Running 2 heads into 1 cabinet.

[Lazybassplr56]

Is this possible? I want to run a old peavey head with a sunn concert bass head into one cabinet. One head alone doesn't seem to do it for me. Would this help. Would it blow out my cabinet. It's a Ampeg PR-410HLF cab which is a pretty good cab.

[lemoley]

what do you mean "doesn't seem to do it"?

[Lazybassplr56]

Well, my Peavy head has the thickness I'm looking for but I cant really get the tone I get with my Sunn head, but Sunn head doesn't really cut through the mix well. Not ballsy enough for me.

[lemoley]

hmmm, have you considered maybe selling those two heads and investing in one which will deliver the tone you want?

[Lazybassplr56]

Yah I know, I was already planning on doing that, the thought just poped into my head today. Maybe I'd stumble over something cool if I did this. I just wanted to know if I would blow up my cab or not.

[lemoley]

so what head would you plug your bass into?

[Lazybassplr56]

The Sunn.

[lemoley]

next question...

how would the Peavy know what your playing - theres no signal running through it meaning all your going to hear is the Sunn head

???

[Lazybassplr56]

bass -> sunn -> peavy -> cab? I don't really know. I'm just experimenting.

[edgebass5]

Your "diagram" looks like you'd be plugging the power section of the Sunn into the preamp section of the Peavey. You can say goodbye to your Peavey if you go that route.

Does either amp have a preamp out? If it does you could run that into the input of the other and you'd end up with a combination of the two preamp signals. With some amps an effects send would work for this purpose.

[Lazybassplr56]

They do have a preamp section.

[lemoley]

edge knows his ****, listen to him

*gracefully bows out of thread*

[Lazybassplr56]

So would I plug my bass into the input of the Sunn head, then the preamp into the input of the peavy. Then the peavy into the cab?

[muthafunkabass]

They do have a preamp section.
No... preamp out. It is a female plug going out of your head that says "preamp out".:thumb:

[Lazybassplr56]

Yes they have that.

[edgebass5]

They do have a preamp section.

They'd better have preamp sections, otherwise they'd just be power amps.

Bass -> Instrument input of Amp A -> Preamp out of Amp A -> Instrument input of Amp B -> Power amp/speaker out of Amp B -> cabinet(s)

This will run your signal through both preamps, but only one power amp.

[Lazybassplr56]

When I'm going from the preamp of Amp A to instrument input of Amp B do I have to use a speaker cable or just a regular instrument cable?

Also, will my cabinet be able to handle this? I don't want to blow it out.

[Mr. Pickle]

By running the heads that way, you aren't getting the power of both heads, only the one that is plugged directly into the cab, so as long as your cab works fine with that head, you have no worries of blowing anything due to this extra preamp.

As long as its a 1/4" cable, it will work.

[edgebass5]

Use an instrument cable. The only time to use a speaker cable is when you're going from the power amp section of your head to your cabinet.

Your cabinet will be able to handle it just fine. You'll only be using the power amp section of Amp B. The power amp section of Amp A isn't part of the equation if you're using the preamp out to drive Amp B.

[Lazybassplr56]

Cool, I had problems with my Sunn head alone. My cab's minimum impedance is 4 ohms and my head is only rated a 2 ohm or whatever the correct way of saying that is. Anyways it over-heated and smoke came from the back of the head and it shut down. I opened it up and nothing looked fried or burned but I noticed these two pieces (don't know what they are but they're plastic and connect wires) were cracked. Im taking the amp in soon to repair damage.

Will I have the same problem again with impedance or will I be okay since I'm using the power amp section of Amp B (peavy)?

Thanks for all the help by the way.

[edgebass5]

The first bit makes NO sense :p Your cab doesn't have a minimum impedance it has a rated impedance load of 4 ohms. Your head does have a minimum impedance which apparently is 2 ohms. I remember reading about old Sunn heads having a problem running at 2 ohms despite the rating. That said, you're running at 4 ohms so you SHOULD be fine. For what its worth, you can run a head/power amp at its minimum impedance or above, but NOT below.

The total impedance will still be the same since you're only using one cabinet. The cabinet dictates impedance, not the the head/power amp.

[Lazybassplr56]

I see, I've never understood this. Thanks for your help.

[edgebass5]

read this document and you'll know all you need to understand amp operations: http://www.talkbass.com/ampfaq

[Polyamarous]

Ampfag!

[edgebass5]

:angry: HOW DARE YOU! :mad:

That document is my holy bible! :p

[rh15951]

Duct tape the whole thing together. Sorted :thumb:

Or just read the ampfag bible :p

[edgebass5]

Or just read the ampfag bible :p

If anybody needs me I'll be in the clock tower with a high powered rifle... :upset:

[muthafunkabass]

I see, I've never understood this. Thanks for your help.
NO NO NO! Don't listen to Edge. He is an idiot. Your minimum impedance is determined by your cable. How long is your cable?

25 feet= 4 ohms
15 feet= 2 ohms
35 feet= 3 ohms
Wireless= 16.75 ohms cubed
2 feet= Pi
55 (or 13) feet= 3.66 ohms

The formula is E=MC^2
Or: Edge= Idiot.







JK! :Lol: I know I'm funny but you can tell me/rep me anyway.
Anywho, you should really read this essay that I made which is in fact completely and totally original.

3. Necessary physics



3.1. Voltage, current, power and resistance



If we're gonna talk about electric basses, some basic knowledge about electricity is essential. Although we could stick to the absolute essentials, we don't. For all electric equipment to operate, a voltage is needed. For example, a wall outlet supplies 120 volts (US) or 240 volts (Europe), and a car battery 12 volts. Once a device is connected (and switched on) current will start to flow, and power is generated. The amount of power is dependent on the voltage of the source and the resistance of the load. Before we start to describe everything in detail, we first we have to make some conventions:






Symbol


Unit


Unit symbol

Voltage


V


volt


V

Current


I


amp�re


A

Power


P


watt


W

Resistance


R


ohm


Ω = Greek capital Omega, rarely the symbol W is used

Frequency


f


hertz


Hz



In the equations these standard mathematical operators are used:

x�y����������� multiplication

x/y����������� division

xy������������� x raised to the power of y

10log x����� the 10-base logarithm of x (scientific calculators have this function built in under the "log" key)



Right, now we've established that, here we go. The relationship between voltage, current and power is best illustrated by an equation:



P = V�I��������� Power equals Voltage times Current. Math is easy ain't it?



Example 1: What's the current through an electric heater that's 115 volts, 1800 watts? Filling in the equation P = V�I gives 1800W = 115V�I, I = 1800W/115V = 15.65A (Note: this is very close to the trip point of the circuit breaker)

Example 2: What's the maximum power generated by a car generator, rated 14.4V, 70A? P = V�I = 14.4V�70A = 1008W. Must be serious car, then.



Resistance is another basic aspect of electricity. It determines how much current will flow though a certain load at a certain voltage level. To be more precise: 1 Volt is generated when 1 Ampere flows through a 1 Ohm load. In an equation it looks like this:



V = I�R��������� Voltage equals current times resistance.



Example: What's the current through a 10Ω resistor connected to a 12V battery? V = I�R gives 12V = I�10Ω, I = 12V/10Ω = 1.2A



If we combine the two above equations we get:



P = I2�R������� (P = V�I, substitute V by I�R)

P = V2/R������ (P = V�I, substitute I by V/R)



A useful help, when working with these equations, is to put them in triangles, like this:



����������������������� P�������������������������� V�������������������������� P�������������������������� V2



����������������� V��� ����� I���������������� I���������� R�������������� I2�������� R�������������� P��������� R



Knowing two variables out of three, you can find the missing one by multiplying or dividing them. If the two known numbers are horizontal, multiply them. If they're vertical, divide the top number by the bottom one.



When values become very large or very small, we can decide to put a prefix in front of the unit symbol, to indicate the magnitude of the number. That way we don't have to write down all the zeros that precede or trail the value. The most common prefix is kilo, as in kilogram (kg), kilowatt (kW), kilohertz (kHz). Below is a list of all prefixes and their multipliers.



Prefix


Symbol


Magnitude multiplier

Power


Absolute

atto


a


10-18


,000000000000000001

femto


f


10-15


,000000000000001

pico


p


10-12


,000000000001

nano


n


10-9


,000000001

micro


μ


10-6


,000001

milli


m


10-3


,001

centi


c


10-2


,01

deci


d


10-1


,1

-


-


100


1

deca


D


101


10

hecto


h


102


100

kilo


k


103


1,000

mega


M


106


1,000,000

giga


G


109


1,000,000,000

tera


T


1012


1,000,000,000,000

peta


P


1015


1,000,000,000,000,000

exa


E


1018


1,000,000,000,000,000,000



Example: What's the current through a 10kΩ resistor connected to a 250mV (line level) source, and how much power is generated?



V = I�R����������������������������������������������� ������������ P = V�I

I = V/R������������������������������������������������� ���������� P = .25V�.000025A

I = .25V/10,000Ω������������������������������������������� P = .00000625W = 6.25μW

I = .000025A = 25μA



3.2. Calculating resistances



Calculating series resistances isn�t very hard. You can just add up their values to find the substitute value Rs:



Rs = R1+R2+R3+�.+Rn�������������������������������� (n resistors in series)



With parallel connections it gets more complicated: say you have two resistors R1 and R2. Their substitute value Rs will be:



Rs = (R1�R2)/(R1+R2)������������������������������������� (2 resistors in parallel)



The following is a more general equation, in which you can put as many resistors as you like in parallel:



Rs = 1/(1/R1+1/R2+1/R3+�.+1/Rn))������������� (n resistors in parallel)



3.3. Decibels



Due to of the characteristics of our hearing (explained in more detail in chapter 4), sound pressure levels (SPL) are expressed in decibels, symbol dB. Initially, this scale was introduced as the Bel, but the deciBel seemed more practical (1 B = 10 dB). The dB is not a unit, like, for instance, the volt. Rather, it describes the magnitude ratio of two numbers along a logarithmic scale, meaning adding dBs actually means multiplying the numbers. This is in such a way that adding 10 dB means multiplying the sound pressure (and the necessary amplifier power) by 10. Consequently, adding 20 dB is a multiplication by 100, 30 dB is times 1,000 etc. As said before, only power ratios (P1/P2) can be expressed. There's always need for a reference level. For instance, when calculating the maximum SPL of a loudspeaker, the reference level would be the efficiency rating (sometimes called "reference efficiency"). In mathematical terms (told you so, we don't stick to the basics):



PdB = PdBref+10�10log(P) and back: P = 10((PdB-dBref)/10)����������������� Reference in dB



PdB = 10�10log(P/Pref) and back: P = Pref�10(PdB/10)����������������������� Reference in watts



Example 1: you have a speaker with a reference efficiency of 96 dB (1W at 1m), and a maximum input power of 250 watts. The maximum achievable SPL is: SPL = 96+10�10log(P) = 96+10�10log(250) = 96+24 = 120 dB. So, the speaker produces 96 dB at 1W of input power. This is the reference. We want to multiply the power by 250. This adds 24 dB to the total SPL.

Example 2: an amplifier's maximum output is 300 Watts. You need more power and decide to purchase a 500 watts amplifier. That way you have 500/300 = 1.67 times as much power available. SPL will increase by: SPL = 10�10log(500/300) = 2.2 dB.



A list of dB conversions:



Increase in dB


Power multiplication factor (rounded)

3


2

6


4

9


7.9

10


10

12


15.8

14


25.1

20


100

24


251

27


501

30


1000

100


10,000,000,000



It was mentioned before: power equals voltage square divided by resistance (P = V2/R). Because of the square in the equation, signal levels, which are voltages, require another method of determining dBs. With signal levels, a rise of 20 dB (instead of 10) is multiplication by ten. This is how we recognize the squared voltage on a logarithmic scale. The number in dBs increases twice as much, compared to power dBs.



VdB� = 20�10log(V/Vref) and back: V = Vref�10(VdB/20)������������������������ Reference in volts



VdB = VdBref+20�10log(V) and back: V = 10((VdB-dBref)/20)������������������ Reference in dB



For audio signal levels, the reference (Vref in the equation) is .775 V. This is a standard value, called 0 dBm.



Example 1: An effects pedal accepts a signal level of �10 dB. This is a very common value and is called "line" level. Converting this value to a voltage gives: V1/V2 = .775�10(-10/20) = .775 � .316 = .25 V

Example 2: A rackmount effects unit accepts a signal level of +4 dB. Again, this is a very common value, sometimes called "studio" or "professional" level. In volts this is: V = .775�10(4/20) = .775�1.585 = 1.23 V

Example 3: When measuring a test sinewave with a digital voltmeter, the display reads 1.00 volts. Determining the signal level in dBs: VdB = 20�10log(1.00/.775) = 2.21 dB

And lengthy. But original.

/Sarcasm is not my strong point

[edgebass5]

My only comment on all of that (glanced through it as I'm at work and don't really have time to read the whole thing) is that in the instrument amplification world, an output voltage of 250mV isn't what I would consider line level voltage. To me an ideal line level voltage is around 600mV, but could go as low as 400mV. Anything lower than that I would consider mic level voltage. This is relegated to the musical instrument amplification world of course...Good talk :thumb:

[HELLonWHEELS]

I say sell both heads and buy something new.