0
\$\begingroup\$

MOSFET: http://ixdev.ixys.com/DataSheet/99175.pdf

Driver: http://www.irf.com/product-info/datasheets/data/ir2183.pdf

What factors should one consider in selecting a good MOSFET/Driver combo, and why?

Why would my selection be wrong or bad in accordance with those factors if I'm operating at 150kHz, 170VDC, 3-4A.

(I think this is a more time-independent/communally useful rephrase of my old question)

\$\endgroup\$
13
  • \$\begingroup\$ Who told you they are a bad choice? \$\endgroup\$
    – Golaž
    Commented Apr 20, 2015 at 16:33
  • \$\begingroup\$ @Golaž My measurements. It's 185nC which means that at 10V(gs) it'll be like 18.5nF. Which means that at a typical driver resistance, with no added R(g) you get 10Ohm * 18.5nF as t. And 2.3t for 90% charge is like 425.5nS. (I actually don't know how to calculate this, this seems right though according to my graph and the really, really bad rise times) \$\endgroup\$
    – ARMATAV
    Commented Apr 20, 2015 at 16:39
  • \$\begingroup\$ What are the current rise/fall times you're getting? What are you aiming for? \$\endgroup\$
    – Golaž
    Commented Apr 20, 2015 at 16:53
  • \$\begingroup\$ 1us rise and 1us fall without fully turning off. Like here; i.sstatic.net/yhZ2r.jpg I'm aiming for just a very square wave, so maybe half of that or something. I don't know. But this waveform with 1K load, 100K load, 0K load doesn't change. It's always very sloped and always doesn't turn the bottom MOSFET off because of the circuit construction. \$\endgroup\$
    – ARMATAV
    Commented Apr 20, 2015 at 16:58
  • \$\begingroup\$ My thinking is that if I boost the rate at which it can turn on/turn off it will reach 0V on the high side, instead of always being a little over Vcc and having a spike. That being said, I'm probably going to map out a new PCB since this one was so terribly crafted, as you can see in the waveform, and so I needed suggestions for better drivers/MOSFETs and also what constitutes a good one in either of those cases. \$\endgroup\$
    – ARMATAV
    Commented Apr 20, 2015 at 16:58

1 Answer 1

1
\$\begingroup\$

Here is an approach to quantitatively screen for a MOSFET most likely to match requirements. Equations used here will be based on those from the thread "micro, MOSFET, and DC motors", but will be rearranged and reformulated to better reflect MOSFET datasheet parameters.

Basic Static Criteria:

  • \$V_{\text{DS}}\$ ~ \$1.5 V_{\text{s-max}}\$ : \$V_{\text{DS}}\$ shouldn't be less, but also shouldn't be much higher than 1.5 times supplied voltage.

  • \$V_{\text{Drv-min}}\$ > 2\$V_{\text{th-max}}\$: If peak drive voltage is less, the FET channel conduction will not be fully enhanced.

  • \$\text{$\Delta $T}_{J-A}\$ < 50C : In the approach that will be shown, temperature rise and part thermal resistance will be used to set overall power criteria. The aim is to keep FET junction temperature less than 120C, which a temperature rise of 50C will do even if the ambient temperature is 70C. For a more reasonable ambient temperature of 50C a \$\text{$\Delta $T}_{J-A}\$ of 50C, of course, results in a junction temperature of 100C, which is what we'll use in the selection criteria.

Total power dissipated in the FET will be temperature rise divided by thermal resistance:

\$P_T\$ = \$\frac{\text{$\Delta $T}}{\Theta _{\text{JA}}}\$= \$P_{\text{cond}}\$ + \$P_{\text{sw}}\$ = \$R_{\text{ds}}\$ DC \$I_d^2\$ + \$ I_d V_s F_{\text{PWM}} \tau\$

where DC = duty cycle and FET switching time \$\tau\$ = \$\frac{2 R_g Q_{\text{mp}}}{V_{\text{drv}}}\$,

I will state, without proof, that the lowest power will be attained by having \$P_{\text{sw}}\$ = \$P_{\text{cond}}\$. Therefore in the following equations, both \$P_{\text{cond}}\$ and \$P_{\text{sw}}\$ will be replaced by \$\frac{\text{$\Delta $T}}{2 \Theta _{\text{JA}}}\$, or 1/2 \$P_T\$.

Dynamic Selection Criteria:

Then selection equations for \$R_{\text{ds}}\$ and \$Q_{\text{mp}}\$ can be written as:

\$R_{\text{ds}}\$ = \$\frac{ \text{$\Delta $T}}{3 I_d^2 \text{ DC } \Theta _{\text{JA}}}\$ : Recall that \$R_{\text{ds}}\$ here is scaled by 2/3 to account for junction temperature of 100C and positive temp coefficient of \$R_{\text{ds}}\$

\$Q_{\text{mp}}\$ = \$\frac{3 \text{$\Delta $T} V_{\text{drv}}}{4 I_d F_{\text{PWM}} R_g \Theta _{\text{JA}} V_s}\$

Example:

For this case the defining parameters are:

  • \$V_s\$ = 170V
  • \$F_{\text{PWM}}\$ = 150kHz
  • \$I_d\$ = 3Amperes
  • \$V_{\text{drv}}\$ = 10V
  • \$\Theta _{\text{JA}} \$ = 62C/W (for TO-220 or TO-263)
  • \$R_g\$ = 20 Ohms
  • DC = 0.5

These yield search parameters:

  • \$V_{\text{DS}}\$ = 250V
  • \$V_{\text{th-max}}\$ < 5V
  • \$R_{\text{ds}}\$ =\$\frac{\text{50C}}{\text{(3)(9)(0.5)(62C/W)}}\$ = 59.7mOhms
  • \$Q_{\text{mp}}\$ = \$\frac{\text{3 (50C)(10V)}}{\text{4 (62C/W)(3A)(150kHz)(10Ohm)(170V)}}\$ = 1.28nCoul

Here is the search result from Digikey

The best match was an IPP600N25N3 , which had a \$Q_{\text{mp}}\$ of 3nCoul, so in order to meet power dissipation requirements either \$F_{\text{PWM}}\$ would have to be lowered to about 50kHz, or a heat sink would be needed to lower thermal resistance.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.