1
\$\begingroup\$

I will drive a load low side with IGBT. IGBT gate-emiter (Threshold) voltage will be 12 V. This voltage level is enough for my load current. I will use a gate driver. When ı search gate driver for low side Igbt always I saw current rating. What is it? I will use only voltage on gate of IGBT and any current will not be on gate of IGBT but I see sink or source current always.

Question two: I will use IR4426 (Infineon) for gate driver. Typical application was attached. According to this product, if I use microcontroller voltage level on input pin, I will to take 12 volt on output? Is that true?

Question three: Can I use this gate driver for this IGBT? IGBT will carry about 500 V and 5 A for 1-2 seconds.

Link: https://www.infineon.com/dgdl/ir4426.pdf?fileId=5546d462533600a4015355d60b491822

enter image description here Sorry for my basic questions. I'm a new hardware design engineer.

\$\endgroup\$
0
\$\begingroup\$

1) The current rating of a gate driver is how much current the driver can drive into the gate. During switching, the gate draws current - the more current you can supply, the faster it will switch. You can think of it like a capacitor - you need to charge it to reach the threshold voltage. The more current you supply, the faster the voltage will increase.

Think of it this way:

schematic

simulate this circuit – Schematic created using CircuitLab

The more current the gate driver can supply, the faster "Gate Voltage" will reach the Gate Drive voltage.

2) Yes, but you have to supply the 12V. You must connect it to the Vs pin. Some gate drivers have internal boost circuitry to create the gate drive voltage, but this one doesn't. You need to create 12V somewhere, and connect it to the Vs pin so the gate driver can pass it through to the OUT pins. In the schematic you've drawn you've connected the 12V to the GND pin. If you connect it to the Vs pin instead, you'll be on the right track.

3) Yeah sure. The gate driver doesn't really care what is passing through the collector/emitter junction. It's job is just to supply the necessary current and voltage to the gate. If you're happy that it will switch it on and off fast enough, then it is good enough.

Bonus answer, because it would be unethical not to mention it: switching 500V at 5A is not a task for "a new hardware design engineer". That's more than enough to kill you, your customer or an innocent bystander. Please make sure your design is reviewed by someone with more experience before you switch power on.

\$\endgroup\$
  • \$\begingroup\$ Thank you for answers. These are most useful for me and I'm showing my designs to my team leader. Thank you again. \$\endgroup\$ – Alican Ozpay Aug 20 at 13:34
0
\$\begingroup\$

IGBTs need quite some energy to switch, you need a high voltage to push in all the electrons needed to charge the insulated gate, which has quite same capacitance, thus a high current is needed to do that quickly. Of course this current is only needed until the gate is on it's desired potential, no static current will be drawn.

\$\endgroup\$
  • \$\begingroup\$ Thank you for answers. These are most useful for me. \$\endgroup\$ – Alican Ozpay Aug 20 at 13:35
0
\$\begingroup\$

When ı search gate driver for low side Igbt always I saw current rating.

You probably mean the: Output high short circuit pulsed current

You're right in that a gate doesn't draw current when it is on or off. But what happens when you switch it on to off and off to on? Note that the gate of an IGBT (or any other device with a MOSFET input) will behave as a capacitor.

The switch the gate voltage this capacitor has to be charged or discharged. That requires a current. A higher current means a faster charge or discharge. So that current rating relates to how fast you can switch the IGBT on/off.

If I use microcontroller voltage level on input pin, I will to take 12 volt on output ? Is that true ?

You mean: is the 5 V logic level at the input (from a microcontroller for example) levelshifted to 12 V levels?

Yes it is, that is an important function of this IC.

\$\endgroup\$
  • \$\begingroup\$ Thank you for answers. These are most useful for me. Also thank you for correcting me. \$\endgroup\$ – Alican Ozpay Aug 20 at 13:36
  • \$\begingroup\$ Every when I switch it on to off or off to on, gate of IGBT want to current ? How can ı calculate this current ? Is there any formula? I need to calculate, how much current do ı need for this ıgbt ? Because ı will select a connector for this ıgbt drivers. I will chose connector according to these currents. \$\endgroup\$ – Alican Ozpay Aug 20 at 17:56
  • \$\begingroup\$ You're making too much fuss about this current! To limit the current, use gate resistors just like shown in the datasheet of the IR4426. A value of 10 ohms is common. That will limit the gate peak current and also helps for EMI (radio frequency emissions). You don't need to use any special connector just for this current. Almost any connector will do as the current peak is extremely short (nano seconds), any connector can handle that. As a beginner designer, look more to what others do and try less to "calculate everything". \$\endgroup\$ – Bimpelrekkie Aug 20 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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