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DC110V to DC48V

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EmmKay

New Member
I have a dc PSU which gives 110V rectified at its output. I need 48V from that power supply. There is no space for step down transformers.. can I use a transistor such as 2N3055 based circuit to get the required output..?? the out doesn't need to be regulated just in the vicinity of 48V +-3V. The current load will be around 10Amps. plus it will be on for atleast 8 hrs in one go. Fan based cooling is already available in PSU case + space for a 3 x 4 x 1.25 inch power board.!

I do not have access to high side voltage regulators.. but can get any type of power Transistor (Well mostly any type!).

What I had in mind was something like a voltage divider feeding the base of a small pn transistor and that inturn driving a large NPN or preferably a PNP power transistor (or number of transistors in parallel) to get the required voltage output.

Am I going in the right direction or or going to gaga land..!!!!!!

Any +ive ideas, suggestions and/or schematics are welcomed.

Cheers all

Emmkay
 

Hero999

Banned
Is that 110VDC? Or is it the DC side of a rectifier fed be 110VAC?

What's the nature of the load?

You could just use a resistor.

The problem is, if you go down the linear route, you have to burn 620W of power.

If the load is purely resistive, you could simply PWM it which will use very little power.

I think a switching regulator is your best bet?
 

EmmKay

New Member
I should have been more clear, the PSU supplies 110VDC at the outputs and drives a spooling DC motor on a machine. The PSU is rated for 40Amps the main motor is rated at 100V and 10 amps ... I need to drive another motor rated at 48V through an H bridge. The 2nd motor is rated for 48 +-3V at 10amps. I do not want to construct another PSU and wish to use the existing PSU as it has ample capacity to support the 2nd motor.

I just have to make certain that I can get 48V @ 10 amps with generating LOTS and Lots of heat!..
 
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Hero999

Banned
Is the h-bridge rated for 110V?

You could just limit the duty cycle so the motor operates at the same speed as it would if it were operated off 48VDC.

I just have to make certain that I can get 48V @ 10 amps with generating LOTS and Lots of heat!
Is that a typo, don't you mean without generating lots of heat?
 

fernando_g

New Member
As HERO999 mentions, if you limit the duty cycle to no more than 43.6%, you'll be able to drive the motor directly from the 110 VDC supply.

Of course this trick will only work with brush-type permanent magnet DC motors. Do not attempt this on a brushless DC motor, as the drive electronics whic performn the commuting function will not tolerate such a high voltage>
 
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Hero999

Banned
I would say you need a duty cycle of (48/110)² = 19%.
 

crutschow

Well-Known Member
Most Helpful Member
I would say you need a duty cycle of (48/110)² = 19%.
For a buck regulator operating in the continuous mode the output voltage is proportional to the duty-cycle. Why did you use the squared term? The duty-cycle is only slightly affected by the power output (as determined by the internal loses).
 

Hero999

Banned
It's not a buck regulator.

IYou get the squared term from RMS voltage of a squarewave which comes from Ohm's law.

Suppose you have a 12V 6W heater you want to run from 24V. You think, I know I'll reduce the duty cycle to 50%, no it doesn't work out that way as I'll explain.

Assuming the heater is Ohmic, the power dissipated by the heater will double when run off twice the voltage at half the duty cycle.

R = V²/P = 12²/6 = 24Ω

Increase the voltage to 24V:
P = 24²/24 = 24W

Average power at 50% duty cycle:
P = 24/2 = 12W

So your 6W heater will fail because it's now dissipating 12W when it's only designed to dissipate 6W. Therefore you need to set the duty cycle to (12/24)² = 25% in order to power it from a 24V supply.

I know the motor isn't an Ohmic load so in this case I'd recommend starting with a duty cycle of 19% and increasing it if it's too slow.
 
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crutschow

Well-Known Member
Most Helpful Member
OK. I understand your reasoning. I was assuming the motor was an inductive load. Certainly, an ohmic load requires the squared term. As you noted, it depends upon what the motor load looks like (how inductive or resistive).
 

EmmKay

New Member
@hero999 the H-bridge driver that came with the motor is rated for a max of 55V thats where the problem is .. and yes that was a typo.. the idea is NOT to generate lots and lots of heat :)

Wouldn't brushed motor appear as an inductive load rather than an ohmic load..! or would it depend on the function that the motor is performing!!!!!

cheers!
 

fernando_g

New Member
A DC brushed motor has rotor winding resistances which are ultra-low. What actually regulates the amount of current consumed by the motor is the counter-EMF produced by the rotation of the armature in the magnetic field.

In equation form:

Ia = (Vdc - Vemf)/ Ra

Ia Armature curren
Ra Armature resistance
Vdc your supply voltage
Vemf counter-EMF voltge, proportional to rotational speed.

Which means that the motor appears as a variable voltage battery opposing the supply voltage. Like all real circuits, there will also be some stray inductance, which you can use to your advantage to smooth out the current peaks of the PWM-modulated voltage. PWM switching frequency will be necessarily be high.
 
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