Ways to efficiently reduce voltage with a small form factor

Thread Starter

jlawley97

Joined Oct 5, 2019
18
Hi I have been trying reduce from 115Vac to 15Vdc to power a chip. I have found some cleaver methods like the one i am have attached with drops it down to 15V(to power a chip not shown) and also then be dropped further to 5V for other components. I am getting 36mA into the drain and so when I test the actual circuit the MOSFET is getting hotter than Id like. I need atleast 20mA to power an LED down the line which is why I dont drop the amperage.

I just wanted to put this out there to see if anyone had so more cleaver methods I can try. Space is definately the main issue here as I need to fit into a space 31mm by 50mm by 22mm so I cannot use transformers or inductors that are large enough for buck converters.
thanks!
 

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Thread Starter

jlawley97

Joined Oct 5, 2019
18
While transformerless power supply circuits now seem to be allowed on this site, many of us won't help with them due to safety issues.
Well I don't know if this will help but I am not a hobbist but rather an engineering working in a firm with many the safety precautions being mandated before testing anything.

Also there will be isolation elsewhere but that is not my jurisdiction

thanks for the quick response either way
 

Ian0

Joined Aug 7, 2020
3,280
This forum can be so hypocritical about high voltage circuits.
Off-line flyback is OK. Inverters which give 230V out are OK, electric fences which give several kV, and valve amplifier are OK.
but capacitive and resistive droppers aren't.

Have a look at:
https://www.power.com/products/linkswitch/linkswitch-tn2
which are designed to replace capacitive and resistive droppers. The efficiency is MUCH higher as @Papabravo says.

I think a 20A MOSFET might just be overkill, and 1uF across the output of the TL431 might make it into an oscillator - the compensation capacitor goes between output and input (cathode and adj).

If a linear supply is good enough and doesn't get too warm, have a look at depletion-mode MOSFETs.
http://ww1.microchip.com/downloads/en/AppNotes/AN-D66.pdf
because there is no resistor required to the high-voltage side, which is less opportunity for something nasty to get in and blow the gate oxide on the MOSFET.
 

Thread Starter

jlawley97

Joined Oct 5, 2019
18
I do have a SLIGHT problem with my current mosfet getting warm(IRF1644G) that has a 65C/W thermal resistance. Considering the voltage drop of about 100V and 40mA I am still getting 4W. Why would a depletion type be more useful in my case?
I was going to use a depletion type for another part of my circuit but it seemed like no one used them so I just shied away.

I may have the space to do the offline dc-dc converter the only thing holding me back right now is again the inductor but it may work, I just have to some component tetris

thanks guys
 

schmitt trigger

Joined Jul 12, 2010
464
A simple way to reduce AVERAGE DC input level and thus average dissipation is half wave rectification with minimum filtering.
The key is to allow te ripple voltage to drop very close to the regulator’s drop out.

Otherwise use the Linkswitch TN2 as suggested previously, and utilize every cubic milli meter of available space.
 

Ian0

Joined Aug 7, 2020
3,280
You’ve already got a big advantage being on a 120V supply.

To see the advantage of the depletion mode, look where the gate resistor goes.
Depletion mode - to the regulated supply, which is quiet.
enhancement mode - to the incoming noisy mains.
there’s no thermal advantage either way.
 

ronsimpson

Joined Oct 7, 2019
1,475
I may have the space to do the offline dc-dc converter
1619443879979.png
1619444218260.png
If you want to use a switcher and know how to be safe with non isolation then this is a example. LinkSwitch This is the PWM equivalent of your linear regulator.
Certainly this power supply is as unsafe as a capacitor a capacitor divider supply that we can not talk about on this forum. But if you go that rout use the circuit with a full wave bridge in it. (not the one diode version)
 

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Thread Starter

jlawley97

Joined Oct 5, 2019
18
question about the LNK320 family.
I am unsure what is controlling the voltage output. From what I gather is that 2 resistors are in the divider configuration and this is what stabilizes the voltage out and the inductor just needs to be sufficiently large. If this is correct I assume the IC is doing something with PWM is it changing the frequency or the duty cycle?

Also according the data sheet one can use a 680uF inductor at 180mA. So I ended up using the formula for calculating inductance and I wrote a little code to calculate it for me so I don't make any mistakes, but I obviously have made some mistake as the inductance I get is orders of magnitude larger than what it should be. If anyone sees anything egregious lmk
 

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Ian0

Joined Aug 7, 2020
3,280
During the "power switch off" part of the cycle in a buck regulator, the voltage across the inductor is the same as the voltage at the output. D2 produces an identical voltage across C10, which is now referenced to the source terminal of the regulator, which is the reference point for its feedback resistors.
 

BobaMosfet

Joined Jul 1, 2009
1,850
Hi I have been trying reduce from 115Vac to 15Vdc to power a chip. I have found some cleaver methods like the one i am have attached with drops it down to 15V(to power a chip not shown) and also then be dropped further to 5V for other components. I am getting 36mA into the drain and so when I test the actual circuit the MOSFET is getting hotter than Id like. I need atleast 20mA to power an LED down the line which is why I dont drop the amperage.

I just wanted to put this out there to see if anyone had so more cleaver methods I can try. Space is definately the main issue here as I need to fit into a space 31mm by 50mm by 22mm so I cannot use transformers or inductors that are large enough for buck converters.
thanks!
There are many methods. Your lack of understanding how to control current suggests to me that your not all that experienced in electronics. Nothing wrong with that. Here are some useful tips-

In the DC realm, resistors control current generally speaking. In the AC realm however, which is a frequency/time domain, you use capacitors. Learn how to control AC current using capacitors. Normally you want to use SAFETY capacitors when dealing with MAINS or any type of high-energy to person (aka step & touch) circuit because if the capacitor fails, you want it to fail in a way that helps make the circuit 'safer', not more dangerous.

Here is AN954, which is an excellent reference on such things and should help you a great deal.
 

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DrBearEE

Joined Feb 3, 2020
2
I'd recommend just buying the smallest USB phone charger cube you can find, they usually make 1 Amp at 5 V. You can change the connector and the enclosure carefully when nothing is plugged in, so it can be done safely. Like everyone else here, I'm going to remind you that mains voltages can kill you, so work carefully and be sure you know what you are doing to make connections safely.
 

Thread Starter

jlawley97

Joined Oct 5, 2019
18
I'd recommend just buying the smallest USB phone charger cube you can find, they usually make 1 Amp at 5 V. You can change the connector and the enclosure carefully when nothing is plugged in, so it can be done safely. Like everyone else here, I'm going to remind you that mains voltages can kill you, so work carefully and be sure you know what you are doing to make connections safely.
I ended up using a linkswitch-tn2 as a buck converter and my power supply (AC """mains""") in this case is limited to .4 amp output so its safe, I havent been shoving it into a wall socket !
 

DickCappels

Joined Aug 21, 2008
7,743
This forum can be so hypocritical about high voltage circuits.
Off-line flyback is OK. Inverters which give 230V out are OK, electric fences which give several kV, and valve amplifier are OK.
but capacitive and resistive droppers aren't.

(Some text removed for clarity)
Allow me to bring you up to date. The prohibition against transformerless power supplies was dropped from the User Agreement over a year ago. Capacitive droppers etc. may be discussed.

May I offer an idea. This does not isolate the load from the AC so you have to be careful of to what you connect it. No transformer to design and no EMI testing required.

1624430299059.png

The way this works is that the 100 uf capacitor (feel free to use other values) charges from the start of the positive going zero crossing until the revers emitter-base voltage of the 2N2222 reaches avalanche, about 6 volts. This causes the 2N2907/2N2222 pair to latch ON, turning the source follower off and keeping it off until the next positive-going zero crossing. This way the MOSFET does not dissipate power when the input voltage is high and everything can run cool. The clipping voltage may be raised easily.

1624430329161.png

Many variations are possible.
1624430854275.png
This one regulates the voltage used to charge the capacitor.

1624430969029.png
This one uses a triac to control the clipping action so the capacitor is charged after both zero crossings.

Just make sure this does not power an earth-referenced circuit, including those that include people!
 

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ronsimpson

Joined Oct 7, 2019
1,475
You can (must) pull 35mA from C2. The two Zener diodes limit the voltage at 5.1V & 15V.
C1 must be a "power line" capacitor. R3 is (more or less) fuse. R1 discharges C1.
Full wave bridge. 200mA to 1A diodes. Only sees 15V.
C2 probably should be 1000uF for less ripple.
Total current is 35mA set by C1 and 60hz. With no load the Zener diodes take the current. If you have a 10mA load on the 15V then the diodes will take 25mA.
1624497317430.png
What is the load on 15V? Load on 5V? current & voltage for LED?
1624497608571.png
Most of the power supplies we talked about live on the power line and will hurt you it you are not careful. The "0V" output is not at ground but it is at line voltage. This works well for light bulbs where we do not care if the LEDs are at the power line voltage.
 
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