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 I have an AC/DC adaptor with an output of 12V - 10A. My new EQ6-R mount :icon_biggrin: requires DC 11-16V 4A and the dew heaters require a maximum of 12V 5A. I think this should be ok and that I’ll have enough amps but just in case, can I run them both off the same adapter without damaging anything?

Thanks

Edited by rubecula
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Should be ok. The mount will only draw 4A during a high speed slew. The rest of the time it will be much less. Any problems in that scenario just turn off the dew heater till you are done slewing. The worst that can happen is that you will get a voltage drop that might cause the mount to complain.

Make sure your cabling is adequately sized to deal with those currents. You dont want the wires to get hot.

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You should be OK, but with the heaters on during a fast slew you could be drawing 90% of the capacity of the power supply. 

This would be OK for short burst, but I wouldn't advise it for very long.  With the heaters drawing 50%  of the capacity on a pretty continuous basis, I would use a larger power supply or a second one so that the heaters and the mount have separate power. But that's just me.  :happy8:

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Are you sure you're actually using 5A for dew heaters? The 5A may be the maximum current the controller can handle, assuming you're using a controller that can drive several dew bands. For an 8" scope like in your sig the the Astrozap heater band draws 1.18A maximum.

Alan

Edited by symmetal
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The advice to use separate power supplies is the best you can get.
That will prevent powerdrops (brown-outs) which your mount will not like.
 I also think 5 A  for dew heaters is quite a lot... 

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19 minutes ago, symmetal said:

Are you sure you're actually using 5A for dew heaters? The 5A may be the maximum current the controller can handle, assuming you're using a controller that can drive several dew bands. For an 8" scope like in your sig the the Astrozap heater band draws 1.14A maximum.

Alan

I must admit I thought exactly the same thing but assumed this was a measured figure or he is using several.  I just logged on to my obsy and put my Esprit 100 and guide scope heaters on full and below is a screen shot showing less than 1A, albeit that is at 13.2V, so it would be a bit more at 12V.  I'd be surprised if the C8 is a great deal more but mine isn't plugged in so I couldn't check that.

Capture.JPG.b2f74458ae60470e3500be74cd6034f2.JPG

I also agree with separate supplies if at all possible, but power wise this one would be fine.

Edited by RayD
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3 minutes ago, kens said:

Don't know if I'd trust something that reports "Amps per hour" or "Wats per hour" :hmh:

That's a total consumption roll up for the power graphs.  The instantaneous current is to the right side of each dew heater slider.

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that depends if you use 1 or 2 power supplies and how many connections you need.

And you need soldering skills of course. But that is easy to learn. 
If you give me more information I will make a list and plan for you.

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23 hours ago, RayD said:

That's a total consumption roll up for the power graphs.  The instantaneous current is to the right side of each dew heater slider.

Doesn't inspire confidence in the calculations!

The integral of amps with time is amp-hours and of watts is watt-hours.

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13 hours ago, Stub Mandrel said:

Doesn't inspire confidence in the calculations!

The integral of amps with time is amp-hours and of watts is watt-hours.

 

The purpose of my post wasn't for a critical review of the equipment I use, it was to provide a simple practical and graphical indication to the OP that he probably doesn't need to be concerned about his heaters drawing 5A, as it is highly unlikely they will be assuming he is using two.  Also to note my opinion that if imaging, given that he has concerns over the size of his current supply, 2 supplies are probably better but the one he has should be fine.

I'm not sure I understand what the purpose of yours and @kens posts is in relation to the OP?  Are you saying that because the text written on the unit is not worded to identify it as the correct integral I shouldn't be taking the instantaneous current readings (the ones I was referring to) as correct so shouldn't be sharing it?

Edited by RayD
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9 hours ago, RayD said:

 

The purpose of my post wasn't for a critical review of the equipment I use, it was to provide a simple practical and graphical indication to the OP that he probably doesn't need to be concerned about his heaters drawing 5A, as it is highly unlikely they will be assuming he is using two.  Also to note my opinion that if imaging, given that he has concerns over the size of his current supply, 2 supplies are probably better but the one he has should be fine.

I'm not sure I understand what the purpose of yours and @kens posts is in relation to the OP?  Are you saying that because the text written on the unit is not worded to identify it as the correct integral I shouldn't be taking the instantaneous current readings (the ones I was referring to) as correct so shouldn't be sharing it?

I just would not trust the output of a program written by someone who clearly does not understand electrical fundamentals. The displayed values are misleading as even you stated that the current is less than 1 amp (presumably based on the average of 0.26A) whereas the instantaneous current is showing as 1.26A.

What the OP needs to know is both the peak current in different situations and the average current or power over a longer period. For example, a high peak current may cause overcurrent protection to kick in . A high average current or power could cause overheating of the power supply or wiring.

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9 hours ago, RayD said:

I'm not sure I understand what the purpose of yours and @kens posts is in relation to the OP? 

Sorry if the tone of my original post was bit flippant.

Because the OP is trying to better understand how volts and amps work and calculate power usage, it seems wise to use the right units.

The integral is the 'area under the graph' so the units are multiplying the two axes together e.g. amps x hours rather than dividing (amps per hour) .

Knowing that your readout means 4.86 amp hours is useful because that relates to the battery capacity you need (also given in amp hours).

The watt hours are less useful but allow easy comparison between setups with different voltages. Obviously as kilowatt hours they mean a lot in the context of an electricity bill!

Edited by Stub Mandrel
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  • 2 weeks later...

If your only drawing 9 amps total and 5 amps continuous and you've got a 10amp transformer, then that should be fine. As long as it's fused, there shouldn't be a problem unless it has voltage stabilisation. If it has, you might experience a current drop.

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On 28/10/2017 at 11:59, Stub Mandrel said:

Sorry if the tone of my original post was bit flippant.

Because the OP is trying to better understand how volts and amps work and calculate power usage, it seems wise to use the right units.

The integral is the 'area under the graph' so the units are multiplying the two axes together e.g. amps x hours rather than dividing (amps per hour) .

Knowing that your readout means 4.86 amp hours is useful because that relates to the battery capacity you need (also given in amp hours).

The watt hours are less useful but allow easy comparison between setups with different voltages. Obviously as kilowatt hours they mean a lot in the context of an electricity bill!

Amps x Volts = Watts               1 watt x  1 hour = 1 Watt/hr 

 

Best to look up in the handbook for electrical specification of mount, then you can do the math for kilowatt/hrs but you'll need to guestimate a duty cycle of the full load capacity plus the low load capacity. For example, if the mount is say running flat out (max slew) for 6 mins in the hour, then it has a 10% duty cycle per hour, which you can calculate the power (watts) per hour.

 

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10 hours ago, Delasaurus said:

Amps x Volts = Watts               1 watt x  1 hour = 1 Watt/hr 

 

Best to look up in the handbook for electrical specification of mount, then you can do the math for kilowatt/hrs but you'll need to guestimate a duty cycle of the full load capacity plus the low load capacity. For example, if the mount is say running flat out (max slew) for 6 mins in the hour, then it has a 10% duty cycle per hour, which you can calculate the power (watts) per hour.

 

The app in the example does that calculation, which is useful, but labels it wrongly, which may confuse...

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4 hours at 16 a would be 64 Ah... not 64 Wh.
To get watt hours you need to know amps, volts and time. So 16 A at 12 V over 4 hours, would be 16 x 12 x 4 = 768 Wh.

Edited by Waldemar
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