Is there a simple way to go from 40vdc to 5vdc without the need for transformers or inductors?
I was going to use an LM317, but then found out that its datasheet specifies a maximum of 30vdc at its input.
Maximum power drawn will be about 20mA @ 5vdc....
Or maybe I should just use a zener for this?

 

Is there a simple way to go from 40vdc to 5vdc without the need for transformers or inductors?
I was going to use an LM317, but then found out that its datasheet specifies a maximum of 30vdc at its input.
Maximum power drawn will be about 20mA @ 5vdc....
Or maybe I should just use a zener for this?

you could 'series' a number of rectifiers such that their collective Ef 'drop' brings the input EMF within 'range' of the linear regulator...

Although this may achieved with a single reverse-biased Zener, the power (and, hence, reverse current) specification may be 'disappointing'...

Best regards
HP

 

There is a suffix H that goes to higher voltage.

 

The TI datasheet says that the LM317A has a max input-to-output differential of 40V, so you are within that.

 

The TI datasheet says that the LM317A has a max input-to-output differential of 40V, so you are within that.

Yes -- Be certain, however, that you are within the 'safe area' RE: [Ei-Eo] @ I_out...

Best regards
HP

 

I would go with the LM317HV, that #12 suggested, to give you some margin, particularly if you accidentally short the output.

 

Or you could use an 'amplified diode' to drop some volts to supply the standard LM317.

 

I would not discard the use of a switching buck regulator. There are lots of them available with quite simple applications. Here is just one: http://www.ti.com/lit/ds/symlink/lm2825.pdf

It doesn't even use an external inductor. NB: That is actually quite an old device. So, I suspect similar and smaller regulators are now available.

John

 

Hi John,

I would not discard the use of a switching buck regulator.

I was thinking this way also until I noticed the "no inductor" requirement. Power dissipation in the regulator will be low at the current the OP is talking about.

 

Is 40V your maximum input voltage, or is that the nominal? If nominal, what is the maximum? What is the minimum input voltage?

 

If you using zener then count the current about 30mA including the current of zener and load, the formula as below:
Vin = 40V
Vout = 5V, 5.1V zener
I_total = 30mA
R_limit = (Vin-Vout)/I_total
= (40V-5V)/30mA
= 35V/30mA
= 1.167 K
Choosing 1.2K
R_Watts = 35V*30mA = 1.05W, choosing the current limiting resistor as 1.2K/5W.

 

Is there a simple way to go from 40vdc to 5vdc without the need for transformers or inductors?

One reason I usually go for Toroidal transformers when building supplies, only takes a small overwind of ~10 turns for a 5v supply.
Max.

 

Is 40V your maximum input voltage, or is that the nominal? If nominal, what is the maximum? What is the minimum input voltage?

it's exactly 40V, regulated

 

it's exactly 40V, regulated

Ah. In that case, either a Zener or the LM317HV would do.

 

Ah. In that case, either a Zener or the LM317HV would do.

I'll check into it, thanks!

 

lt1074 buck chip...


http://cds.linear.com/docs/en/datasheet/1074fds.pdf

 

lt1074 buck chip...
http://cds.linear.com/docs/en/datasheet/1074fds.pdf

That's a 5 amp switcher with an inductor needed. Doesn't quite match the request for, "no inductor".

 

I see no reason to use a switcher for only 20mA of current unless it's a battery source and you need maximum efficiency.

 

Quite a while back I had to design some equipment for a mobile generator application running 240vdc.
The Op-amp circuit was probably no more than 20Ma, I ended up using a resistor, zener and electrolytic there are many of these units in place with no problems to date.
Max.

 

Quite a while back I had to design some equipment for a mobile generator application running 240vdc.
The Op-amp circuit was probably no more than 20Ma, I ended up using a resistor, zener and electrolytic there are many of these units in place with no problems to date.
Max.

Ok... maybe I've just been "drowning in a glass of water" as we like to say down here.

Here's the complete picture:
I want to power up a LED in an optoisolator (H11L2M) that works at a maximum of 30mA. I've used this opto before with a 5V power source and connected it to a 270 ohm resistor to limit its current to below 18mA, and it's been working just fine all this time. I don't know what the emitter's voltage drop is for this device, but I'm guessing it's not too much.

Anyway, I'd now like to feed the same optocoupler's emitter with a 40VDC regulated supply (never mind for now what the conditions on its output are going to be, I've got that covered). So I've been thinking about simply using a resistor voltage divider.

Now according to this site, I've plugged in the values of

• Vin=40, R1=2200, R2=1800, RL=330.

And the results are:

• Vout, no load = 18V
• Vout, with load= 4.5V (sounds good to me)
• Current through the load = 14mA (sounds good too)
• Power dissipation at the load resistor = 0.061W

So it seems I'm safe, if I just use this circuit considering the previous values, and I connect the LED in series with the load resistor. I know that the current through the LED will be a little less due to its voltage drop, but I doubt it will reach a level below 10mA, which is its threshold current, according to the opto's datasheet.

Any thoughts anyone might want to share?