using inline diodes to drop regulator input

[brownt]

I have a AMS1117 5 volt regulator powering a system that draws about 90mA.
The only option for input to the regulator is a 12V plug pack. The regulator is good for input voltage up to 15 volts, but the drop across the regulator is creating a lot of heat. Too hot to touch. The regulator is mounted on a small PCB with a copper pad that acts as a small heatsink.

I have put three 1N4001 diodes in series with the supply, dropping it to 9.9 volts, and the heat on the regulator is OK.

Is there a problem with dropping the voltage that way? Or is there a better solution given the circumstances.

[TERRA Operative]

I think it comes down to that you are still dropping the same voltage in the end, so you are basically still generating about the same heat, just spreading it out over the diodes as well. Not a bad thing as each part sees less heat. Just means your cost and number of parts goes up..

You could put a buck converter in if possible to increase efficiency and reduce heat, or maybe epoxy a small heat sink onto the vreg.

[Siwastaja]

Not a problem. A very typical way to share heat dissipation away from the linear reg. Resistors are used, as well, but diode offers more constant voltage drop, i.e. varies between about 0.5V-1V depending on load, logarithmically.

You may want to consider SMPS for efficiency, energy saving reasons, and it would simplify thermal design.

[brownt]

what sort of epoxy would you use for this?

[digsys]

what sort of epoxy would you use for this?

If it's surface mount, then your only easy choice is to solder copper foil "tabs" along the edges of the reg, say 10mm H (careful not to short IGO)
The idea is to move the heat off the pcb and into thermals. You can also use 3 diodes in series in a tent formation. That will drop the Vin even better.

[Conrad Hoffman]

If the load is reasonably constant, it's also common to put a power resistor across the regulator so most of the current is passed through that, rather than the regulator, which then just has to do minor adjustments to hold the voltage steady. No matter what you do, other than going to a switcher, the total power dissipated will be the same, just moved to different locations.

[Zero999]

Don't connect a power resistor across the regulator, unless you know the current will always remain high enough for it to drop enough voltage, otherwise the output voltage will rise, when the load is disconnected.

Use a series resistor, to drop the voltage before the regulator. The regulator you're using has a drop-out voltage of 1V, a 1A probably much less at 90mA (I haven't read the data sheet in great detail). You can drop 6V across the resistor, which at 90mA would require:

R = V/I = 6/0.09 = 66²/₃Ω resistor.

So use the next E12 value down, which is 56R.

The resistor will need to dissipate at least:
P = I²R = =0.9²*56 = 0.9*0.9*56 = 0.0081*56 = 0.4536W = 453.6mW

So a 500mW resistor would do, but go for 750mW or more, just to be on the safe side.

Note that there should be a capacitor connected across the regulator's input, after the series resistor: go for 1µF and 100nF in parallel.

[Siwastaja]

Series resistor before the regulator input works well even with non-constant current draws, because at lower currents, the reg can handle the dissipation anyway. You just size the resistor so that the highest current still maintains the minimum dropout voltage for the resistor. This, of course, doesn't help if there is a problem with the maximum voltage rating of the reg. In this case, diodes are much better, but beware of the simplification to Vdrop=0.7V/diode - it varies more, and can be down to 0.4V with extremely low currents.

[duak]

I haven't used series diode strings for any production designs, but I have use them with laptop supples to run other pieces of equipment without having to do more complicated things.

The 1N4001 diodes have a negative temperature coefficient so as their temperatures increase, their forward voltage drop decreases.  At 90 mA, this is a minor concern unless the diodes are packed together in a small volume with minimal heat sinking.  Too bad there isn't a power version of a band gap reference (eg. LM385) that has a more defined voltage drop.  I think one of the older temperature sensors (LM35?) comes in a TO-220 package but its current limit is far too low.  It does have a positive temperature coefficient, its primary function.

There is a simple transistor circuit called the Vbe multiplier that uses a voltage divider around a bipolar transistor to make a crude voltage reference.  It's still temperature sensitive though.  It wouldn't make sense to use it to replace 2 or 3 diodes, but it might be a better choice for higher voltages.   A small power transistor could easily handle the dissipation.

Cheers,

[ogden]

I have a AMS1117 5 volt regulator powering a system that draws about 90mA.
The only option for input to the regulator is a 12V plug pack. The regulator is good for input voltage up to 15 volts, but the drop across the regulator is creating a lot of heat. Too hot to touch. The regulator is mounted on a small PCB with a copper pad that acts as a small heatsink.

Too hot to touch is 60 degrees Celsius and above which for regulator is "mildly hot". LM1117 and "derivatives" are coming in various packages up-to TO-252 or even TO-220. Both are capable to dissipate more than 0.6W your case requires. I would just pick proper regulator rather than introduce diode or resistor as "heat dissipation helper".

Unless you are in mass-market where every penny counts, better consider switching regulator. 0.6W is lot of heat. Especially for small PCB.

[edit] Power dissipation graph for LM1117 packaged in SOT-223 (@ ti.com) attached

[Zero999]

I have a AMS1117 5 volt regulator powering a system that draws about 90mA.
The only option for input to the regulator is a 12V plug pack. The regulator is good for input voltage up to 15 volts, but the drop across the regulator is creating a lot of heat. Too hot to touch. The regulator is mounted on a small PCB with a copper pad that acts as a small heatsink.

Too hot to touch is 60 degrees Celsius and above which for regulator is "mildly hot". LM1117 and "derivatives" are coming in various packages up-to TO-252 or even TO-220. Both are capable to dissipate more than 0.6W your case requires. I would just pick proper regulator rather than introduce diode or resistor as "heat dissipation helper".

Unless you are in mass-market where every penny counts, better consider switching regulator. 0.6W is lot of heat. Especially for small PCB.

[edit] Power dissipation graph for LM1117 packaged in SOT-223 (@ ti.com) attached

I agree, that's a valid point, but we don't know if the original poster has used a large enough copper pad, as the heat-sink. If the regulator isn't shutting down and is regulating properly, then it's most likely fine.

[TERRA Operative]

what sort of epoxy would you use for this?

For this application, practically any good quality epoxy would work.
How big is your copper pad in the PCB, and can you make it bigger?

[brownt]

Like Araldite? or tarzan grip?

copper pad is barely larger than the fin on the regulator. Its one of those you buy already mounted on a pcb with caps.

[TERRA Operative]

Araldite would be fine. Use a clamp or clothes peg or similar to squish it down nice and tight so the epoxy is as thin as possible.

JB weld might be better for thermal conduction, and Arctic Silver epoxy would be best, but normal Araldite would be fine in this application.