Discuss Output Max Amps in the UK Electrical Forum area at ElectriciansForums.net

What output current will I get if I have an input voltage of 700 volts at 1 amps and if I step the voltage down to 230 volts what would be the max amps I'll get.
Erin to start with what are you wanting to put 700 volts in what to get 1amp I guess there’s not enough information
 
Didn't think about that yet. 700 volts at 1 amp is single phase using a generator. I want single phase 230 volts. I want to know what max amps I'll get if I want to step down to 230 volts. Please advise
 
Is it a 3 phase transformer and what’s the KVA rating of the transformer
What is your normal voltage
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Didn't think about that yet. 700 volts at 1 amp is single phase using a generator. I want single phase 230 volts. I want to know what max amps I'll get if I want to step down to 230 volts. Please advise
So if you start your generator you get 700 volts
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What is your normal voltage
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So if you start your generator you get 700 volts
If you want to step down to 230 volts you will need a transformer
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What is your normal voltage
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So if you start your generator you get 700 volts
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If you want to step down to 230 volts you will need a transformer
I’m curious where did you get a generator that puts out 700 volts
 
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I'll be connecting 100 motors in series (model of motor is 775 motor). All I know is that the winding of the month can handle around 1 amp current. Assuming 1 amp and 700 volts if I want to use a transformer to step it down to 230 volts what would be the max amps I'll get.
 
Hi - if it was a perfect transformer with no losses its a simple calculation 700/235 Amps. As this transformer doesn’t exist you’ll get a bit less.
 
Out of interest why are you connecting 100 motors in series?
 
I got one question how do you wire 100 motors in series, I would think that would be impossible
Anything is possible , but not necessarily a good design .
More points of failure ,
Labour to assemble ,
Hiding inefficiencies in multiple places = harder to measure !
( using whats available cheaply -adhoc .. feels like todays strategy )
--Rather dangerous if no earthing is present--
 
Ignoring transformer losses, Vout/Vin = Iin/Iout.
1 x 700/230 = 3.0 as Kidsolo states above.
Also as above, devices rated for 7V each won't normally have insulation capable of withstanding 700V. They would need to be insulated from each other and from ground adequately.

However...

You mention motors, but they are seemingly being used as generators. I'm picturing 100 small moving parts driven by wind or water or something, from which you want to harvest the total output. If you want to transform the output of a series connected string of mini AC generators they must all be in phase; 100 randomly moving AC generators wired in series will give near zero output volts. If they are synchronous, they must be mechanically locked together (in which case you could use one large generator). If they are asynchronous, they will need to draw magnetising current from the AC supply; this works for a single induction generator but I can imagine chaos with a large number of units in series.

Small AC generators are very rare and specialised. Are you sure they are not small permanent-magnet motors? If so, the outputs will still be DC when used as a generator, and a transformer will not help you. One approach would be to series-connect them in blocks of say 48V, which is safe and their insulation will probably be OK with, then use DC-DC converters with paralelled outputs to convert up to a suitable bus voltage for a DC-input AC inverter.

Some significant electronics / generator knowledge will be needed to find an effective solution.
 
Hi,Are these "motors" being used as generators?
....is there wind or water involved?
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Darn it....on silly phone,and never read luciens post?
 
A weak point for series DC , the same current flows thru all ,if they are not generating a matching current (at same instant) they will become motors again. wasting what others have made.
(before embarking on an expensive combiner -need to understand the dynamics of loading generators , add an individual load to each one ,and test if currents are in sync and can be combined )
Open circuit voltages mean little without a load !
 
As SZ mentions, an undriven or weakly-driven PM generator is a problem in simple series-connection. It will behave like a constant torque source, so if the drive is uncoupled it will motor in the reverse direction at unlimited speed. Thankfully, this problem can be solved by a parallel diode; back-biased when generating but forward-biased while motoring, which would reduce the available motoring voltage to a fraction of a volt and hence avoid power loss and damage.

Parallel connection has the inverse effect, behaving as a constant speed source that will drive its prime mover in the same direction if the torque falls. This is typically not harmful but is wasteful. A series diode here would cure it, but unlike the parallel diode may incur losses of a few percent under normal operation. Preventing the dynamo motoring was the reason for the 'cutout' relay in generator control boxes used for battery charging. I could go on at length about the Lucas RB340 at ths point...
 
erwin1991c: You intend to use these dc motors:
https://www.robotshop.com/media/files/PDF/rbban70-banebots-DC-motor-specs.pdf

Using a source of torque (wind? water?) you intend to spin them in order that they generate an emf. The first thing to say is that you will be generating a direct current. The second thing is you will need to spin them very quickly in order to generate single digit voltages. Looking at the specification, the motor can be considered as a dc generator with an output voltage of 1205rpm/Volt in series with a resistance of 12V/120A = 0.1 Ohm, where 120A is the stalled motor current so there will be no back emf. So, to produce a no-load emf of say 10 Volts, the rotor will need to turn at 12000 rpm. Is this practical in the application you intend to use the motors in?

The peak efficiency as a motor is when the rpm is 15000 and the current is 17A. In this state, the power output is 15000/1200 x 17 = 12.5 x 17 = 212 Watts. The power input is this figure plus the Ohmic loss which is 17 x 17 x 0.1 = 29W; 212 + 29 = 241W, an efficiency as a motor of 212/241 = 0.88 x 100% = 88% But this figure ignores friction and windage - the specification quotes 70% efficiency in conversion of electrical energy into mechanical work. Considering the motor as a dc generator similar reasoning applies. What the specification does not say is what are the continuous ratings for the motor to avoid it becoming too hot and being damaged. Don't expect the motor/generator to be able to deliver circa 200W and dissipate 30W for long periods.

So, some basic considerations are what speed can you spin the motor, what mechanical power do you have available (in terms of torque x rotational speed) and what electrical power output do you anticipate per 775 and require for your application? What is the range of torque-speed from your source of mechanical power? Is the peak mechanical power at the same torque-speed as those for the peak efficiency of the motor/generator?

Look up the 'maximum power transfer' theorem.

Maximum power transfer theorem - Wikipedia - https://en.wikipedia.org/wiki/Maximum_power_transfer_theorem
 
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I'd like to thank everyone for replying back to me. I've gone through all your replies and decided to drop the idea of using 100 775 motors as a generator. Thanks for enlightening me. I'd like to know what if I replaced the 775 motor with a modified ceiling fan (will install some neodymium magnets so it'll act as a generator). I'll get 230 volts by easily spinning it. I'd like to know what would be the output amps at 230 volts and how many I'll have to connect in series/parallel to get 100/200 amps.
 
Modifying a ceiling fan motor into a PM generator seems to be going about it the wrong way. Why not specify the torque and speed range of the driving mechanism, then find a generator that suits?

P = ωT = 2π*n*T
E.g. if each of your mechanical drives is capable of 1.5 Nm torque at 250 rpm, which I am guessing to be a representative ceiling fan maximum output:
P=6.28*250/60*1.5=39W mechanical.
If you take the generator efficiency as 70% then
100*230/0.7/39 = 842 ceiling fans to produce 100A at 230V AC 50Hz.

Of course, with a PM AC generator the drives have to be synchronous which imposes some limitations on the driving mechanism. It's difficult to make suggestions without knowing more of the background.
 
Are these going to be used outside, ie. wind driven? They would need to withstand the elements.

I'd also like to see the salesman's face at the local electrical store when you order 842 ceiling fans.

As Lucien said, we could do with knocking more of the background of this project.
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Auto correct strikes again - obviously it should say 'knowing' in my post above.
 
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I'd like to thank everyone for replying back to me. I've gone through all your replies and decided to drop the idea of using 100 775 motors as a generator. Thanks for enlightening me. I'd like to know what if I replaced the 775 motor with a modified ceiling fan (will install some neodymium magnets so it'll act as a generator). I'll get 230 volts by easily spinning it. I'd like to know what would be the output amps at 230 volts and how many I'll have to connect in series/parallel to get 100/200 amps.
Are these going to be used outside, ie. wind driven? They would need to withstand the elements.

I'd also like to see the salesman's face at the local electrical store when you order 842 ceiling fans.

As Lucien said, we could do with knocking more of the background of this project.
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Auto correct strikes again - obviously it should say 'knowing' in my post above.
What Lucien said good luck
 

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