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Re: MiniDsp  Let's break it down
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pdxrealtor wrote:
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You lost me. Let's use the DCX for example, or any device that only accepts balanced inputs. Let's call the rating 10 volts RMS max in. Now let's say I have take an unbalanced line and jump  and S, as is typically done.
Question: does that change the input volt rating?
Question: does that change the input volt rating?
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If I did state that it was a typo apologies. The minidsp specs a phoenix connector MAX INPUT VOLT RMS as 4 volts unbalanced, and 8 volts balanced. This is VOLTS RMS. If using unbalanced the instructions state to jump  and S.
That would seem to contradict what is consistent with all the others you are aware of.
That would seem to contradict what is consistent with all the others you are aware of.
Thanks for indicating that this was rms. It does not change the principal involved. I just wanted to get a sense of whether it is similar to other specs I have seen and I do think this is pretty typical for consumer gear.
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Again they instruct an unbalanced connection to be connected as + and S and the S and  be jumped. I don't know what would happen on the minidsp board (2x8 or 4x10) if you connect unbalanced to +/.
I know that on the OUTPUT of the minidsp I myself have to float the S (float pin 1) because as far as I can tell their boards are grounded to the PCB, not the chassis. This creates, in my system, a hum. So, I have no idea what affect floating pin 1 would do on the INPUT RE: voltage limits. These are the kind of things I'd like to know.
I know that on the OUTPUT of the minidsp I myself have to float the S (float pin 1) because as far as I can tell their boards are grounded to the PCB, not the chassis. This creates, in my system, a hum. So, I have no idea what affect floating pin 1 would do on the INPUT RE: voltage limits. These are the kind of things I'd like to know.
For general information about connections schemes including unbalanced to balanced just google "Rane Notes".
Now concerning the 2x4 and your initial problem:
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...I hope I addressed all your questions.
MiniDSP suggested a pot at the 2x4 input and that would reduce the level at the input. More gain would still be needed then at the SW amp and it sounds like that may be possible for you. That would then seem to be one viable solution. An RCA 10dB attenuator from someplace like Parts Express should also work then and be more convenient.
If the 2x4 is rated at 2Vrms and the 4x8 is rated at 8Vrms balanced then using the balanced ± input connections in a 4x8 will provide an extra 12 dB of headroom in the MiniDSP. So your intended move to that unit should also be enough. [Still assuming there are no special reasons not to use the ± connection scheme.]
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Re: MiniDsp  Let's break it down
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pdxrealtor wrote:
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Minidsp suggested I put a potentiometer on the sub line in to the minidsp.
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HifiZine wrote:
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Actually. I think *I* suggested that, not miniDSP. However, it is an example (a fairly straightforward one, if you don't mind my saying) of how to deal with gain structure issues  increase or decrease gain as necessary according to your devices in the chain.
Re: MiniDsp  Let's break it down
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jtalden wrote:
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Yes, with the input connection to the DCX as you stated the input rating is the half the value of its balanced rating. So if for example we say it's rated at 10 Vrms balanced then with the connection you cited (+ and S) the input rating would be reduced to 5 Vrms. We would be using 1/2 the input circuit instead of using the full circuit.
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I think both input connections options can be used on most consumer units. I believe it is more commonly recommended to connect to the ± terminals and allow the S to float. If one type is not shown by the manufacturer as an option however I would confirm with them or make sure that others users are successfully doing it with that unit.
I searched real quick on hooking up unbalanced lines to a DCX and sure enough the voltage is halved because typical hookup is +/S and jumping /S.
This is also how most if not all RCA to XLR premade cables are made. + to pin 2, S to pin 1 and 3 (jumping S and ).
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For general information about connections schemes including unbalanced to balanced just google "Rane Notes".
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Now concerning the 2x4 and your initial problem:
Well I think I understand that you have indeed tried it with the AVR SW offset at or near the minimum level setting possible in the AVR. At this setting you still have enough gain in the SW amp available, but are still clipping the 2x4 MiniDSP if large filter boosts are applied. If that is correct, at least, I now understand the issue.
MiniDSP suggested a pot at the 2x4 input and that would reduce the level at the input. More gain would still be needed then at the SW amp and it sounds like that may be possible for you. That would then seem to be one viable solution. An RCA 10dB attenuator from someplace like Parts Express should also work then and be more convenient.
If the 2x4 is rated at 2Vrms and the 4x8 is rated at 8Vrms balanced then using the balanced ± input connections in a 4x8 will provide an extra 12 dB of headroom in the MiniDSP. So your intended move to that unit should also be enough. [Still assuming there are no special reasons not to use the ± connection scheme.
Well I think I understand that you have indeed tried it with the AVR SW offset at or near the minimum level setting possible in the AVR. At this setting you still have enough gain in the SW amp available, but are still clipping the 2x4 MiniDSP if large filter boosts are applied. If that is correct, at least, I now understand the issue.
MiniDSP suggested a pot at the 2x4 input and that would reduce the level at the input. More gain would still be needed then at the SW amp and it sounds like that may be possible for you. That would then seem to be one viable solution. An RCA 10dB attenuator from someplace like Parts Express should also work then and be more convenient.
If the 2x4 is rated at 2Vrms and the 4x8 is rated at 8Vrms balanced then using the balanced ± input connections in a 4x8 will provide an extra 12 dB of headroom in the MiniDSP. So your intended move to that unit should also be enough. [Still assuming there are no special reasons not to use the ± connection scheme.
There you suggest using the + /  inputs again. If the + is allowing 4 Volts and the  is allowing 4 Volts that must mean the S is passing Volts??
Re: MiniDsp  Let's break it down
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pdxrealtor wrote:
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So the MAX Volt RMS spec doesn't technically change based on +/S connection or a +/ connection. But the headroom is halved if not using both legs?
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Now you've got me wondering how one could hook up a S to  if the source is unbalanced.
Doesn't the  leg of a + /  / S (balanced) connection need a signal? S is just ground and therefor has no signal. Is that right?
Doesn't the  leg of a + /  / S (balanced) connection need a signal? S is just ground and therefor has no signal. Is that right?
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I searched real quick on hooking up unbalanced lines to a DCX and sure enough the voltage is halved because typical hookup is +/S and jumping /S.
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This is also how most if not all RCA to XLR premade cables are made. + to pin 2, S to pin 1 and 3 (jumping S and ).
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I was looking at those inline attenuators but most of them neuter <20 hz response.
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There you suggest using the + /  inputs again. If the + is allowing 4 Volts and the  is allowing 4 Volts that must mean the S is passing Volts??
Re: MiniDsp  Let's break it down
This from a post over on the minidsp forum. Not sure if this helps anyone. Wayne???
This is the reason I say technically 8 Volt max spec. I got this info from an EE who says it's not possible to get 4vrms or 8vrms based on a balanced or unbalanced connection.
It's not possible to have an RCA connect to and XLR as you mention above. RCA only has two cables so on the XLR end you must jump one of the wires to a pin, or leave it open. Maybe you misunderstood what I was saying?
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To the other engineers asking for more details, the internal voltage on the board is +/15V via DCDC flyback stage for a split rail supply to the op amps.
This is the reason I say technically 8 Volt max spec. I got this info from an EE who says it's not possible to get 4vrms or 8vrms based on a balanced or unbalanced connection.
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I just looked at the input voltage spec of the 2x8 and it is contradictory. I'll explain why below.
A balanced linelevel input uses a circuit called a differential amplifier (diff amp), usually implemented with an IC opamp and four resistors. I'm going to do a little math on this thing, because it's the only way I know to explain it without being ambiguous myself. In typical operation, the diff amp takes two input voltages that vary with time, neither of which is zero. Let''s name these voltages as follows:
Let input voltage #1 be called v1(t)
Let input voltage #2 be called v2(t)
The output voltage of the diff amp is a single groundreferenced voltage that is applied to the input of the A/D converter chip. Let this voltage be named as follows:
Let the diff amp output voltage (= input voltage to the A/D) be called vO(t)
Now let the gain from input 1 of the diff amp to the diff amp output, with input 2 grounded be called A1. Also, let the gain from input 2 of the diff amp to the diff amp output, with input 1 grounded be called A2. In linear operation, superpositon applies, which means vO(t) is determined by a simple formula:
vO(t) = A1 * v1(t) + A2 * v2(t)
We're almost there. To get a formula that has a better relationship to reality, we must make a change of variables. We define the differencemode voltage vDM(t) as the difference between v1(t) and v2(t), and we define the commonmode voltage vCM(t) as the average of v1(t) and v2(t). In formulas, it looks like this:
vDM(t) = v1(t)  v2(t)
vCM(t) = 0.5 * (v1(t) + v2(t))
Now we can calculate vO(t) in terms of the differencemode and commonmode input voltages. I'll skip the math in between and go straight to the result:
vO(t) = (A1 + A2) * vCM(t) + 0.5 * (A1  A2) * vDM(t)
Now, here is the trick: If we design A2 so it's equal to the negative of A1, then by the formula above, the signal vO(t) applied to the A/D depends only on the difference between the two input voltages (the differencemode voltage vDM(t)) and all dependency on the average of the input voltages (the commonmode voltage vCM(t)) disappears! This is called commonmode rejection. You may have heard of that term.
In a typical input diff amp of a balanced system, A1 = 1 and A2 = 1, giving:
vO(t) = vDM(t)
This follows from the longer formula above, with A1 = 1 and A2 = 1.
So, how does this relate to the miniDSP input voltage spec? (finally!). Well, the 2x8 spec says the maximum balanced input voltage is 8 Volts RMS (= 20 dBu). But remember, a properlydesigned diff amp responds only to the differencemode voltage at the inputs, and rejects the commonmode voltage. So this 8 Volt RMS balanced input voltage in a typical balanced system is achieved by having two equalamplitude 4 Volt RMS signals at the input that are 180 degrees out of phase with each other. But there's another way, and that is to make one of the input voltages zero, and the other 8 Volts RMS. That also gives the same 8 Volt RMS differencemode input voltage.
The conclusion? If the 8 Volt RMS balanced maximum input voltage is correct, then that means the maximum unbalanced input voltage is also 8 Volts RMS (because the diff amp only responds to the difference of the two input voltages, and within limits, it doesn't matter how that difference is achieved). But the spec says 4 Volts RMS! Retardedness.
Now there is one caveat, which deals with an unspecified parameter that relates to nonlinear operation: the maximum commonmode input voltage. From the math above, the commonmode input voltage is the average of the voltages at the two inputs, For a fully balanced system, the average of the input voltages is always zero for all time, because the two input voltages are negatives of each other. For a singleended input signal of 8 Volts RMS, the commonmode input voltage is 4 Volts RMS (the average of 8 and zero). In a proper design, with a typical opamp, the circuit will be safe with this commonmode input voltage.
So if all is well, it looks like the 2x8 has 4x the input voltage capability of the 2x4 (8 Volts RMS vs. 2 Volts RMS), assuming the 8 Volt RMS balanced maximum input voltage spec is correct.
Sorry for all the math, but if you take the time to understand the above, you will know more than miniDSP tech support and the people who wrote the miniDSP specs.
A balanced linelevel input uses a circuit called a differential amplifier (diff amp), usually implemented with an IC opamp and four resistors. I'm going to do a little math on this thing, because it's the only way I know to explain it without being ambiguous myself. In typical operation, the diff amp takes two input voltages that vary with time, neither of which is zero. Let''s name these voltages as follows:
Let input voltage #1 be called v1(t)
Let input voltage #2 be called v2(t)
The output voltage of the diff amp is a single groundreferenced voltage that is applied to the input of the A/D converter chip. Let this voltage be named as follows:
Let the diff amp output voltage (= input voltage to the A/D) be called vO(t)
Now let the gain from input 1 of the diff amp to the diff amp output, with input 2 grounded be called A1. Also, let the gain from input 2 of the diff amp to the diff amp output, with input 1 grounded be called A2. In linear operation, superpositon applies, which means vO(t) is determined by a simple formula:
vO(t) = A1 * v1(t) + A2 * v2(t)
We're almost there. To get a formula that has a better relationship to reality, we must make a change of variables. We define the differencemode voltage vDM(t) as the difference between v1(t) and v2(t), and we define the commonmode voltage vCM(t) as the average of v1(t) and v2(t). In formulas, it looks like this:
vDM(t) = v1(t)  v2(t)
vCM(t) = 0.5 * (v1(t) + v2(t))
Now we can calculate vO(t) in terms of the differencemode and commonmode input voltages. I'll skip the math in between and go straight to the result:
vO(t) = (A1 + A2) * vCM(t) + 0.5 * (A1  A2) * vDM(t)
Now, here is the trick: If we design A2 so it's equal to the negative of A1, then by the formula above, the signal vO(t) applied to the A/D depends only on the difference between the two input voltages (the differencemode voltage vDM(t)) and all dependency on the average of the input voltages (the commonmode voltage vCM(t)) disappears! This is called commonmode rejection. You may have heard of that term.
In a typical input diff amp of a balanced system, A1 = 1 and A2 = 1, giving:
vO(t) = vDM(t)
This follows from the longer formula above, with A1 = 1 and A2 = 1.
So, how does this relate to the miniDSP input voltage spec? (finally!). Well, the 2x8 spec says the maximum balanced input voltage is 8 Volts RMS (= 20 dBu). But remember, a properlydesigned diff amp responds only to the differencemode voltage at the inputs, and rejects the commonmode voltage. So this 8 Volt RMS balanced input voltage in a typical balanced system is achieved by having two equalamplitude 4 Volt RMS signals at the input that are 180 degrees out of phase with each other. But there's another way, and that is to make one of the input voltages zero, and the other 8 Volts RMS. That also gives the same 8 Volt RMS differencemode input voltage.
The conclusion? If the 8 Volt RMS balanced maximum input voltage is correct, then that means the maximum unbalanced input voltage is also 8 Volts RMS (because the diff amp only responds to the difference of the two input voltages, and within limits, it doesn't matter how that difference is achieved). But the spec says 4 Volts RMS! Retardedness.
Now there is one caveat, which deals with an unspecified parameter that relates to nonlinear operation: the maximum commonmode input voltage. From the math above, the commonmode input voltage is the average of the voltages at the two inputs, For a fully balanced system, the average of the input voltages is always zero for all time, because the two input voltages are negatives of each other. For a singleended input signal of 8 Volts RMS, the commonmode input voltage is 4 Volts RMS (the average of 8 and zero). In a proper design, with a typical opamp, the circuit will be safe with this commonmode input voltage.
So if all is well, it looks like the 2x8 has 4x the input voltage capability of the 2x4 (8 Volts RMS vs. 2 Volts RMS), assuming the 8 Volt RMS balanced maximum input voltage spec is correct.
Sorry for all the math, but if you take the time to understand the above, you will know more than miniDSP tech support and the people who wrote the miniDSP specs.
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?? All the ones I bought were 1S/+2/3 and I thought that was typical. 1S/+2/3 is the way Rane Notes has their recommendation. It's a lot of work to convert them so when I got 18 new cables wired that way I just used them as is.
Re: MiniDsp  Let's break it down
pbxrealtor,
It is probably just best to discount all my above comments.
I finally did some testing and it is clear I was wrong on the input headroom issue and confused in some spots regarding some the connections schemes I have used.
Sorry for the misdirection.
It is probably just best to discount all my above comments.
I finally did some testing and it is clear I was wrong on the input headroom issue and confused in some spots regarding some the connections schemes I have used.
Sorry for the misdirection.
Quote:
jtalden wrote:
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pbxrealtor, It is probably just best to discount all my above comments. I finally did some testing and it is clear I was wrong on the input headroom issue and confused in some spots regarding some the connections schemes I have used. Sorry for the misdirection.
I asked about your suggestion of connecting an rca + /  and was told it could work with a 1:1 center tapped transformer.
This from another EE who seems to think the input limits do change with balanced vs unbalanced (on minidsp).
Re: MiniDsp  Let's break it down
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pdxrealtor wrote:
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Mind sharing what testing you did?
I asked about your suggestion of connecting an rca + /  and was told it could work with a 1:1 center tapped transformer.
This from another EE who seems to think the input limits do change with balanced vs unbalanced (on minidsp).
I asked about your suggestion of connecting an rca + /  and was told it could work with a 1:1 center tapped transformer.
This from another EE who seems to think the input limits do change with balanced vs unbalanced (on minidsp).
Re: MiniDsp  Let's break it down
lol... at this point I'm just trying to learn. I'm sure the 4 volts, or 8 volts, of input headroom on the 4x10 unit I ordered will serve my purpose just fine either way.
I have no idea what the hell a center tapped transformer is, but thought Jtalden might.

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