# Thread: Power supplies for UHU

1. If you were using geckodrives, a switch between the capacitors and the gecko would be a very expensive idea.....goodbye gecko.

2. Thanks for updating me G' ,

meanwhile an electrical engineer friend and colleague gave me some horrific facts about my powersupply.

1. For the capacitance(25000uF 150V) I am using the initial surge current would go upto 1kiloamps.
2. The bridge bkpc 5010 has a surge current handling capacity of 400 amps, beyond which the bridge will blow, like it did in my case.
3. He calculated that the bridge can take upto 11000uF capacitance at 110V.
3. The capacitance might burst in use bcos is just 40V above supply voltage.
4.he wasn't sure why the 8 amps fuse blew near the UHU servo and asked me to run the motor again in the lower PS and confirm if mosfet have survived.

well he filled up a full sheet with calculations and I think he is correct, he wanted me to use some sort of pi filter which he would design with a new powersupply.

I have asked him to design supplies individually for each motor so that I can switch on each one individually? I donno if I am doing right or wrong.........

then I just mentioned to him that design the best economical solution and he wasn't sure if a single PS would be economical or separate ones?

What ever the case I am looking again at around 200USD (8000INR) + investment on this.

hope some one else has better ideas!

RGDS
Irfan

3. Hi Irfan,

What you need to do is limit the inrush current.

Attached is a schematic of a simple way of doing it that I think will work in your case.

The relay, K1, should be a 24V one with contacts rated at the expected maximum current on the primary side (10A will probably be fine). To calculate the value of R6 take your DC supply voltage, subtract the relay coil voltage and divide by the relay coil current.

Let's assume:
DC voltage: 110V
Relay voltage: 24V
Relay current: 55mA

110V - 24V / 0.055 = 1560ohm, (1.5k will be just fine)

The power rating of the resistor is calculated as (I^2 X R) so with the above numbers you'll get: 0.055*0.055*1500=4.5W (use atleast 10W)

So, how does it work.... When you power up the circuit the realy isn't energized so the current to the transformer need to go thru the 47ohm resistor, this limits the inrush current to around 5A. The voltage in the capacitor bank starts to build up and when it reaches a certain point the relay will energize and bypass the resistor, enabling the full power of the supply.

There are more advanced ways of doing this with timing relays etc but I think this should work in your case. If the relay you use is has two poles you could put a second resistor between the rectifier and the capacitors and bypass it with the other contact in the relay. Make sure it will handle the secondary current though.

/Henrik.

4. Henrick,

If i decide to put in a resistor in between the bridge and the capacitor bank, then what sould be its value?

RGDS
Irfan

5. Hi Irfan,
It's not at all critical, somewhere between 10-100ohm would do it. Remember, it's just used to limit the initial inrush current, when the capacitors are empty.

Use wirewound resistor of 25W or more. With a 47ohm resistor you'll probably get a delay of around 0.5-1s before the relay activates - that should be more than enough to limit the inrush current you're seeing.

Good luck!
/Henrik.

6. Thanks Henrick you have helped me to sound sleep today, will carry on the modifications tomorrow and update all.

RGDS
Irfan

7. Originally Posted by contactirfu
Meanwhile one of the zone friend Manjeet suggest to me that If I use a switch between the Servo boards and the power supply after the capacitors then there will be an open secondary and I can have a delayed switching after the surge has stabilized, so I am going to have a DPST switch rated 20 amps in between the UHU servo power supply and the UHU boards. In future I will use a delay switch to switch on the PS for the UHU - hope it works!
DON'T Do That!!!, it will ruin your HP UHU. Opening the switch will create a high voltage surge due to the inductance of the cables.

8. Originally Posted by H.O
What you need to do is limit the inrush current.
Attached is a schematic of a simple way of doing it that I think will work in your case.
You could add a simple way to unload the capacitors when the circuit is switched OFF.
I have not tried this circuit.

Works like described in previous message.
And when mains is switched ON, the relay K2 comes up.
Which disconnects the resistor with K2-P1.
When mains is switched OFF relay K2 comes down.
And connects resistor via K2-P1 to the capacitors.
The capacitors unload via the resistor.

Does this work ?
Or did i make a mistake somewhere ?

See the circuit image below.
The zip file also contains the eagle .sch circuit. (in case someone wants to improve it :-) )

Vroemm

9. Here is a other way to unload the capacitors:
http://gsst.wikispaces.com/UHUpowersupply
From where comes the VCC in this circuit ?

Which method is the best ?

Vroemm

10. To be honest, I did not check the value of the capacitor you had spec'd in.
If you design for 5% ripple at your voltage/current, which is usually very adequate, the value would be 10,000µf, for this value there is no need to limit inrush at all.
Al.

11. ## Calculating Brake Resistor Value and power rating.

Here it is. This is a preliminary release, if you find errors, let me know.

Thanks,

Kreutz.

12. Hello Henrik, Kreutz, Al the man, Vroeem,

Thanks for all inputs, Vroemm the wiki talks about the torroids having the huge inrush current requirement which I feel is very true in my case. Today I am going to acquire the components required to alter the PS

here is what I am gonna get

1. Capacitors rated at 250V probably will take snap on types
2. 4x 45 to 100 ohm 25W resistors - just in case I require more
3. 3x 1.5K 10W resistors for switching the 24v relay
4. Double pole relay 24v rated 240v 30 amps on switching side.

I will work on Kreutz calculations - as my brain need to still get accostomed to the wordology of the content.

hope will have good news soon.

RGDS
Irfan

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