## AP-6

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Any electronic device's long-term reliability is directly related to how
the overall system design
allows it to dissipate any generated heat properly. Likewise, the long
life and reliability of UltraVolt
high-voltage power supplies (HVPSs) will be enhanced through proper thermal
management.

Due to the laws of physics, the process of converting one voltage level to another cannot be done without energy loss; that loss, the power used to operate the power supply itself, is usually dissipated as heat. UltraVolt HVPSs have excellent efficiency ratings, ranging from 75% to 92% at full output power (depending upon the model), delivering full rated voltage at nominal operating conditions. The remaining fraction of the HVPS's input power is converted to heat. Methods of properly dissipating this heat can be found in this application note.

The operating efficiency of a power supply — one of the key variables in thermal calculations — can be determined in different ways. The Acceptance Test Procedure (ATP) that came with your UltraVolt power supply contains a few common operating points and related data. The ATP data can be used to determine the supply's operating efficiency at various points. Alternately, the efficiency of a particular UltraVolt HVPS under specific operating conditions can be obtained from our Customer Service Department. Typical operating efficiency is also listed in the DC Efficiency vs. Input Voltage Range graph of the power supply's data sheet. This graph shows only typical values for the various models in that series of power supplies.

The analysis and several sample calculations in this
application note are divided between *Units
without user-supplied heat sinks (and -H option units)*, and *Units
with user-supplied heat sinks
(and other cooling methods)*.

__Units without user-supplied heat sinks (and -H option
units):__

An UltraVolt 2A12-P4 is continuously delivering its
nominal 2mA at 2kV (*PDCout* = 4W) with a
nominal 12VDC input. The input current is reported as 390mA (*PDCin* = 4.68W),
based on the ATP
at this operating point (nominal output power, nominal 12V input voltage).

We can determine the heat energy dissipated as

*Pheat* = *PDCin* – *PDCout*

where

*PDCin* = *VinIin*

and the unit's efficiency is

*PDCout* / *PDCin*.

Solving for *PDCin* yields

PDCin = |
VinIin =
((12V)(0.39A)) |

= |
4.68W. |

So, Pheat = |
PDCin – PDCout =
(4.68W) – (4W) |

= |
0.68W. |

Thus, efficiency
= |
PDCout /
PDCin = (4W) / (4.68W) |

= |
86%. |

Therefore, this power supply will dissipate 0.68W of heat when operated as per the discussed operating conditions. At this operating point, its DC efficiency is 86%.

For an “A” Series UV HVPS (for example, the 2A24-P4) with the standard plastic case, no extra heat sinking, and all surfaces exposed to free air, calculate the temperature rise of the power supply's case under these conditions. Assume the HVPS is mounted in sockets on the printed circuit board (PCB).

First, the total surface area (*SA*) of the standard 4W “A” Series
must be calculated. The *SA* can
be calculated from simple geometry, since the HVPS has 6 sides consisting
of 3 sets of 2 equal
sides. (Dimensions can be found in the “A” Series datasheet,
where *l* represents Length, *w* is
Width, and *h* is Height.) Solving for surface area we find,

SA = |
2 (lw + lh + wh)
= 2 ((3.7in)(1.5in) + (3.7in)(0.77in) + (1.5in)(0.77in)) |

= |
19.11in^{2}. |

It is universally accepted that 1W will raise 1in^{2} approximately
100°C
for all surface area exposed
to free, still air (assumptions made for simplicity). Since we are
dissipating 0.68W,

∆T = |
(Pheat / SA)(100°C • in^{2} /
W) = (0.68W / 19.11in^{2})(100°C • in^{2} /
W) |

= |
3.56°C. |

Finally, assuming the enclosure within which the UV HVPS
is placed has a continuous operating
temperature of 40°C (*Tambient*), we have a final
UV HVPS case temperature of 43.56 °C ( *∆T* +
*Tambient*). Thankfully, this value is comfortably below the recommended
maximum “A” Series
operating temperature of 65°C. In practice, the actual case temperature
may be slightly different
than the expected value, depending upon the mounting of the power supply
and the heat
dissipation properties of that mounting.

The power supply will also have certain areas that are
warmer and other areas that are cooler
because certain components within the unit dissipate more heat than
others. These warm areas
are generally referred to as “hot spots.” Note, flush-mounting
this power supply onto a PCB will
hinder any convection cooling along the surface of the power supply
that faces the circuit board.
To account for this, the *SA* calculation of this particular
example is carried out by summing the 5
exposed sides, ignoring the cooling effect of the mounting surface.
Alternately, if the power supply
is PCB-mounted using sockets, then the mounting side will allow slight
cooling by convection; this
should be accounted for in the calculations using a scaling factor.
A discussion of mounting methods for UltraVolt high-voltage power supplies
appears in Application
Note #3.

As seen in the calculations above, a 4W unit running at full rated power rarely needs any supplemental heat sinking. However, a 20W, plastic-encased “A” Series running continuously at full power will dissipate approximately 3.5W, assuming 85% efficiency (which is a rough approximation used only for this example; do not use this number for your own calculations).

Without any supplemental heat sinking and with one surface of the power supply mounted onto a PCB, the case temperature will rise 26°C, an unacceptable temperature increase in applications with more than a 40°C ambient temperature. The design of 20W “A” Series, as well as all other UV modular HVPSs, facilitates easy interfacing to a heat-dissipation medium. The top surface of the 20W, plastic-case “A” Series is actually a 0.062”-thick aluminum plate, which will efficiently conduct heat to an external heat-sink medium, such as a chassis wall or other customer-supplied heat sink. This aluminum plate also helps regulate internal power supply hot spots, which occur because of the heat dissipations of different components within the power supply itself (as mentioned previously).

A 4A24-P20-H (a 4kV, 20W supply) delivering 2500V at 3mA
(much less than the rated 20W out),
draws .46A from its 24V supply. Solving for *Pheat*,

Pheat = |
PDCin – PDCout =
((24V)(0.46A) – (2.5kV)(3mA)) |

= |
3.54W. |

Therefore, this power supply will dissipate 3.54 watts of heat when operated, according to the operating conditions (68% efficiency due to the light load). For comparison purposes, a 4A24-P20-H running at full output voltage has been analyzed in the appendix of this Application Note, providing a typical 89% efficiency example.

For the previously mentioned power supply, flush-mounted to a PCB, with the -H (heat sink) option, and with all sides except the mounting side exposed to free air, calculate the temperature rise of the power supply's case.

First, the **exposed** surface area of the standard 20W “A” Series
must be calculated. From simple
geometry (dimensions from “A” Series datasheet), the SA of
the sides exposed to free air
(neglecting the mounting and the heat sink sides) can be calculated as

SA1 = |
2(lh + wh) = 2((3.7in)(0.77in)
+ (1.5in)(0.77in)) |

= |
8.01in^{2}. |

The -H heat sink adds approximately another 13in^{2} of
surface area, so the total exposed still-air
surface area is

SAtot = |
SA1 +
SAhs = (8.01in^{2}) + (13in^{2}) |

= |
21.01in^{2}. |

It is again assumed that 1W will raise 1in^{2} of surface area
approximately 100°C (in still air, which
we will assume here for simplicity).

Since we are dissipating 3.54W, we can calculate *∆T* as

∆T = |
(Pheat / SAtot)(100°C • in^{2} /
W) = (3.54W / 21.01 in^{2})(100°C • in^{2} /
W) |

= |
16.85°C. |

If the enclosure within which the UV HVPS is placed has a continuous operating temperature of 40°C, we will have a final UV HVPS average case temperature of 56.9°C. (Again, due to the differing heat dissipations of certain components within the power supply, certain regions of the unit will be warmer than others.) Of course, this value is below the recommended maximum “A” Series operating temperature of 65°C.

It is important to consider that all of our calculations have assumed the power supply is supplying a continuous amount of power. Low duty-cycle or pulsed operation of a UV power supply may dissipate a considerably smaller amount of average heat than the calculations here have yielded. In these situations, the HVPS may not need supplemental heat sinking. For example, a power supply operating at a 10% duty cycle will dissipate a small fraction of the continuous full-output dissipated heat (and hence can be operated without supplemental heat sinking).

Note, should these units be operated at less than full-rated output power, their operating efficiency will be less than their potential maximum efficiency. UltraVolt power supplies are designed to operate most efficiently when delivering their full-rated output power.

In summary, when dealing with an HVPS unit that does
not have any supplemental heat sinking
(or has only a -H heat sink), a few basic rules can be followed to
simplify calculations. When
calculating the surface area of the unit, ensure only sides that have
access to free, open air are
included in the calculations. If a side is in close proximity to a
surface, such as a high-voltage
power supply mounted with sockets onto a PCB, a fraction of the actual
surface area of the
hindered side(s) should be included in the calculations. If a -H heat
sink is used, be sure that its
side is summed into the equation for surface area only once (for that
particular side, the 13in^{2}
surface area of the -H heat sink is the surface exposed to air).

__Units with user-supplied heat sinks (and other cooling
methods)__

Through testing, an UltraVolt 6C24-P125 is determined to draw 5.9A from its 24V supply while delivering 6kV into its rated load.

Under the above operating conditions,
determine whether a fully enclosed metal container of
dimensions 12” x 6” x 4” (288in^{2} total
surface area) will sufficiently cool this HVPS if it is mounted
to the wall inside the chassis. (Only the exterior of the container
is exposed to the outside free air.)
All sides of the metal container are exposed to free air. Assume the
container is a good thermal conductor and has a low thermal-resistance
interface between the HVPS and the container.

Therefore,

Pheat = |
PDCin – PDCout =
((24V)(5.9A)) – 125W |

= |
16.6W (88% efficiency). |

Here, since the HVPS is mounted inside the
enclosed container, the surface of the HVPS will not
directly dissipate its heat to the outside air. Instead, it will transfer
its thermal energy to the
chassis. The only surface that can dissipate heat to the outside air
is the chassis' surface. The
cooling provided by the HVPS's surface inside this container is negligible
compared to the cooling provided by the heat transfer from the HVPS
to the chassis wall. We approximate the dissipating
surface area to be 12” x 12” x 2” = 288in^{2}.

Since we are dissipating 16.6W,

∆T = |
(Pheat / SAtot)(100°C • in^{2} /
W) = (16.6W / 288in^{2})(100°C • in^{2} /
W) |

= |
5.7°C. |

A chassis temperature increase of 5.7°C should be quite acceptable under nearly all circumstances. So in conclusion, the HVPS would operate quite satisfactorily from a thermal standpoint.

In the above example, we made some assumptions that
will now be discussed. First, it was
assumed that a low thermal-resistance interface was used between the
HVPS and the chassis
wall. This implies the HVPS can transfer its heat to the heat sink
- the chassis in the above
example - without hindrance. In practice, this can be accomplished
using an excellent heat conduction

medium, such as thermal elastomer (0.010” or 0.020” are
common thicknesses),
double-sided thermal tape, or thermal grease. This heat-conduction
medium is then placed
between the HVPS and the heat sink, and the HVPS is securely mounted
to the heat sink using
the proper bracket and/or mounting fasteners. These fasteners are then
torqued sufficiently to
guarantee the correct contact pressure on the heat-conduction medium.
For example, the #8 studs on a 60W unit should be torqued to 8ft • lb
of torque to ensure the proper ‘cold flow’ of
thermal-elastomer heat-conduction medium. Remember, in order to keep
the thermal resistance low between the HVPS and its heat sink (allowing
the greatest transfer of heat), the HVPS should
have as much surface area as possible facing the smooth heat-sink medium
(via the heatconduction
medium).

Second, it was assumed the heat-sink medium — the chassis wall in the above example — was a good conductor of heat. This assumption simplifies calculations by allowing us to utilize the entire open-air, exposed surface area of the heat sink as a dissipation medium (since the heat to be dissipated is easily conducted to the exposed surface area of the dissipation medium). Not surprisingly, heat sinks are usually made from metals known for their excellent heat conduction properties (such as aluminum).

When cooling an HVPS, many design variables enter into the equation and can determine how actual implementation is carried out. Airflow is one very important variable, and is, unfortunately, too complex to fully analyze here. Generally speaking, forced-air cooling from a simple fan can drastically reduce the size of a required heat sink, since the flowing air removes the dissipated heat more efficiently (by transferring the heat to a larger volume of air and not relying solely on convection). The ambient temperature of the HVPS's surroundings is also a very important variable. A 15°C temperature increase may be acceptable if the ambient is 25°C, but may be totally unacceptable if the ambient temperature is 60°C (due to other devices within the same chassis dissipating a large amount of heat, for example). The mounting method is another crucial design variable.

If the HVPS is mounted to a large chassis wall, then it is unlikely a supplemental heat sink will be required.

Oddly enough, one of the most important considerations when it comes to cooling design is to allow for a large margin of safety. Even common dust coating a heat sink can drastically reduce its cooling effectiveness. Obstructed or dust-coated fan vents can seriously reduce the cooling efficiency of the fans themselves. The importance of air-vent filters and regular cleaning becomes obvious. Also, load changes (due to slight design changes) can require more power to be delivered from the HVPS and, hence, change its actual heat dissipation. In summary, include a large margin of safety in your heat calculations to ensure your high-voltage power supply cools sufficiently under all conditions. This can be achieved simply by assuming the power supply will operate in an ambient temperature higher than previously calculated.

Although one can carry out thermal calculations using
thermal junction analysis and actual
thermal resistance values, we have not opted for that method here.
While they will yield more
accurate results, the methods described here will sufficiently approximate
the cooling
requirements for your HVPS, given an adequate safety margin. Complex
cooling requirements,
such as very limited space restrictions with forced-air cooling may
require a more in-depth
analysis than proposed here. In these cases, we recommend you consult
both a text on thermal
management and UltraVolt's Customer Service Department. Remember to
keep your UltraVolt
HVPS cool, and it will reward you with a long, trouble-free life.

Surface Area of UltraVolt Models | |
---|---|

UltraVolt
Model |
Total Surface Area (in square inches) |

“A” / with -F / with
-C / with -F-C |
19.11 / 25.26 / 28 / 37.69 |

“10A” / with -C |
20.46 / 29.2 |

“15A” / with -C |
25.26 / 35.4 |

“20A” / with -C |
31.5 / 43.2 |

“25A or 30A” / with
-C |
39.93 / 57 |

“C” / with -C |
19.11 / 28 |

60W/125W “C” |
54.28 |

250W “C” |
98.88 |

“8C - 15C” |
98.88 |

“20C - 25C” |
98.88 |

__Appendix:__

A 4A24-P20-H (a 4kV, 20W supply) delivers 4kV at 5mA (the
rated 20W out), and draws .94A
from its 24V supply, operating at 89% efficiency. The heat energy dissipated
is

Pheat = |
PDCin – PDCout =
((24V)(0.94A)) – ((4kV)(5mA)) |

= |
2.56W. |

This power supply will dissipate 2.56W of heat when operated as described above.

For the previously mentioned power supply, flush-mounted to a PCB, with the -H (heat sink) option, and with all sides except the mounting side exposed to free air, calculate the temperature rise of the power supply's case.

First, the **exposed** surface area of the standard 20W “A” Series
must be calculated. From simple
geometry (dimensions from the “A” Series data sheet), the *SA* of
the sides exposed to free air
(neglecting the mounting and the heat sink sides) can be calculated as

SA1 = |
2(lh + wh) = 2((3.7in)(0.77in)
+ (1.5in)(0.77in)) |

= |
8.01in^{2}. |

The -H heat sink adds approximately another 13in^{2} of surface area,
so the total exposed still-air
surface area is

SAtot = |
SA1 +
SAhs = (8.01in^{2}) + (13in^{2}) |

= |
21.01in^{2}. |

It is again assumed that 1W will raise 1in^{2} of surface area approximately
100°C
(in still air, which
we will assume here for simplicity).

Since we are dissipating 2.56W, we can calculate *∆T* as

∆T = |
(Pheat / SAtot)(100°C • in^{2} /
W) = (2.56W / 21.01in^{2})(100°C • in^{2} /
W) |

= |
12.2°C. |

Assuming the enclosure within which the UV HVPS is placed has a continuous
operating temperature
of 40°C, we have a final UV HVPS average case temperature of 52.2°C.
This value is comfortably
below the recommended maximum “A” Series operating temperature
of 65°C.

Rev. D1

***** END OF APPLICATION NOTE #6 *****