18v and 9v active circuit

[munkonthehill]

so whats the diffence between 18v and 9v active circuits????

About 9v............. Ba Boom



All joking aside, is the extra voltage purely to keep the circuit active as it needs 18v or is it to give the circuit extra power?

Im a lawman myself so have no real or very little knowledge of electronics (except having an NC in electronic engineering, however that was 12years ago!!!!!!!!!!!! god am getting old )

So just checking this bit of info,,,,,the bongo im using just now seems to overload my preamp a bit more than the stingray I had or either of my fenders. This may of course be down to EQ settings, however im sure some BC'ers may have more info for me.

Cheers
Steven

Edited September 9, 2010 by munkonthehill

[Randy_Marsh]

I've always wondered about this actually. I just got the idea that the EQ would be more powerful and have a larger output

[JakeBrownBass]

Pretty sure it's to do with headroom, as Vin can't exceed Vout. So with a 18v preamp, Vin is bigger thus Vout can be larger.

[mart]

[quote name='J.R.Bass' post='950166' date='Sep 9 2010, 03:53 AM']Pretty sure it's to do with headroom, as Vin can't exceed Vout. So with a 18v preamp, Vin is bigger thus Vout can be larger.[/quote]

That's the most obvious reason for using an 18v supply. But a preamp designed for 18v probably produces a hotter signal than a 9v preamp, because by doubling the supply voltage there's ample room for both increasing the headroom and giving a larger signal, and a larger signal will give a better signal-to-noise ratio. The downside, of course, is that you may overload the preamp stage of your amp. That's why you have a volume control on your bass.

[flyfisher]

I'm guessing here, but I can't see it being an "output power" issue - that's not (normally) the purpose of a pre-amplifier.

I've always assumed it makes it easier to provide positive and negative voltage rails to the preamp circuitry, which allows the output signal to sit at 0volts and therefore not require a dc blocking capacitor.

In this configuration, the batteries are connected as -9 , 0 , +9.

There are other ways of achieving +/- voltage rails from a single +V source, but it makes things more complicated and might introduce unwanted noise into the output signal.

Could be wrong though.

[munkonthehill]

[quote name='mart' post='950215' date='Sep 9 2010, 08:30 AM']That's why you have a volume control on your bass.[/quote]

volume control!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

am not with you!

[munkonthehill]

[quote name='flyfisher' post='950278' date='Sep 9 2010, 09:33 AM']I'm guessing here, but I can't see it being an "output power" issue - that's not (normally) the purpose of a pre-amplifier.

I've always assumed it makes it easier to provide positive and negative voltage rails to the preamp circuitry, which allows the output signal to sit at 0volts and therefore not require a dc blocking capacitor.

In this configuration, the batteries are connected as -9 , 0 , +9.

There are other ways of achieving +/- voltage rails from a single +V source, but it makes things more complicated and might introduce unwanted noise into the output signal.

Could be wrong though.[/quote]

yeah I thought it may have something to do with the current that needs pushed around the circuit hence the extra 9v. some sort of Thévenin's theorem or ohmes law, but as I said its been a while since I studied this i cant be sure on the answer

Edited September 9, 2010 by munkonthehill

[munkonthehill]

anyone?????????????????

[Mr. Foxen]

More power available gives you more options when designing the circuit. More headroom and more dynamic range is probably the main thing. I think higher voltages are more efficient, like when power lines are high voltage, but not really sure how that would apply to bass. Oh, lower noise floor/signal to noise ratio.

[stingrayPete1977]

I think all of the above is wrong but not sure why? but I think someone with the right knowledge will be along soon to explain if not I will have to see for myself.The 18v circuits are much more high tech and Im guessing take more drain through resistances etc it could be a simple as the type of chips used and very little to do with output etc remember the passive has nothing so even 9 volts is enough to destroy an amp if it went wrong so I dont think thats the issue. There will be a reason but like I said it might just be because that style of chip requires it,but gives the added bonus of more to play about with.See I told you I didnt know!

[JakeBrownBass]

Well as i said before i'm pretty sure its headroom.

The less a system's power supply has to work at providing power to a certain component the better that component can do it's job.

[stingrayPete1977]

Im not an expert but I think cmos chips only run up to 15v (like my Ray preamp etc) headroom would not be found if you put 18v round it! So like I said they are obviously different chips if they are 18v circuits.

[stingrayPete1977]

This may have something to do with it?
[url="http://en.wikipedia.org/wiki/78xx"]http://en.wikipedia.org/wiki/78xx[/url]

As it would make sense then as you would have a really good 9/12volts or whatever you require for the circuits in the preamp but for each IC you would need as it says a spare 2 volts.
Its sort of simple if you know what I mean, If not then it will make nil sense at all and would take forever to explain and I type with 1 finger!

[lowdown]

Does 18 volt go up to eleven so its that bit louder?




Garry

[Dood]

[quote name='J.R.Bass' post='950166' date='Sep 9 2010, 03:53 AM']Pretty sure it's to do with headroom, as Vin can't exceed Vout. So with a 18v preamp, Vin is bigger thus Vout can be larger.[/quote]


Excellent place to start. I agree.

Now - the thing is, with modern pickups - is that their outputs can be much higher as well. For example, I use active EMG pickups with a very high output. According to the EMG specs can be 1.4 volts peak each. If you then add a few dB's of EQ boost or what have you, then you could be looking at say 2 volts each.

Now - plug a 9v battery in to an onboard preamp - lets say for the sake of arguement that it consumes 2v from the battery as well.

Ok, so here's my example: If you are playing and peaking at your 2v and the preamp is pulling a good 2v on peaks as well, maybe more in those dynamic moments - you can say that 4v is being used up out of your 9v available. I'd say that was a fairly large chunk of consumption!

What happens now if you are using say, a rechargeable battery? - Well - they don't charge to 9v anyway. 8.4 at the most, but less efficient ones, you're actually looking at 7.2v!! - so that's a little more of your available power gone. Or, obviously - the battery is a bit on the flat side? 6v if it hasn't been changed for a while?

So running both the pickups and preamp at 18V now seems a far more sensible thing to do in my mind. (The products you are using must be able to run at 18v mind lol!) because even if the electronics were able to pull 9v at any given moment of performance - you still have another 9 to play with.

In the real world? - Well, I have actually upgraded two systems from 9 to 18V. I noticed two differences. I think the other differences were too subtle to say for sure that the upgrade made a marginal impact. The first was that my top end had a more pristine 'sheen' to it. Especially recorded through quality monitors. I also noticed that when I slapped hard it sounded less choked. - or indeed 'compressed' - Which in turn meant my bass appeared to have more bottom end 'thump' to it. Always a good thing eh?

Finally - I went to look up for consumption and capacity figures for a 9v battery - quite happy to see that my figures were pretty close! - but it's worth noting that some 9v batteries just don't have the capacity that you might hope for. To put it in to perspective - it's like running a mini at 90Mph vs a Jaguar!

I've tried not to be too boring and not go in to too much detail about current etc!

[Dood]

[quote name='lowdown' post='952770' date='Sep 11 2010, 02:51 PM']Does 18 volt go up to eleven so its that bit louder?




Garry[/quote]


lol. Actually no - having an 18v system doesn't actually make your bass any louder. *unless* the preamp is specifically designed to add gain.

[markstuk]

Found the following over on BGRA - may have some perspective...

"Bartolini:

"NS2TMB-183", 18 Volts. Bass

This is Bill Bartolini's latest and most refined design. It is quieter and more transparent than anything else out there (108 dB S/N, THD .003% @ 100Hz) and really sounds good. To quote from the spec sheet "The (bipolar) +9V/ -9V system allows a completely DC coupled signal path. There are no capacitors to diminish in any way the low end punch, clarity and frequency response of the instrument". Bill has used wide bands that overlap slightly to avoid the honky sound of other midrange controls. The 250Hz mid could be useful as a cut option when recording."

Now all we need to know are the benefits of a completely DC coupled signal path :-)

[Dood]

[quote name='stingrayPete1977' post='952732' date='Sep 11 2010, 02:17 PM']Im not an expert but I think cmos chips only run up to 15v (like my Ray preamp etc) headroom would not be found if you put 18v round it! So like I said they are obviously different chips if they are 18v circuits.[/quote]


Bass preamps won't be CMOS. That specification is more likely to be found in control systems, not audio.

[Dood]

[quote name='markstuk' post='952788' date='Sep 11 2010, 03:13 PM']Found the following over on BGRA - may have some perspective...

"Bartolini:

"NS2TMB-183", 18 Volts. Bass

This is Bill Bartolini's latest and most refined design. It is quieter and more transparent than anything else out there (108 dB S/N, THD .003% @ 100Hz) and really sounds good. To quote from the spec sheet "The (bipolar) +9V/ -9V system allows a completely DC coupled signal path. There are no capacitors to diminish in any way the low end punch, clarity and frequency response of the instrument". Bill has used wide bands that overlap slightly to avoid the honky sound of other midrange controls. The 250Hz mid could be useful as a cut option when recording."

Now all we need to know are the benefits of a completely DC coupled signal path :-)[/quote]


Over complicating things now! I'm trying to stay away from being too tech - but the [i]idea[/i] being discussed above is that the more passive components you add to a signal path (such as resistors and capacitors) the greater likelihood that distortion, phase change or noise will be added to the signal passing through them. If you can design a circuit with fewer of those components then *hopefully* you will have a cleaner siganl path. The bipolar design tries to deal with that issue by balancing the powersupply around a reference 0v ground.

[lowdown]

[quote name='dood' post='952783' date='Sep 11 2010, 03:10 PM']I noticed two differences. I think the other differences were too subtle to say for sure that the upgrade made a marginal impact. The first was that my top end had a more pristine 'sheen' to it. Especially recorded through quality monitors. I also noticed that when I slapped hard it sounded less choked. - or indeed 'compressed' - Which in turn meant my bass appeared to have more bottom end 'thump' to it.[/quote]


I have an 18 volt system - and i agree with that.


Garry

[stingrayPete1977]

[quote name='dood' post='952790' date='Sep 11 2010, 03:13 PM']Bass preamps won't be CMOS. That specification is more likely to be found in control systems, not audio.[/quote]

Im not being funny Dood but you are talking tripe mate!
Like I said before I dont know enough to make a proper statement on this but I know enough from your last few replies that nor do you? Im not going to get too involved in this but for a start all 2EQ Stingrays contain a cmos LM4250CN IC chip.Voltage does not work like you say either (I am an electrician!)and a cmos chip will work down to 5V which I as just said ARE found in these circuits and I imagine most.Im going to leave it at that as I dont feel either of us know enough on this one and it will become a silly flawed debate.

[Dood]

[quote name='stingrayPete1977' post='952890' date='Sep 11 2010, 05:03 PM']Im not being funny Dood but you are talking tripe mate!
Like I said before I dont know enough to make a proper statement on this but I know enough from your last few replies that nor do you? Im not going to get too involved in this but for a start all 2EQ Stingrays contain a cmos LM4250CN IC chip.Voltage does not work like you say either (I am an electrician!)and a cmos chip will work down to 5V which I as just said ARE found in these circuits and I imagine most.Im going to leave it at that as I dont feel either of us know enough on this one and it will become a silly flawed debate.[/quote]


Well, I'll certainly concede over the CMOS part - that's fine - but as I said in my post, I was endeavouring to not be too techy as not to get in to too much detail. However to suggest I don't know what I am talking about is a wee bit silly as you don't really know anything about my background. I certainly haven't done the same to you, have I? I'm just trying to offer a [b]simplistic response[/b] to the OP.

[stingrayPete1977]

Sorry I do appologise that did come out a bit arsey! But your background to me is not important if you are telling with some authority all about this stuff but do not think a preamp contains a cmos chip? Its not a life and death matter really is it but someone who knows there onions would be able to wrap this up.As I said I know a little and a little knowledge is dangerous so Im staying out of it so you can carry on as you were,Ta.

[Dood]

[quote name='stingrayPete1977' post='952985' date='Sep 11 2010, 06:31 PM']Sorry I do appologise that did come out a bit arsey! But your background to me is not important if you are telling with some authority all about this stuff but do not think a preamp contains a cmos chip? Its not a life and death matter really is it but someone who knows there onions would be able to wrap this up.As I said I know a little and a little knowledge is dangerous so Im staying out of it so you can carry on as you were,Ta.[/quote]

Yup, you certainly did! I forgive you though.

CMOS aside, I felt that my examples of how raising the supply rail voltage can affect headroom and some real world examples were satisfactory to describe at least two benefits. Again, I expected to remain in layman's terms and without disecting the other posts too much did not feel it necessary to display any electronic prowess in my post.

ok CMOS - yes, it IS possible to get terms confused - I'm certainly no expert, but then - never assert that anyway lol lol. However, to my defence - an audio chip does not have to be CMOS. - for the sake of moving on - I'll leave that discussion to the rest of the internet as it's not really relevant to the 9/18v OP.

[flyfisher]

[quote name='markstuk' post='952788' date='Sep 11 2010, 03:13 PM']Found the following over on BGRA - may have some perspective...

"Bartolini:

"NS2TMB-183", 18 Volts. Bass

This is Bill Bartolini's latest and most refined design. It is quieter and more transparent than anything else out there (108 dB S/N, THD .003% @ 100Hz) and really sounds good. [b]To quote from the spec sheet "The (bipolar) +9V/ -9V system allows a completely DC coupled signal path. There are no capacitors to diminish in any way the low end punch, clarity and frequency response of the instrument". [/b]Bill has used wide bands that overlap slightly to avoid the honky sound of other midrange controls. The 250Hz mid could be useful as a cut option when recording."

Now all we need to know are the benefits of a completely DC coupled signal path :-)[/quote]

[quote name='flyfisher' post='950278' date='Sep 9 2010, 09:33 AM']I've always assumed it makes it easier to provide positive and negative voltage rails to the preamp circuitry, which allows the output signal to sit at 0volts and therefore not require a dc blocking capacitor.

In this configuration, the batteries are connected as -9 , 0 , +9.[/quote]

I rest my case.


Think of it this way. The waveform from the pickup is an AC signal and it will alternate around zero volts (ground). An amplifier - either a bass power amp or a guitar pre-amp - must therefore amplify both the +V and -V portions of the waveform. It is not possible to amplify the -V portion of the waveform if there is no negative voltage rail to supply the amplified negative signal voltage. With an 18V system, the two 9volt batteries are connected together to give 18 volts across the voltage rails BUT the connection between the batteries will be connected to the ground of the power amp, which means that the pre-amp voltage rails are actually +9V and -9V with respect to ground. This configuration allows the AC pickup waveform to be correctly amplified on both its positive and negative cycles. Such a design allows the pre-amp to be DC-coupled to the power amp and eliminates the requirement for a DC-blocking capacitor, which is a good thing as far as frequency response is concerned (for the reasons described in the above quote).

With a single 9V battery as a power supply, the pre-amp output can only swing between zero and +9v, so how can the -V portion of the pickup waveform be amplified? The answer is to apply a DC bias voltage to the input of the pre-amp so that the AC pickup waveform is centred around 4.5V instead of zero volts. The pre-amp can now amplify the AC waveform and the output will be a larger AC signal, but still centred around 4.5 volts.

The bass amp input will be expecting an AC signal centred around zero volts (ground) so, to remove the 4.5V DC offset of the pre-amp output, the pre-amp is connected to the bass amp via a capacitor, which will only allow the AC signal to pass through it and will block the 4.5V DC voltage offset (hence the term 'blocking capacitor').

Unfortunately, this capacitor will have some effect on the AC waveform, as mentioned in the above quote, which may not be desirable. Hence the advantage of an 18V system, more accurately referred to as a -9V, zero, +9V system, which allows direct DC-coupling of the pre-amp output to the bass amp input.

Apologies for the ramble but although simple in principle it's not easy to describe purely in words.