Reducing voltage without losses?

[FilipSebik]

Hey, I am making my own 37V 8A variable power supply. The problem is that my IC for regulating the voltage handles only 40V input voltage but the output from 30V transformer after rectifying is 43V. Is this a big problem or should I put some diodes for the voltage input pins of the IC to reduce the voltage to 40 or 39V?

Here is the scheme

And the input voltage pins for the IC are 11 and 12.

IC Datasheet

[mariushm]

You have a transformer that outputs 30v AC which is then rectified to :

 

Vdc peak = Vac x sqrt(2) - 2 x Vdiode  =  ~ 1.414 x 30  - 2 x ~ 1v  = ~ 40v    (Vdiode = voltage drop on the bridge rectifier diodes)

 

But assuming you may have variations in the mains voltage which will raise the output of the transformer a bit, and that the transformer's output is higher at low currents, let's go with 43v peak DC voltage.

 

That's peak voltage, the capacitors after the bridge rectifier are there to smooth the output and make it flat.

 

The capacitance required after the bridge rectifier can be approximated with the formula :

 

C = Current / [ 2 x FrequencyAC x (Vdc peak - Vdc min) ]  - where Vdc min is the minimum voltage you're willing to have in your circuit.

 

From the above formula :  

 

2 x FrequencyAC x (Vdc peak - Vdc min) = Current / C

Vdc peak - Vdc min = Current / ( 2 x FrequencyAC x C)

 

You have 7000uF in the schematic, but let's say you go with 10000uF or 0.01 Farads and if you're in US where you have 60Hz mains frequency.

 

Then 43v - Vdc min = 8A / 120*0.01 = 6.67    therefore, the minimum DC voltage at 8A using 10000uF will be : 43 - 6.67 = 36v

 

edit: but I see you're in Europe where we have 50Hz mains, so 43-Vdc min = 8A / 100*0.01 = 8/1 => Vdcmin = 43v - 8v = ~ 35v minimum voltage

 

So unless you raise capacitance a bit, you won't get your aim of 37v, especially since you'll have some additional voltage drop on the pass transistors as well (at least around 0.6v)

 

-

 

Yeah, you could use diodes in series but at 8A of current, each diode in the series will dissipate around 8 watts. For safety, you'd want at most around 38-39v so you'd probably have to use 4 or 5 such diodes in series.

 

Really, it would be better to just go for a 24v AC transformer and use more capacitance and make it a 30v DC max power supply

 

Or just consider not using LM-723 , plenty of power supply schematics that don't even use ICs , only some opams and references like *431

 

I'll attach a bunch of them below, they're for max 5A but can be extended to 8A easily

 

 

 

 

Q 1760_DickSmith_powersupply_circuit 30v 5a.pdf

Schematic Adjustable Linear PSU 1x30v 5A Circuit Specialists 3005x.pdf

Schematic Adjustable Linear PSU 1x50v 1A Circuit Specialists 3003x.pdf

[FilipSebik]

1 hour ago, mariushm said:

-SNIP-

The IC doesn't draw 8A so the diodes will be fine and I will use 4.7mF 63V cap and 3x 680uF 50V caps to smooth out the voltage (using what I have)

And I already bought all the parts. so the IC will be fine with 4-6 1n4007 diodes that will lower the voltage.

[Unimportant]

4 hours ago, FilipSebik said:

The IC doesn't draw 8A so the diodes will be fine and I will use 4.7mF 63V cap and 3x 680uF 50V caps to smooth out the voltage (using what I have)

And I already bought all the parts. so the IC will be fine with 4-6 1n4007 diodes that will lower the voltage.

Should be fine, although I'd have gone with a reverse biased zener diode for a more stable voltage drop.

Off course, if you lower the supply voltage of the LM723 the maximum output voltage of your power supply will drop by the same amount.

[FilipSebik]

11 hours ago, Unimportant said:

Should be fine, although I'd have gone with a reverse biased zener diode for a more stable voltage drop.

Off course, if you lower the supply voltage of the LM723 the maximum output voltage of your power supply will drop by the same amount.

Yeah, now I powered the transformer on with FBR and smoothing cap and I see its 45.4V so I will need 9-10 1n4007 diodes to drop the voltage

[mariushm]

Just add a small linear regulator that can handle voltages up to 60v DC - the chip itself won't use more than a few mA.

 

Something like this would be super easy to use, and has a 12v dropout voltage so with 45v in at best you're gonna get 35v out... but you can just use 2 resistors to set the output voltage: http://uk.farnell.com/microchip/lr12k4-g/linear-volt-reg-0-1a-1-2-88v-to/dp/2448520

 

LM317AHVT would be half the price and can also handle up to 57v : http://uk.farnell.com/fairchild-semiconductor/lm317ahvt/ic-ldo-volt-reg-57v-1-5a-to-220/dp/1564497

 

[FilipSebik]

On 9/15/2017 at 0:06 PM, mariushm said:

Just add a small linear regulator that can handle voltages up to 60v DC - the chip itself won't use more than a few mA.

 

Something like this would be super easy to use, and has a 12v dropout voltage so with 45v in at best you're gonna get 35v out... but you can just use 2 resistors to set the output voltage: http://uk.farnell.com/microchip/lr12k4-g/linear-volt-reg-0-1a-1-2-88v-to/dp/2448520

 

LM317AHVT would be half the price and can also handle up to 57v : http://uk.farnell.com/fairchild-semiconductor/lm317ahvt/ic-ldo-volt-reg-57v-1-5a-to-220/dp/1564497

 

I could use LM317HVT and but it costs 2.63 Euros. But its a better option than using 10 diodes. thanks