. If you're determined to keep one of the Source bypass caps (C7, C9), make it the first one (as the signal travels). Always put greater gain as early as possible; otherwise later stages amplify the noise of the early stages. Frankly, I'd like to see you try it without either, but that won't work if you keep the 4k7 and 10k Source resistors. If I can find an LTSpice model for your 2N5457's, I can simulate the present gain, then choose Source resistors to get pretty close to that same gain. Gimme another day or two if you can.
- You won't be able to measure voltages at the Gate pins with that meter, but we can just ply some arithmetic on the divider. In the future just measure it at the divider, the other side of the 1 meg's.
- Since you're willing to double your battery expenses for added 'headroom'(*1), and you've mentioned it several times, it seems like it matters to you. These high-value Source resistors are costing valuable headroom.
On a general note, there must be plenty of places around the web to find better narrative and diagrams than I can cobble together. But as a start concerning the Drain and Source resistor values: Since in normal operation, the current MUST be the same through each, as you increase the Source resistance to approach that of the resistor providing the Drain load, the gain approaches unity. At that point more than half the headroom is being expressed anti-phase the Source; even if the transistor was a simple switch, it couldn't be more on than 0 ohms, or more off than infinity. If you decrease the Source resistor to say, 1k8, then you can have (1k8 / (1k8 + 10k)) * B+ (if it was a switch) for swing at the Drain -- a huge increase in dynamic range (as long as that's where you're getting the signal from), and gain.
- I'd still like to see some Drain voltage measurements -- from ground. In this circuit configuration that voltage says more about headroom than just about anything else.
*1) Unless you need it for output swing, there's really nothing to be gained by an 18V power supply. You would need to be driving a line-level input to make use of the ~5,5V RMS signal it could deliver. There is no instrument amp input that wouldn't be swamped (hard clipped).
Cheers
- The 'gain structure' is backwards. I'm not keen on the term, but a subset of it applies here -- earlier stage(s) should have more gain. If you're determined to keep one of the Source bypass caps (C7, C9), make it the first one (as the signal travels). Always put greater gain as early as possible; otherwise later stages amplify the noise of the early stages. Frankly, I'd like to see you try it without either, but that won't work if you keep the 4k7 and 10k Source resistors. If I can find an LTSpice model for your 2N5457's, I can simulate the present gain, then choose Source resistors to get pretty close to that same gain. Gimme another day or two if you can.
- You won't be able to measure voltages at the Gate pins with that meter, but we can just ply some arithmetic on the divider. In the future just measure it at the divider, the other side of the 1 meg's.
- Since you're willing to double your battery expenses for added 'headroom'(*1), and you've mentioned it several times, it seems like it matters to you.
These high-value Source resistors are costing valuable headroom.On a general note, there must be plenty of places around the web to find better narrative and diagrams than I can cobble together. But as a start concerning the Drain and Source resistor values: Since in normal operation, the current MUST be the same through each, as you increase the Source resistance to approach that of the resistor providing the Drain load, the gain approaches unity. At that point more than half the headroom is being expressed anti-phase the Source; even if the transistor was a simple switch, it couldn't be more on than 0 ohms, or more off than infinity. If you decrease the Source resistor to say, 1k8, then you can have (1k8 / (1k8 + 10k)) * B+ (if it was a switch) for swing at the Drain -- a huge increase in dynamic range (as long as that's where you're getting the signal from), and gain.
- I'd still like to see some Drain voltage measurements -- from ground. In this circuit configuration that voltage says more about headroom than just about anything else.
*1) Unless you need it for output swing, there's really nothing to be gained by an 18V power supply. You would need to be driving a line-level input to make use of the ~5,5V RMS signal it could deliver. There is no instrument amp input that wouldn't be swamped (hard clipped).
Cheers
Darn it -- I messed up that formula in line 16 of post #38. Bet there's somebody out there that can fix it quicker than me. This might be it:
B+ * ((10k / (10k + 1k8))
Ooops, still forgot to address the linearity issue.
Unbypassed resistance in the Source leg improves linearity, but reduces gain. It allows the Source to attempt to follow the signal, giving local negative feedback.
Best Regards
B+ * ((10k / (10k + 1k8))
Ooops, still forgot to address the linearity issue.
Unbypassed resistance in the Source leg improves linearity, but reduces gain. It allows the Source to attempt to follow the signal, giving local negative feedback.
Best Regards
That's ok Rick no probs
Ok, voltages towards Ground are:
1st stage: Source 1.66V, Drain 16.44V
2nd stage: Source 1.62V, Drain 15.08V
Voltage at divider (the other side of 1Meg's): 1.48V
Ok got it, gain always in the earliest stage possible. I've tried without both caps and adjusting the Source resistors close to the values you are mentioning, i'm getting the needed gain good output but current draw jumps from 0.5-0.6mA to above 2mA, that's why i've put one cap back in. (not sure that i could hear any difference though without the caps which actually should be more linear and with one cap in).
No problem, if you're thinking that taking that extra step will help i don't know what to say except that i'm very thankful for your time and will you put into this Rick! But feel free to skip that if it's gonna take a lot of time.
Measured this and it's up with the other measured V.
Doing this for 2 reasons:- Since you're willing to double your battery expenses for added 'headroom'(*1), and you've mentioned it several times, it seems like it matters to you.
1. I'm about to use both preamp in one bass so it's easier to switch from same power supply rather having one battery drain unevenly on one preamp, while the other will work longer on 18V.
2. This system is implemented by Aguilar in the same preamp, which i have, and using 9V vs 18V on the same does give a lot more headroom... so i don't know, that's one of my guides for the 18V system, but yeah i'm ok with it, you replace the batts once maybe even less in a year so it's not that bad.
Ok, I think i know where you want to get things with this, but a lot of it i don't understand properly... :/
Is that dynamic range difference you're reffering to audible?
What is output swing if i may ask? Sorry Rick, i feel like i'm overwhelming you with questions, but i don't know some of the things you're refering to
I know something about the Voltage level at the output though
Maybe there isn't anything to gian from 18V, i don't know, just following Aguilar's thing here where 9V and 18 do make a difference, unless their system is flawed too but works nicely audibly....... The other reason is that as i mentioned above, i'll supply both preamp from one power supply and makes it easier to use same V on both.... but yeah at the end i wouldn't know the "actual benefits" from 18V for this circuit.
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