output from transformer is coming not as expected

[VickySalunkhe]

Sir
Readings on
Post no. 98 was taken by rebuilding the circuit part by part as you suggested
By which i got an increase of output current by 1amp
But still it wasn't upto expectation

I have provided you all the true readings
I don't know
Whether i have been wrong with the readings you all asked for or not.

I think brumby got confused as soon as i posted readings of ac voltage to find out the ripple

Can any one plz tell me little bit clearly what was he actually asking for
And from where should i measure it and with which parts connected to thr transformer


I have taken all those reading while all parts (cap, rectifier,motor) were connected

And took the readings from the points where i have taken the readings of dc voltage
For ac i just switched the dial of multimeter to ac voltage

Is that wrong  or something is?

[Brumby]

I am drawing up something that will make the measurement process clear....  I have some pressing priorities, so please be patient.

[Gregg]

I guess I thought I could slip in a lesson in basic physics for you to see hands on what happens when trying to parallel rudimentary DC power supplies.  Also I was curious as to how much the mains voltage dropped when this system was attempting to power your full load and to see if there was significant voltage drop across any of the twisted wire connections or bridge rectifiers with a bad diode.

Basically the power supply with the greatest potential (voltage) will try to power the entire load.  Other power supplies at lower potential won’t contribute anything unless the highest potential one sags to the next supply’s potential and even then the load won’t be shared equally.

This is very similar to charging lithium batteries and need for balancing every cell (only in reverse). 

You could make actively controlled rectifiers using things like SCRs and balancing control circuitry to effectively force each power supply to the same potential with current control for dissimilar transformer output capacities but that is way beyond the scope of this thread and hardly worth the time and effort considering a single transformer of suitable capacity would be much cheaper in your case.

The telco industry uses nominal 48 volts DC supplied by huge batteries and multiple chargers (called rectifiers in telco language) that all parallel and share the load.  If the share wiring between the chargers gets compromised then they fail to share but they have the safety of current limiting where some rectifiers go to full load and others go to zero load.  They also have at least one more rectifier on line at any given time than absolutely necessary (N+1 redundancy) 

In your system the transformer with the highest potential provides current until magnetic saturation and there is no current limiting or voltage regulation to save it.

[Brumby]

OK...  Reference schematic....



How I envisage measurement requests...
V_AB = Voltage between points A and B
V_JK = Voltage between points J and K
I_L = Current flowing past point L
I_J1 = Current from Section 1 flowing past point J
I_J2 = Current from Section 2 flowing past point J
I_J = I_J1 + I_J2  (Kirchhoff's current law)

V_KQ = Voltage drop from point K (supply) to point Q (controller)

.. and how they would be reported.


I hope I have it right.  All equipotential points within the supply have been considered to be exactly that.  Only for the longer wires has there been an allowance to measure voltage drops.

Hopefully this will allow for unambiguous measurement reference points - which I hope will be a start in clearing some confusion.  We still need to be conscious of clear communication with regards to specifying AC and DC ... and anything else we need to be aware of - such as the specific conditions (eg load) under which these measurements are made.


Apologies if I don't participate as significantly over the next several days... some other priorities I cannot ignore.

[VickySalunkhe]

Hello sir
First of all thankyou very much for helping me out

Just wanted to tell you that i have told you all the readings from the points you have mentioned above
Expect for the current from point L

I don't understand there one thing
I'm connecting a +ve and -ve wire for a series connection
How can i measure current from there ,   can i ?

And how can i measure this
VKQ = Voltage drop from point K (supply) to point Q (controller)

I am sorry but it got me little confusing.

Thanks

[Brumby]

Just wanted to tell you that i have told you all the readings from the points you have mentioned above
Expect for the current from point L

I expect you have ... but there is the question: Which ones are they exactly?  This may sound strange, but what you think is clear and what others think is clear - may not be the same thing.  Most of the time, it would be straightforward - but as we get deeper into this, being exact is important.

I don't understand there one thing
I'm connecting a +ve and -ve wire for a series connection
How can i measure current from there ,   can i ?

Unless you have a clamp meter that can do DC current (which most beginners won't have) you measure current the old fashioned way - break the circuit at that point an insert an ammeter.

And how can i measure this
VKQ = Voltage drop from point K (supply) to point Q (controller)

Just use the reference as a definition - such as this:
V_KQ
 - Meter on Volts (probably DC)
 - Red lead of meter on point K
 - Black lead of meter on point Q
That is how.

It may seem strange to measure the voltage from one end of a piece of wire to the other - but it is not.  It is important to make sure that when you put your meter lead on point K, that it is as close to the power supply negative rail as possible.  Likewise, point Q should be as close to the controller as possible.

The measurement, however, will be dependent on the current flowing.  If there is no current flowing, the answer will be zero, but if there is a heavy current flowing - there will be a measurable voltage drop.  In this example, expect the answer to be negative, since point K is on the negative supply rail and will be below the potential of every other point of the DC circuit.

If your connections are good and the wire gauge is heavy enough - then these "end-to-end" measurements on the wires should be very low under load.  You are getting large voltage drops which shouldn't be happening according to what we are understanding - so there must be some part of the circuit that is NOT behaving as expected.

[VickySalunkhe]

How I envisage measurement requests...
V_AB = Voltage between points A and B
V_JK = Voltage between points J and K
I_L = Current flowing past point L
I_J1 = Current from Section 1 flowing past point J
I_J2 = Current from Section 2 flowing past point J
I_J = I_J1 + I_J2  (Kirchhoff's current law)

V_KQ = Voltage drop from point K (supply) to point Q (controller)

V_AB = 30V AC
V_JK = 54V DC
All current values are max value i got by motor running at full speed
I_L = 1.42A
I_J1 = 1.42A
I_J2 = 1.42A
I_J = I_J1 + I_J2  (Kirchhoff's current law)
I_J = 2.84A

V_KQ = 0V
No change of voltage occurred when motor was at 0 speed and full speed

And sir i have replaced the LONG WIRE
so don't consider the voltage drop for that
The total wire distance back and forth between transformers and controller is 1.9meter

[Brumby]

V_AB = 30V AC
V_JK = 54V DC
All current values are max value i got by motor running at full speed
I_L = 1.42A
I_J1 = 1.42A
I_J2 = 1.42A
I_J = I_J1 + I_J2  (Kirchhoff's current law)
I_J = 2.84A

V_KQ = 0V
No change of voltage occurred when motor was at 0 speed and full speed

And sir i have replaced the LONG WIRE
so don't consider the voltage drop for that
The total wire distance back and forth between transformers and controller is 1.9meter

OK ... Now we have some clear measurement points.  If we all use this nomenclature, then there will be much less confusion as to what is being measured.

However, from the above figures, someone could ask ... So what is the problem?  V_JK looks good and the current measurements seem reasonable.

From these figures, V_JK is always 54V ... which, from previous discussion, does not seem to be the case.  This means you need to specify the conditions under which any given measurement is taken.

For example:
V_JK = 54V DC  @  I_J = 2.84A
 -or-
V_JK = 54V DC  @  I_J = 0A
 -or-
V_JK = 54V DC  @  some other condition

Knowing what the circumstances are for any given measurement are just as important as the measurement itself.  Power supply and high current circuitry are classic examples - so are electric motors.

Without even thinking, I can name 4 states for an electric motor that will completely change the load characteristics...
 - disconnected
 - unloaded  (free spinning - just the motor shaft with nothing connected to it)
 - typical load  (doing the work it would need to do to move a scooter with someone on it)
 - stalled  (shaft prevented from rotating)

These are arranged in ascending order of the electric load on the power supply - and they will drastically affect the current demands from the power supply.  The mechanical load condition of the motor is, therefore, critical in understanding the electrical load put on the power supply.  This information MUST be provided if we are to be able to help.

[VickySalunkhe]

Voltage readings taken
In 4 different stages as you suggested

INPUT AC 234 V

DISCONNECTED
V_JK = 55.2V @ 0A

UNLOADED  (free spinning - just the motor shaft with nothing connected to it)
V_JK = 53.6V @ 0.05A
V_JK = 46V @ 1A
V_JK = 45V @ 1.19A MAX
Max speed of motor was achieved at 1.19A i think thats why motor didn't ask for more current

According to that i think motor will draw more current to displace the load present on the vehicle


TYPICAL LOAD  (doing the work it would need to do to move a scooter with someone on it)
V_JK = 53.6V @ 0.05A
V_JK = 39V @ 2A
V_JK = 37V @ 2.6A MAX

STALLED (shaft prevented from rotating)
V_JK = 53.6V @ 0.05A
V_JK = 37.1V @ 2.7A MAX

As soon as the voltage reaches to 37V the motor stars all over again from 0  and gives a sudden jerk

[IanB]

If I understood correctly you have a 24 V transformer in series with a 12 V transformer which would give a 36 V supply. And you have a voltage that drops to about 37 V when the motor is loaded. So your measurements seem entirely reasonable. I think you can ignore the measured 55 V when the motor is not drawing a load--this is a "fake" voltage produced by the capacitors.

It might be interesting in fact to remove the capacitors from the circuit completely and repeat the measurements. Does the motor controller still work with no capacitors in the supply? I would guess the motor itself might work if connected directly to the supply without the controller.

If you want to supply enough voltage to the motor and controller you will need a 48 V supply instead of a 36 V supply, which means you will need to put two 24 V transformer secondaries in series rather than 24 V and 12 V. But if you do this with the capacitors you will get nearly 70 V unloaded. You have to be careful about not exceeding the maximum voltage on the controller.

[VickySalunkhe]

Sir you have understood it correctly

I cannot connect the psu to the motor directly as it is an brushless dc motor

I have tried two 24V transformers in series in earlier stages of testing the psu
And you guessed it right it gave 74V as ouput
And as soon as i connected that psu with controller

No power was given to the motor
that's why i have used 24V and 12V in series

May be some security is present in the controller

[IanB]

Have you tried the existing arrangement of 24 V and 12 V with no capacitors to find out if it works or not?

[VickySalunkhe]

Not yet
Will do it asap
But wouldn't it create a problem as too much ripple will be present in power supply

[IanB]

But wouldn't it create a problem as too much ripple will be present in power supply

Maybe, maybe not. There is no way to know without trying it.

[Brumby]

With clear measurements, the issues are also much clearer.

For the conditions specified, the power supply is performing pretty much as you might expect.  The problem is that this power supply is not appropriate for the need.

When you look at the transformers, the voltages are given as 12V and 24V - which are AC voltages.  AC voltages like this are typically given as RMS.  If I skip quickly through the math, this means that you should only expect DC voltages of 12V and 24V when your power supply is operating at its rated load.  Considering diode voltage drops and other circuit parameters, this supply is actually operating within reasonable expectations.

The first point of understanding is that the "no load" voltage of V_JK = 55.2V @ 0A relates to the peak voltage.  The capacitors will charge up to this and keep the voltage high - but it is meaningless.  It's like a weightlifter lifting only the bar - with no weights on it.

The under load measurement of V_JK = 37V @ 2.6A is a more realistic reflection on the demands of the load and the capabilities of the supply.  However, as you have observed, at this voltage, the controller has a problem.

This is the obvious solution -

If you want to supply enough voltage to the motor and controller you will need a 48 V supply instead of a 36 V supply, which means you will need to put two 24 V transformer secondaries in series rather than 24 V and 12 V.

but with one significant caveat...

But if you do this with the capacitors you will get nearly 70 V unloaded. You have to be careful about not exceeding the maximum voltage on the controller.

The answer is that your power supply needs to be voltage regulated.  The regulator must keep the voltage to a respectable level across a range of current - and allow the required current to pass.

You might be asking yourself ... So why is this so hard? - and the answer lies in the normal mode of operation for which the scooter equipment was designed to operate - batteries.  Four 12V, high capacity batteries will produce a reasonably stable 48V (nominal) over the range of current drawn by the motor.  A simple unregulated supply will not do this (unless it was massively overrated).


To continue your project, you will need to regulate your supply - or run off batteries.

[VickySalunkhe]

Sir
At the very start only
I thought to make a regulated power supply
For which i have created a post for it also
Here is the link
https://www.eevblog.com/forum/beginners/design-of-circuit-for-0-24v-5amp-regulated-power-supply/

This is for half part of my psu
In this i have tried the regulated psu
It was running perfectly with loads other than motor like led, fan etc...

But as soon as i have connected this psu with motor by switching the voltage level to 53V
as max voltage levels were 73V
No power was given to the motor
Because of which i thought to go for a simple way i.e, without regulated psu
But now which comes to a conclusion i have to move back to regulated psu

SCHEMATIC IS ATTACHED FOR HALF PORTION
lm338 voltage regulator is used

[Brumby]

That circuit is very basic.  Since you want to parallel two of these, there are more things to consider.

Personally, I would take a different approach - but rather than muddy the waters with my thoughts, I will defer to those who have had more experience in building custom power supplies.

[VickySalunkhe]

Sir
I would really like to know your approach

As i have searched a lot for regulated power supply for 48V 10amp
But i found no results for it
And certainly results which were showing up were 24V 1amp
That's where this circuit belongs to just some changes with cap and voltage regulator were done.

[IanB]

As i have searched a lot for regulated power supply for 48V 10amp
But i found no results for it

That's because it is not reasonable or practical to build such a supply from commonly available inexpensive parts or from simple instructions. In simple terms, it is not possible to make one.

You can buy a switched-mode supply with those specifications that has been designed and engineered by a manufacturer with the necessary expertise, but you cannot make one yourself as a hobby project. I can't do it and you can't do it.

So, to get what you require you need to obtain four 12 V lead-acid batteries and wire them in series to make 48 V. That will work, you can do it yourself with readily available parts, and it will perform exactly as required.

[VickySalunkhe]

Sir
But wouldn't it be a problem to lead acid battery to provide current of 13amps
As I have read the configuration of lead acid battery which are available there output is about 1amp

Can i draw more current from them without harming them?

[IanB]

It depends on the size of the battery. A car battery or a motorcycle battery can provide very high currents, up to hundreds of amps for short periods. Any moderately sized battery can provide 10 amps.