H-Bridge Driving Circuit Help using IR2181

[xt22]

I have designed the attached H-Bridge circuit, but I am having trouble using an IR2181 High/Low Side driver for driving the FETs. Even with nothing on the High-Input of the IR2181, the device is outputting on the high-output and I cannot determine why. Does anyone have any experience with the IR2181's? Any suggestions for troubleshooting?



EE Intern

Stevens Institute of Technology

 

[LionelHutz]

Vcc should be connected to the +20V.

 

[xt22]

It is, there is a +20VDC net label on the VCC pins, and I know this connection is made on my PCB also.

EE Intern

Stevens Institute of Technology

 

[OperaHouse]

"Even with nothing on the High-Input of the IR2181"

In my world nothing ain't nothing. A floating input is often a high.

 

[IRstuff]

And what do you have on the Low input when you have nothing on the High?

TTFN
faq731-376

 

[IRstuff]

What transistor types are Q1 and Q2?

TTFN
faq731-376

 

[xt22]

I would agree, but that doesn't seem to be the issue here. I attempted shorting the High-input to ground, but this did not change the output of the device.

EE Intern

Stevens Institute of Technology

 

[xt22]

Right now all of the inputs to the driver are a logic low, connected to the output of a buffer (74ACT244SC).

The FETs, Q1 and Q2, as well as Q3 and Q4, are FQB55N10. (

http://pdf.zener.ru/90581.pdf?datasheet Fairchild Semiconductor FQB55N10TM)

Something else I have noticed is that Vs is sitting at about 11V. Not sure if this is normal for this device.





EE Intern

Stevens Institute of Technology

 

[IRstuff]

11V is different than a "high." Are both sides behaving the same? Have you tried something like a 10k from output to ground?

TTFN
faq731-376

 

[OperaHouse]

So what is Vb? A bootstrap circuit doesn't magically create voltage. Firing up this circuit without drive can produce unpredictable results. Vs has to be driven to zero in order to charge the bootstrap cap. Not too much to the circuit. This is where you finally get your education. There is always a chance you zonked the part.

 

[xt22]

My mistake, I should have mentioned that I currently have the circuit connected to a 12V supply.

OperaHouse, I agree with you. I found that the low-side FET needs to be turned on in order for the high-side output of the same driver to be turned off. So, for example, if Q3 is PWM'd Q1 will no longer be driven high, and the bridge will act as it is supposed to. The issue that caused me concern was that I was not using the circuit in it's full implementation when I started measuring voltages.

Thanks for the help guys.


EE Intern

Stevens Institute of Technology

 

[Comcokid]

First - You indicate "the device is outputting on the high-output". How are you measuring this high-output? Remember, if you are measuring the high output on a boosted output you need to measure to Vs or the high-MOSFET drain as a reference, and not between the high-output and logic ground. When the high-output is off it will be the same voltage as Vs. Vs is effectively a new 'ground' for those devices on the boosted side of the driver that is at a higher potential than your logic ground.

Second - Watch your layout around half-bridge drivers. Check the IR website - they have app notes for layout of half bridge drivers. This becomes especially true the faster you switch these circuits. You're not indicating it on your schematic but you need a ceramic capacitor bypass from your +20V very near your diode to ground. I always make sure this capacitor is at least 2x to 5x the boost capacitor value.

 

[LionelHutz]

Personally, I think you don't understand the circuit.

C7 and C8 are basically the power supplies for the high side driver circuit. You have to turn-on transistors Q3 and Q4 to charge C7 and C8 before the high side driver will even work. Then, you have to be switching Q3 and Q4 on every x seconds to keep the capacitors C7 and C8 charged so the high side driver will keep working. I can't tell you the number x but how to calculate it or pick the capacitor sizes for C7 and C8 based on your switching duty is in the IR data.

You have to measure the high side output between H0 and Vs. Measuring from H0 or Vs to COM means nothing.

And of course Vs will be around 11V when you first power up the circuit. Basically, it will be a little less than the power supply voltage until you start operating the circuit. The transistors are off so C7 and C8 are not charged and the high side driver places some load across C7 and C8, which keeps the capacitors discharged. So, you have 12V applied to D8 and D9 which passes through the diodes with the diode voltage drop before reaching Vb and then it passes through the IC load before reaching Vs.

 

[OperaHouse]

There are circuits that use a 555 to charge the high side driver so it is always drive voltage even in very static conditions. It is easy to see where someone can get in trouble driving a motor steady state not using PWM. A lot of neophytes get in trouble just using a high side driver in a buck converter. It can be done with some clever positioning of components. Although not an issue in this case, some get in trouble when testing at lower voltage. Some drivers and PWM chips have a low voltage lockout that can become an issue with voltages of 11V.

Wisdom comes from knowledge.
Knowledge comes from experience.
Experience comes from bad decisions.