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I have some questions regarding Transient Voltage Suppressors (TVS).

The specific datasheet I'm referring to is this, from 1SMB22AT3G.

Here is some of its specs:

  • \$V = 22V\$
  • \$V_c = 35.5V\$
  • \$I_{pp} = 16.9A\$
  • \$tp = 10us\$ (protection pulse width)
  • \$P_p = 5kW\$ (power rating)

My questions are:

  • How do I calculate its Junction Temperature at \$T_a = 55°C\$?
  • If the datasheet says the device can dissipate 5kW @10us pulse width, how is it that its \$R_{JA}\$ is 226°C/W?
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Is Pp 5kW or 0.6kW @1,000μS? –  GR Tech May 5 at 5:55

3 Answers 3

up vote 4 down vote accepted

Calculating the peak junction temperature for a short pulse is difficult. The 226°C/W is for steady state or dissipation that is much longer than the thermal time constant of the device.

1SMB5.0AT3 Pulse Rating Curve

The pulse rating curve (Figure 1, above) in the data sheet shows how much power the device can dissipate for a single pulse event. It's probably safe to assume that the pulses along the line cause the junction to heat to somewhere between 120°C and the maximum of 150°C. From that you might be able to estimate a thermal time constant and come up with a general formula for junction temperature for a given pulse.

You can see how that would work in this WikiPedia article: Thermal time constant.

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Thanks for the reply, I found more detailed datasheet here vishay.com/docs/88390/smaj50a.pdf , they are showing impulse impedance it in figure 4. Can you help me understand, is this for peak pulse rating and how to exploit this graph. –  Adi May 4 at 18:13
    
@Adi Sure, find your pulse width on the x-axis. Say it's 1 ms. Read the thermal impedance on the y axis, say it's 2 degrees C per watt. Now if your pulse is 10 watts your junction will rise 20 degrees over ambient. –  John D May 4 at 23:56
    
I am getting something wrong. from figure 5 , 1000microseconds corresponds to .8C/W and from figure 1, I get 0.3KW . the result is 240C. –  Adi May 5 at 7:39
    
The issue is that figure 1 defines a very specific type of pulse (see Fig. 3), which hits the peak power for a short defined time. Though they don't specify the pulse shape on the transient thermal impedance curve I believe it is specified for a pulse that stays at the peak power for the entire duration, a rectangular power pulse. Therefore for that type of pulse, at 1ms you can only dissipate something like 80W abs max. For the other waveform that only touches the peak power for a short time you can hit a peak of 300W. –  John D May 5 at 13:30
    
Thanks for clarification, but I am getting 240C Tj above ambient...... :-( –  Adi May 5 at 14:18

How about a quick calculation for power dissipation.

Suppose that 5 kW (5,000 W) was going to be dissipated for 10us. And also suppose that the pulse was to be repeated every one second. You could multiply:

$$\frac{5,000W \times 10us}{1,000,000 us}$$

Equivalent is:

$$5,000 \times \frac{10}{10^6} = 0.05W = 50mW$$

Additionally the temperature of the junction would then be \$(226°C/W) \times (0.05W) = 11.3 °C\$ above ambient temperature.

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The datasheet says Non-Repetitive, in your calculation you have repeated it every 1sec . It also shows figure 6 whih again says number of cycles.It says dutycycle 0.01 %. –  Adi May 5 at 7:50
1  
A duty cycle of 10us/1second seems well within the parameter that you cite. –  Marla May 5 at 23:52
    
@ Adi : Against guidelines, but, Glad you have found interest in high voltage and hi energy application. Keep going. This field needs interested persons like you. –  Marla May 6 at 0:10
    
@ Adi : One Unmentioned problem with repetition rate is Thermal Cycle. A temperature excursion from ambient to 200 degrees C (in short time, e.g. 10us) and then back to ambient. Each thermal cycle puts tremendous stress on the junction. I can't say for this device, but for high power IGBT's , a million thermal cycles will tear the junction apart. –  Marla May 6 at 0:16
    
Thanks, that you noticed my interest in High voltage , its definitely very interesting but components are hard to find (not the TVS). –  Adi May 6 at 7:45

TVS diodes are designed to withstand high junction temperatures those they have much larger die area than the conventional zener diodes. So don't try to calculate \$T_j\$ of a TVS diode at surging time, using the conventional thermal equivalent circuit. Even you cannot use the transient power-temperature or thermal time resoponse calculations as in regulator zeners.

For steady-state temperature the junction temperature, and using the minimum recommended footprint will be:

Derating power dissipation @55°C = 0.682W, so

$$Δt=θ_jL \times P_d= 25°C/W \times 0.682W = 15.7°C$$

$$T_j = 55 + 15.7 = 70.7°C$$

Soldering footprind recommended by the manufacturer shown bellow on a FR-4 standard board

enter image description here

The big difference in junction temperature between the two mounted methods is that this diode package is designed to dissipate better from leads than the case.

EDIT

With this new junction temperature, the power capability of the diode reduced to 420W for 1ms pulse, and the new brakedown voltage increased by 1.1 volt because TVS diodes temperature coefficient it is similar to Zener diodes.

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