I'm just throwing this out there, I've had a quick google/forum search, but I would appreciate other input.
A mate came round tonight and asked for a harp stompbox. I think his initial thought was an all in one A/B off/on, equalizer, tube distortion box with XLR cables.
I told him he should buy an equalizer. Which got me thinking that for his distortion (and the equalizer) there could be an effects loop so he would be more flexible.
So I haven't got time to reinvent the wheel if it's not necessary, I'm trying to keep costs down for him (I won't be making any money on it), is something like this out there, can he buy something cheap or if I have to make it - have any of my fellow forumites
already spent time on this?
I'm specifically interested in impedance matching, balanced vs unbalanced lines, optimal use of buffers.
It should look something like this:
Harp>>>XLRcable>>>A/B/Y ish stomp>>>XLRcable>>>Mixer
^
^
Effects injected somewhere in signal path
It occurs to me that a blend pot might be useful to tune in the proportion of distortion/clean signal. He would like LED's to indicate what's active at any one time.
Any ideas appreciated.
This is his Mic
(http://www.highlite.nl/var/StorageHighlite/ProduktBilder/medium/D1380.jpg)
and Mic Spec.
Type: Dynamic, moving coil
Freq. response: 40 Hz - 16 kHz (-3 dB)
Polar pattern: cardioid
Sensitivity: 1,2 mV/Pa -58 dB -
Max. SPL: 140 dB
Impedance: 220 Ohm
Power supply: none
Connector: 3p. XLR male type
Net weight: 276 g
Dimensions: 57 mm (diameter) x L 76 mm
Accessories: Wooden box
lo fi echo or something pt2399 based would work well for that . i have a tendancy to want everything " built in " so ... disregaurd my inane ramblings . i have been using karoake circuits , hanna montanna guitar guts , modded out with additional controls and couldnt be happier with the results , especially considering i get em for 2 bucks at goodwill , but ... :icon_mrgreen:
eq , once again the super easy way out may be to go on ebay and get a 3 bad bass guitar eq mudule , they start at 10 bucks , and they are really small , come with everything included ... seems like a dirt circuit that has wide lo end , say something utilized for a bass guitar would be nice , especially if you can adjust the bass . also a clean blend would make that more flexible , yet pretty simple . thing is for these types of apps , you dont need much gain or distortion , it will ruin the tone / feedback . most guys use sure green bullets with blues jr tube amps . maybe find a popular ( with harp guys ) dirt box , clone that ? lo gain , maybe a rotary switch with say 5 input capacitors would do the trick and possibly eliminate the eq ?
the main issue i would forsee is feedback , i am not aware of a diy anti-feedbacker , but that may be needed ... i think something like a hush circuit would work well for feedack control , and there are more expensive options for that out there . along the way you may find yourself trying a di box as well . i know some guys have had success using guitar multi effects for this , utilizing the cab sims ...
i have tried something in this vein , using a dirt box on a mic , i found the digitech bad monkey was the best i could find , it has treble and bass , gain , plus a cabinet out jack that worked very well to my surprise . we ran a cb mic , and a gas mask with a mic element in it . i still couldnt get a good di tone ( going straight to the board) , i didnt have a di / cabsim and wanted good / cheep so i ended up using a loud mini amp , in a styrofoam insulated ( packed with tshirts ) cooler , with an sm57 in there going to the board . the sound was excellent , but it was a hassle , and i digress , but this was a line in / headphones / studio type app ...
cool idea . also it would seem to me that you wouldn't need something very hi fi for this , kind of the opposite would make for ideal tone and help with feedback , but then headroom ... hmmm . standard 9 volt stuff sould be fine i would think .
ot , and i am sure you know this , but i believe old telephone handsets , sawed off to just utilize the mic portion work very well for harp mics .
good luck , keep us posted on the developments , its an interesting project for sure !
Thanks Petey, some good ideas in there. I'll have to have a look at the Cab Sim, because I need one myself for recording so I could just as well build two.
A bass tone stack/simple eq might do the job.
If I get frisky I might throw in a Little Angel chorus, and let him add in reverb from dsp in the mixer or routed through an amp.
It would be more satisfying to build everything in. :D
check it out :
http://www.ebay.com/itm/Active-Preamp-3-Band-Equalizer-EQ-Harness-Guitar-Bass-Tone-Control-System-/190748210737?pt=Guitar_Accessories&hash=item2c697ab631
they have a few different models , if you read the specs , the freq bands are different so ... look for one thats closer to a higher set of freq ?
careful , they dont like rev polarity ... ha . ( gooped in epoxy ) . i was surprised how well this worked ( on a bass mind you ) .
but then they are selling these too :
http://www.ebay.com/itm/Preamp-5-Band-Equalizer-EQ-Pickup-Tuner-LCD-Chromatic-for-Acoustic-Guitar-Bass-/300692246071?pt=Guitar_Accessories&hash=item4602a78e37
Ha ha, great, I'll look into it. Thanks
The only thing I know about electric harp is that the amps are designed to minimize feedback and to be compatible with the mics they use. It seems that you'd need to be careful about the input impedance of the effect and that the output would be compatible with his harp amp.
Feedback seems to be key, true enough.
I'm working on it.
Thanks
This looks like it's going to be the front end .....
so..... I made up a layout for the first time in ages, now I'm wondering about ground/chassis ground, any thoughts.
Can I just ground everything to the trace to pin 1 or do I need a separate chassis ground for the minus side of C3
Schem:http://sound.westhost.com/project122.htm (http://sound.westhost.com/project122.htm)
(http://i713.photobucket.com/albums/ww136/skruffyhound/MicPreamp_zps98f3514b.jpg)
It looks to me like you don't need to change anything, just tie the chassis ground to the XLR pin 1 or anywhere along the trace that is connected to pin 1. You'll also connect your output ground to the same place.
This pinout confirms that pin 1 should be connected to the chassis. This also makes sense because pin one is connected to the mic shield and the chassis is supposed to also act as a shield. Tying the chassis to pin one makes a continuos shield that includes the mic cable and chassis.
(http://i1160.photobucket.com/albums/q485/jdansti/31F2E220-63E1-4EB4-9452-A0B29BF68B95-13175-00000D825B5CC507.jpg)
Thanks for the input John, I'll give it a go as it is.
It took ages because I'm unfamiliar with DIYLC (was using Stripmagic but wouldn't start this time), and it wouldn't save this layout for me.
Looks a bit messy today, but I was nearly finished by the time I realized I could resize the resistors from giants that take 6 holes down to a more normal 4.
It'll go better next time.
The trouble with 3-band EQs is that: a) the resonant frequencies tend to be fixed and not of one's choosing, b) the shaping is generally fairly shallow - you have to accept/forfeit things to get what you want, and c) they assume you generally want the entire spectrum with a dip/peak here or there.
Bullet mics like the one shown are highly prized because they provide the "right" sort of tonal character for a harmonica blown hard, so dips and peaks are likely to be superfluous. If it were me, I'd go with some kind of adjustable-frequency 2-pole highpass filter, so that low-freq breath noises could be managed, and similarly adjustable 2-pole lowpass filter, such that "warmer" tones could be achieved. That would let you make the passband as wide or narrow as you want, but not interfere with, or colour, what's IN that passband. If one was insistent on getting some sort of resonant peak, you could include variable Q on the lowpass that would produce a resonant peak near the corner frequency.
I should update the state of the project now because things have changed after the many replies to all my threads.
I am side-stepping the filtering issues by giving him an effects loop where he can if he wishes place an EQ (or anything else, chorus,reverb,delay) that will be better than I have time to make.
I will include a Subcaster in the box and hopefully everything will run off +/- 12v so there's more clean headroom.
Switching will be on/off so he can just dump the mic in a corner. Bypass everything and bypass Subcaster.
I can't decide whether to have the Subcaster before or after the effects loop to take best advantage of the potential EQ, but it'll probably be before because we could end up with time based effects in that loop.
I want to keep it down to 3 ish knobs, two for the distortion and one volume/attenuation for effects loops. (00ps there's also one on the mic pre)
Any thoughts are very welcome
From this : http://www.geofex.com/article_folders/cheapgoodprot.htm (http://www.geofex.com/article_folders/cheapgoodprot.htm)
(http://i713.photobucket.com/albums/ww136/skruffyhound/Polarityprotection2_zps927bd284.jpg)
Should stop him from burning my work.
Edit Q2 was incorrectly labeled, I've changed pic now it should read EBC
Good idea. I think I'm going to make a handful of those and start plopping them in my builds.
A more compact version
(http://i713.photobucket.com/albums/ww136/skruffyhound/PolarityprotectionV2_zpse1bd6135.jpg)
I'll be verifying this in the next few days hopefully, until then build at your own risk :icon_eek:
I've been looking at the ICL7660S, bit concerned that it won't deliver enough juice
QuoteThe ICL7660 delivers an open-circuit output equal to the negative of the input voltage to within 0.1%. Capable of producing 20mA, the device has a power-conversion efficiency of about 98% for load currents of 2mA to 5mA.
Thus at 18mA load, the output voltage drops about 1V below the input. Beyond around 40mA, the voltage drop becomes very nonlinear, and the circuit self-limits, thereby protecting itself against excessive power dissipation.
That info was hard to find, had to read carefully through several datasheets, as opposed to these guys who are selling the chip on exactly that point (they also had the most retro funk datasheet I've ever seen).
http://www.digchip.com/datasheets/parts/datasheet/262/LT1054CN8-pdf.php (http://www.digchip.com/datasheets/parts/datasheet/262/LT1054CN8-pdf.php)
The LT®1054 is a monolithic, bipolar, switched-capacitor voltage converter and regulator. The LT1054 provides higher output current than previously available converters with significantly lower voltage losses. An adaptive switch driver scheme optimizes efficiency over a wide range of output currents. Total voltage loss at 100mA output current is typically 1.1V. This holds true over the full supply voltage range of 3.5V to 15V. Quiescent current is typically 2.5mA.
The LT1054 also provides regulation, a feature not previ- ously available in switched-capacitor voltage converters. By adding an external resistive divider a regulated output can be obtained. This output will be regulated against changes in both input voltage and output current. The LT1054 can also be shut down by grounding the feedback pin. Supply current in shutdown is less than 100μA.
The internal oscillator of the LT1054 runs at a nominal frequency of 25kHz. The oscillator pin can be used to adjust the switching frequency or to externally synchronize the LT1054.
The LT1054 is pin compatible with previous converters such the LTC1044/LTC7660.
(http://i713.photobucket.com/albums/ww136/skruffyhound/LT1054_zps25c643de.jpg)
For the price of the three extra components I might build the regulated version.
I've got one of these chips playing up in my ADA flanger (may not be the chips' fault) and they are not cheap, but it still sounds like a better idea than a very squishy -12V with the ICL7660. Thoughts?
After searching every single page of the whole internet...there were the values I needed in the datasheet :icon_redface:
NOTES: 5. All regulation specifications are for a device connected as a positive-to-negative converter/regulator with R1 = 20 kΩ, R2 = 102.5 kΩ,
external capacitor CIN= 10 μF (tantalum), external capacitor COUT 100 μF (tantalum) and C1 = 0.002 μF
I couldn't work out the calculation ^^^, didn't really understand where they got their Vref
also mentioned in the datasheet
capacitor selection
While the exact values of CIN and COUT are noncritical, good-quality low-ESR capacitors, such as solid tantalum, are necessary to minimize voltage losses at high currents. For CIN, the effect of the ESR of the capacitor is multiplied by four, because switch currents are approximately two times higher than output current. Losses occur on both the charge and discharge cycle, which means that a capacitor with 1 Ω of ESR for CIN has the same effect as increasing the output impedance of the LT1054 by 4 Ω. This represents a significant increase in the voltage losses. COUT alternately is charged and discharged at a current approximately equal to the output current. The ESR of the capacitor causes a step function to occur in the output ripple at the switch transitions. This step function degrades the output regulation for changes in output load current and should be avoided. A technique used to gain both low ESR and reasonable cost is to parallel a smaller tantalum capacitor with a large aluminum electrolytic capacitor.
Interesting
Ok now I read the datasheet a couple of hundred times more :D
Vref. is 2.5 volts, thats just what comes out of pin 6
the regulated version of this circuit will give you -5V (slightly under in reality) at Vout if you use the values R1 - 20k (it has to be at least 20k) and R2 102.5k. I used very accurate values and got exactly that.
I still can't get the calculation to work, probably the units I'm using (-5V÷1.21V+1) x 20 = -62.644 ??? Where Vout = -5V and (Vref÷2)- 40mV=1.21
By changing the value of R2 you can adjust the output voltage up to the negative value of your positive supply (or pretty close).
I of course can't get the calculation to work so I had to use a multi-turn pot instead of R2 and I got a value of 220k, the datasheet recommends 100k-300k
I've got Vin at 11.75V and Vout at 11.60V
I'm quite happy but could someone please show me how to get this formular to work.
Another question
Quote"Special care must be taken in LT1054 circuits to avoid pulling [the Vout pin] positive with respect to any of the other pins" ... "Note that most op amps present just such a load"
From this thread http://www.diystompboxes.com/smfforum/index.php?topic=100742.0 (http://www.diystompboxes.com/smfforum/index.php?topic=100742.0)
Do I need to add anything to the basic voltage inverter/ regulator circuit above so this ^^^doesn't happen. If so - what?I think I found the answer to this, if I don't have a load connected between Vin and Vout then it's all good. ✻crosses fingers✻
This is not true ^^^, I do have a load between an external power supply dammit
So I'll have to add this transistor somewhere along the line.
(http://i713.photobucket.com/albums/ww136/skruffyhound/LT1054safetytransistor_zps40e57a53.jpg)
> I'm using (-5V÷1.21V+1) x 20 = -62.644 Where Vout = -5V and (Vref÷2)- 40mV=1.21
The | | marks around "-5V" mean Absolute Value. Throw-away the sign.
5V/1.21V= 4.132
4.132+1= 5.132
20K * 5.132= 102.64Kohms
> I might build the regulated version
WHY??
#1: In audio, *generally*, a squishy 12V beats heck out of a solid but small 5V.
#2: When a datasheet is this obfuscated (_I_ hadda draw meself a picture (http://i.imgur.com/3m7Tc.gif)!) (and where did that 40mV come in??), time wasted trying to put the sense back in the toothpaste tube is time not spent on something useful.
Great! this
QuoteThe | | marks around "-5V" mean Absolute Value. Throw-away the sign.
was pretty important to know. Thank you.
I've learnt more this week than in the last year I think.
Now I have to find out what R1 should be in the diagram
There's more of it here http://www.diystompboxes.com/smfforum/index.php?topic=100797.0;topicseen (http://www.diystompboxes.com/smfforum/index.php?topic=100797.0;topicseen)
Your drawing was good :D
> what R1 should be
You know your V. You can find I on the load-chip's (filter?) spec-sheet. B is 100 for practical purpose (see transistor sheet's minimum hFE at approximate current and aim lower).
Great Paul
(http://i713.photobucket.com/albums/ww136/skruffyhound/2N2222_zps94d11a08.jpg)
2N2222 says ^^^at 10mA and 10V Hfe= 75
at 150mA and 10V Hfe = 100-300
Lets say 100 :icon_rolleyes:
I think my circuit is about a 50-60mA load and I'm running at lets say 12V
N5532 supply current at 12V = about 6.4 mA
TL072IP supply current at12V = about 1.34 mA per amplifier. So 2.68mA ?
6111 supply current at 12V = about 0.5 mA per plate. So about 1mA - not sure about this because that doesn't include the cathode. (the heaters are not part of this circuit) This is less than I previously thought
2 superbright Led's at 12V =
Vin(12V)-VF(3.2V)/R(4.7K)= I (11.99uA) - so about 24uA Although if I limit the current by half then : P=I^2R, 0.012 x 0.012 x 4700= 0.6768 Watts Thats a bit much, might have to give them more current.
(12-3.2)÷0.02 = 440 0.02 x 0.02 X 440 = 0.176 back to quarter watt resistors and 40mA current
So altogether 50mA
(Vout(11.60)xBeta(100))÷Iout(0.05)=R=23.200Ω 23k er.... does that sound reasonable?
The obfuscated data sheet wants this:
QuoteRX should be chosen to provide enough base drive to the external transistor so that it is saturated under nominal output voltage and maximum output current conditions
I don't see why superbright LEDs on a stage as status indicators would need even 2mA each.
Even rounding-up I don't see 15mA there. Perhaps 25mA if all your chips run to the high end of tolerance.
OTOH, the transistor must be "saturated". That means heavy over-drive. So if you did need 60mA, and if a '2222 can be as low as 75 (not to mention the 35 at -55 deg C spec because you won't be playing that cold), then you need nearly 1mA into the Base.
And the specs are for 10V drop. The idea here is "NO" drop. In practice, less than a Volt. The hFE will fall some at very low voltage. They don't like to tell you that. In the Fairchild lit (http://www.fairchildsemi.com/ds/PN/PN2222A.pdf) Fig 12 shows the trend, but not all the way to zero volts. Fig 2 shows low Vce _but_ assumes you drive with Beta of 10. So 1mA into Base for 10mA load, 5mA for 50mA. It is likely you don't have to push it that hard. It is likely that very few modern '2222s have low hFE. But that's how you'd figure it if you were making a million units.
Yeah, use 22K and see what happens. If the Collector won't pull very close to Emitter, try 10K and see if it gets better.
I'm annoyed because this is a Known Issue and they could have built-in 4 extra transistors instead of "fixing it in the application sheet". Aside from the relative cost of transistors versus paper, this is untold hours of engineering salary over the product's lifetime. And a sure way to make the engineer look at other chips for the next product.
I'm going on this http://www.banzaimusic.com/LED-5mm-blue-ultrabright-8000mcd.html (http://www.banzaimusic.com/LED-5mm-blue-ultrabright-8000mcd.html) info, because these are the LED's I have. As you could see though when I trimmed the current down to 12mA I suddenly had to dissipate over half a watt.
Nobody uses 1 Watt resistors around here. I know I'm running at 12V, but anyway this suggests to me that they should be around 20mA. I'll take some measurements. I don't know where the 20mA spec comes from.
I could also just stick a little old timey red LED's in, they have a 2mA spec.
This made me think though. Some fairly big part of the load will be switched on and off. I may need to recalculate this resistor when I figure out the switching. There's no need to include the LED's in this load on the chip (I don't think).
Interesting how this project evolves as I get more insight.
It is annoying that they didn't just stick this transistor in the chip, I can only think that it must be detrimental for one of the many other functions (voltage doubler etc.)
It is trying to be all things for all people this chip, which I guess just leads to compromise.
Don't understand.
> had to dissipate over half a watt.
12V at 20mA 0.020A is 0.240 Watts. Total in R and LED.
Given say 3V in LED, the resistor is 9V 0.020A or 0.180 Watts.
1/4 will do. 1/2W may live forever.
And 20mA in that LED gives 8000mcd, 8 Candellas, a LOT of light.
That's the maximum. _Any_ LED may be run at lower current; and often is, because we don't need that much light.
> old timey red LED's in, they have a 2mA spec
All that I can recall are rated 10mA-30mA (usually 20mA) MAX. Again, often run at a small fraction of rated power because who needs that much?
Ok, I've somehow been thinking about this the wrong way round, the Power calculation above is confusing me.
If I say (Vs - Vf) ÷ If = R (Supply Voltage - LED forward voltage drop) ÷ LED forward current = R (12 - 3.6) ÷ 0.010 = 840Ω
It would be better with a graph on the datasheet because the Vf is stated at a fixed If of 20mA, and the If is what I want to change. Maybe that's what is confusing me. I split the difference to give Vf of 3.6
Then : 12V - (voltage drop from the LED) - (voltage drop from the resistor) = 0
(12V - 3.6) - 8.4 = 0
P = IV 0.01 x 8.4 = 0.084 W
R = 840 (or thereabouts) dissipates 0.084 W.
I use 1/4 watt resistors, no problem. My resistor was a guess (3 posts above), I thought I could remember using 4K7 resistors, but it was just too big.
Thanks Paul
> Vf is stated at a fixed If of 20mA
In solid-state diodes, including LEDs, the forward voltage "hardly changes" over a w-i-d-e range of currents.
(That's why they make OK clippers/fuzzes.)
That 3.6V at 20mA is probably above 3.0V at 1mA.
So it's easy. (And I think a spreadsheet is just a distraction.) You got 12V, and you got a 3V diode. You need to waste-off 12-3= 9V.
OK, spreadsheet:
9V @ 20mA = 450 ohms
9V @ 2mA = 4500 ohms
9V @ 1mA = 9000 ohms
9V @ 0.20mA = 45000 ohms
Yeah, if you compute the *exact* answer for 12V and 3.6V and 450 ohms you get 18.7mA (because the approximation "3.0V" was off). But 20mA, 18mA, who cares?
Anyway, you reach for your 450 ohm resistor and find 470 ohm nearest common value. Now it's only 17.9mA, and who cares?
Likewise the "1mA" approximation of 9K leads to a 10K standard-value, And say the LED is really 3.4V near 1mA. You really get 0.86mA. That's plenty close to 1mA. You don't know that you "need 1mA", that was just a handy guess to get "much less than FULL brightness" (which you rarely need).
The LED dissipation is limited by the rated 20mA MAXimum current. Apparently 3.6V*0.020A= 0.072 Watts (72mW) is all it can take. And if you stay down to 20mA or less, you don't have to compute LED dissipation.
You, as you say, need resistor dissipation. In this case, you start from 12V and at high current the LED won't be under 3V. So 9V in the resistor. V^2/R= power. 9V*9V/450= 81/450= 0.18 Watts.
And at lower currents? The voltage is very-nearly the same, the current is much less, we expect the dissipation will be less. Assume the same 9V across a 4700 ohm resistor. 0.017 Watts. Get very paranoid and 'assume' the LED drop might be zero. The full 12V across 4,700 ohms is 0.03 Watts.
(http://i713.photobucket.com/albums/ww136/skruffyhound/PolarityprotectorV2_zps0cee1519.png)
I can't replace the photobucket pics
this would be the correct version