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Does anyone have an interest in knowing how much wattage a particular piece of equipment uses?


MartinTheMixer

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Hello all, While I was planning the building of my battery packs for Nova, (which I think nobody cared about, ha) I was testing to see how much wattage various components use so that I could build  corresponding battery packs. Does anyone care to know what pulls watt? Pun there.  Maybe I could list the wattage for various components? 

Thank you, Martin

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10 hours ago, Izen Ears said:

I’m curious!  Some time ago I tested the full 120 volt draw of my sound cart, by metering the Powermax power cable, and it was 1.2 amps.  So that’s like 120 x 1.2 = 140 watts?  I’m pretty dumb about this stuff and most likely just embarrassed myself haha!

 

Dan Izen

Dan, 140 watts? Wow. So you used an ammeter on the AC line which has given you that 1.2 amp figure?

Thanks, Martin

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On 11/20/2021 at 9:24 PM, MartinTheMixer said:

Dan, 140 watts? Wow. So you used an ammeter on the AC line which has given you that 1.2 amp figure?

Thanks, Martin

Yes it was a clamp meter over the hot, the cable was a special edison cable with the three wires loose (not in housing) so I could isolate the hot.  Is that a lot?  1.2 amps seems like a low draw.  
 

I’m pretty sure the Powermax switches to 100% AC with a trickle charge to the batts, so that reading should have been accurate.  We did it a few times throughout the day and it was the same.  I forgot to try it without the monitors off.

On 11/20/2021 at 6:38 PM, John Blankenship said:

I = E / R 

and 

P = I x E

 

I is Current (amps)

E is Voltage 

R is Resistance 

P is Power 

 

Once you learn Ohm's Law and know any two of those four specifications, it's pretty much all you need to know to figure this stuff (and so much more in electronics) out. 

 

 

 

 

 

 

 

 

 

So I was right!  Am I actually learning,   or was I wrong?  Watts = P right?

 

I’ve been reading the 1954 Navy books on electricity (Van Valkenburgh) at a very slow pace.  I want to learn how to service and build guitar amps, and Ohm’s law is constantly repeated.  But my self-motivation sucks so I may have to take a cours

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8 minutes ago, Izen Ears said:

Yes it was a clamp meter over the hot, the cable was a special edison cable with the three wires loose (not in housing) so I could isolate the hot.  Is that a lot?  1.2 amps seems like a low draw.  
 

I’m pretty sure the Powermax switches to 100% AC with a trickle charge to the batts, so that reading should have been accurate.  We did it a few times throughout the day and it was the same.  I forgot to try it without the monitors off.

Dan, I wasn't questioning the way in which you arrived at the power figure, I was just surprised at the number on the wattage.  Thank you, Martin

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7 hours ago, MartinTheMixer said:

Dan, I wasn't questioning the way in which you arrived at the power figure, I was just surprised at the number on the wattage.  Thank you, Martin

Really?  Is that so high?  What is your cart?    It’s your standard sound cart, Deva, 2x Venues, 3x hd monitors, Solice, Powermax Ultra, Bst-25, T4, led light bar and that’s it.

 

Okay now I want to know what everyone else’s sound carts draw!

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Imho and if your refer to John useful post, I do not think knowing the "wattage" of a particular equipment draws is relevant but you should rather focus on the current it draws when operating. If you follow the ohm law you will understand that the power (watt) will be dependent on the voltage as much as the current. The current drawn by your gear is constant in operation, the DC voltage is not: your battery start maybe to deliver 16v then goes down to maybe 13v few hours later, so if you only read watt it does not tell you what your gear needs.

Izen, measuring the AC line does not give you the current draw of your gear per se because the gear operates in DC. It gives you the power (or current) drawn from the AC power plug yes (and that is usefull if you want to know how much it is going to affect your electricity bill) but our gear has very minimal impact on the main power plug (compared to light for instance). Furthermore you have a converter in your gear that turns the 230 or 110 AC into usefull DC (I would guess 18v). That transformer, whether a transformer or switch converter also uses power. If you want to know how much total DC current your cart draw you should measure the DC current after the converter. You will read a much higher measure than 1,2 Amp.

If your cart has elements that actually use 110v AC power then I guess you have to do several measures, one for AC and one for DC :-)

Since the original post is about building a battery power pack I assumed Martin's gear is portable and all operating in DC.

Martin you should actually take at least two kind of Amp measures if your build a battery pack: peak and nominal.

Most gear especially digital recorders  draw a lot of Amps at turn on and then get a more or less steady consumption thereafter (actually recorders draw more when recording than idle so you should measure at least at full use: all mike channels armed and with phantom power on and mikes and headphones plugged on).

Once you know the total peak and nominal current draw from your gear you can start build your pack. Most Li-on battery cells or packs have a max peak Amp and a max nominal Amp spec. So you choose accordingly. Most commercial battery pack have higher peak current tolerance than what is advertised. Also if you want to insert a self resetting fuse (polyswitch) in your pack you must take that in consideration, it must not be too sensitive to peak current (not sure about the english words sorry).

 

 

 

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14 hours ago, Fred Salles said:

Imho and if your refer to John useful post, I do not think knowing the "wattage" of a particular equipment draws is relevant but you should rather focus on the current it draws when operating. If you follow the ohm law you will understand that the power (watt) will be dependent on the voltage as much as the current. The current drawn by your gear is constant in operation, the DC voltage is not: your battery start maybe to deliver 16v then goes down to maybe 13v few hours later, so if you only read watt it does not tell you what your gear needs.

Izen, measuring the AC line does not give you the current draw of your gear per se because the gear operates in DC. It gives you the power (or current) drawn from the AC power plug yes (and that is usefull if you want to know how much it is going to affect your electricity bill) but our gear has very minimal impact on the main power plug (compared to light for instance). Furthermore you have a converter in your gear that turns the 230 or 110 AC into usefull DC (I would guess 18v). That transformer, whether a transformer or switch converter also uses power. If you want to know how much total DC current your cart draw you should measure the DC current after the converter. You will read a much higher measure than 1,2 Amp.

If your cart has elements that actually use 110v AC power then I guess you have to do several measures, one for AC and one for DC 🙂

Since the original post is about building a battery power pack I assumed Martin's gear is portable and all operating in DC.

Martin you should actually take at least two kind of Amp measures if your build a battery pack: peak and nominal.

Most gear especially digital recorders  draw a lot of Amps at turn on and then get a more or less steady consumption thereafter (actually recorders draw more when recording than idle so you should measure at least at full use: all mike channels armed and with phantom power on and mikes and headphones plugged on).

Once you know the total peak and nominal current draw from your gear you can start build your pack. Most Li-on battery cells or packs have a max peak Amp and a max nominal Amp spec. So you choose accordingly. Most commercial battery pack have higher peak current tolerance than what is advertised. Also if you want to insert a self resetting fuse (polyswitch) in your pack you must take that in consideration, it must not be too sensitive to peak current (not sure about the english words sorry).

 

 

 

 

Fred, I used watts instead of amps, because I think most people on hear would have thought that I was deriving the amp figure from a 12 volt system, which I don't run and have never run. Mine has always been a 16 volt system. So, if I had listed amperage instead of wattage, I could have mislead someone with what would be a lower number for my setup. Less Amps equals less heat.

 

The current drawn on my DC mixer is not constant, as the voltage goes down, the amperage goes up. This is why I would use watts. Anyone who knows their voltage can then calculate the number that is appropriate to their own setup. 

 

My original post, this one, wasn't about battery packs. I was just measuring some equipment because I was curious how much power they needed. For instance, an ERX3TCD uses 1/3rd of a watt. I think I was surprised how low that was. 

 

My comment about Nova using 15.5 watts with 2 MRX414's running 8 wireless receiver channels did not have analog inputs turned on, because few people will have more than the 8 wireless running, so it didn't make sense to give a value that would not be used much by many people.

 

I have been using lithium packs with my equipment for many years. I have never had a problem. I have never powered my equipment with anything other than lithium packs. Everybody is different,  this is just how I prefer to do it. I like a pack that will run a nominal setup until lunch and weighs very little, and thanks to building it myself, I can make it into a shape that will fit well into the Nova bag. Thanks to Glenn and Howy, it now appears, that I will be able to run 8 wireless and not do a battery switch until after lunch.

 

The only wildcard that will be new with Nova is the fan setup. I can't know what that is going to do to battery consumption in the real world. 

 

I think your english is just great. Thanks for your input.

 

Martin

17 hours ago, Mungo said:

Slot-In receivers that may be operated in Sony camcorders never draw more than 4 watts. Because the Sony slot doesn't provide more. Major problem for manufacturers to design such low energy devices.

The Zaxcom RX-4 running 4 receivers is 5 watts total, 1.25 watts per wireless receiver, if you want to think of it like that. 

 

Thank you, Martin

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7 hours ago, MartinTheMixer said:

The Zaxcom RX-4 running 4 receivers is 5 watts total, 1.25 watts per wireless receiver, if you want to think of it like that.

 

Sorry, misunderstood. I meant compatible to an ENG camcorder receiver slot. Wisy, Sennheiser, Lectro, Audio Ltd, Sony, MiPro, Shure ...

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2 hours ago, Mungo said:

 

Sorry, misunderstood. I meant compatible to an ENG camcorder receiver slot. Wisy, Sennheiser, Lectro, Audio Ltd, Sony, MiPro, Shure ...

Mungo, Well of course an RX-4 won't slide into anything, a receiver slides into it. But, how many receivers do you get for the 4 watts on a Sony slide in slot? 

Thanks, Martin

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On 11/23/2021 at 3:51 AM, Fred Salles said:

Imho and if your refer to John useful post, I do not think knowing the "wattage" of a particular equipment draws is relevant but you should rather focus on the current it draws when operating. If you follow the ohm law you will understand that the power (watt) will be dependent on the voltage as much as the current. The current drawn by your gear is constant in operation, the DC voltage is not: your battery start maybe to deliver 16v then goes down to maybe 13v few hours later, so if you only read watt it does not tell you what your gear needs.

Izen, measuring the AC line does not give you the current draw of your gear per se because the gear operates in DC. It gives you the power (or current) drawn from the AC power plug yes (and that is usefull if you want to know how much it is going to affect your electricity bill) but our gear has very minimal impact on the main power plug (compared to light for instance). Furthermore you have a converter in your gear that turns the 230 or 110 AC into usefull DC (I would guess 18v). That transformer, whether a transformer or switch converter also uses power. If you want to know how much total DC current your cart draw you should measure the DC current after the converter. You will read a much higher measure than 1,2 Amp.

If your cart has elements that actually use 110v AC power then I guess you have to do several measures, one for AC and one for DC 🙂

Since the original post is about building a battery power pack I assumed Martin's gear is portable and all operating in DC.

Martin you should actually take at least two kind of Amp measures if your build a battery pack: peak and nominal.

Most gear especially digital recorders  draw a lot of Amps at turn on and then get a more or less steady consumption thereafter (actually recorders draw more when recording than idle so you should measure at least at full use: all mike channels armed and with phantom power on and mikes and headphones plugged on).

Once you know the total peak and nominal current draw from your gear you can start build your pack. Most Li-on battery cells or packs have a max peak Amp and a max nominal Amp spec. So you choose accordingly. Most commercial battery pack have higher peak current tolerance than what is advertised. Also if you want to insert a self resetting fuse (polyswitch) in your pack you must take that in consideration, it must not be too sensitive to peak current (not sure about the english words sorry).

 

 

 

Great points.  I totally left this info out. We are talking about cart vs bag and those are different beasts.  You’d never put a powermax in a bag haha!  Apologies for the OT, please continue with your discussion.

 

Actually I’m curious, how are you metering these peak vs nominal values?  Aren’t the peaks super fast and hard to measure?  And, why are the peaks important if they’re so momentary?

 

Re: my crap, my uneducated thoughts have always been that at 12 volts my sound cart passes 11 amps, but at 110 volts it’s only 1.2 amps.  Nominal.  But then again I am a dummy and am probably wrong.

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1 hour ago, Izen Ears said:

Great points.  I totally left this info out. We are talking about cart vs bag and those are different beasts.  You’d never put a powermax in a bag haha!  Apologies for the OT, please continue with your discussion.

 

Actually I’m curious, how are you metering these peak vs nominal values?  Aren’t the peaks super fast and hard to measure?  And, why are the peaks important if they’re so momentary?

 

Re: my crap, my uneducated thoughts have always been that at 12 volts my sound cart passes 11 amps, but at 110 volts it’s only 1.2 amps.  Nominal.  But then again I am a dummy and am probably wrong.

Dan, you weren't wrong to chime in. I was just surprised at how much power it takes to run that cart. 140 watts is enough to run 9 Nova's that are running 72 receivers and 9 zaxnets. 

 

As to your question of peak vs. nominal,  that doesn't have any effect in my world. I just design the packs for the real world power that my system will need for the amount of time that I will need it. The battery distribution system is another conversation, there I did have to design for max current. I cover that by a lot more than 10 percent. Of course there is no battery distribution sytem on Nova, so I did not need to redesign a smaller sytem for the Nova. Just from memory, I believe Nomad with 4 QRX running was about 24 watts. Nova is now about 15.5 watts, about 36 percent less power. That's what led me to redesigning my battery packs, because they can now have about 36 percent less capacity,  yet maintain the same run time. 

 

I'm pretty sure you're not a dummy. 

 

Thank you, Martin

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Wow, 140W is a huge amount ... I have to wonder whether you've accounted for power factor?  If you are using an inverter without an active power factor correction circuit (which seems likely, though I'm not super familiar with the current state of circuitry in the types of inverters we commonly use on carts), chances are the power factor is in the 0.5-0.6 range, which would significantly boost the AC current you are measured.  In other words, the actual power use may be closer to 140W x 0.6 = ~84W.  Add in an expected 70-80% efficiency for the inverter itself, and the true DC power draw may be closer to 60W.  Which seems more reasonable based on what I know of the gear we use.

 

Power factor would affect current measurements because the simplest circuits for converting AC to DC don't use the AC sine wave to its full effectiveness, which means the current peaks are much higher than the average, and that will affect the current read by the ammeter that you used (this is similar to measuring peak vs RMS voltage of an audio waveform).  In other words, the metering system you use for measuring AC current matters just as much as it does for measuring an audio signal.

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On 11/24/2021 at 2:28 PM, Izen Ears said:

Actually I’m curious, how are you metering these peak vs nominal values?  Aren’t the peaks super fast and hard to measure?  And, why are the peaks important if they’re so momentary?

 

Hi, you need a high read rate meter (most professional multimeters) with a « «max » memory option. At startup most electronic machines draw a high peak current for some milli seconds to a few  seconds. To know the peak current is useful whenever there is a protection circuit to choose in your project. For example if you are conceiving a power distributor and want to include self reseting fuse (=polyswitch) on outputs: in order to choose the right one it is good to have an idea of the max peak current as if you choose a fuse that has a too short term sensitivity it will block itself at startup cause of the peak current draw, if it has too long term sensitivity you might damage a piece of gear or most likely blow a  unprotected battery pack in case of accidental shortcut.

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