RC Filter in Line with IC Power Supply

Thread Starter

TechWise

Joined Aug 24, 2018
77
I have been studying some datasheets for gate drivers and some feature RC filters in line with the power supply to the low voltage side. I was quite surprised by this as it seemed to go against everything I thought I knew about providing power to an IC. I have always used low impedance power and ground planes and observed good layout practices to reduce inductance. Putting such a large resistor in series with a power supply would appear to go against the conventional wisom of providing an ideal, low impedance supply to a device.
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I expect this would serve the dual purpose of preventing noise from entering the gate driver from the power supply while also preventing noise from the gate driver from propogating back into the rest of the system. The datasheet simply says "An RC filter filters the 3.3V rail". I am concerned that if I select a different IC where the values are not specified that getting them wrong could cause problems. I would also prefer to use a 5V rail.

Can anyone clarify the role of this circuit anymore and how it should be designed? Could a ferrite bead serve a similar purpose?
 

Papabravo

Joined Feb 24, 2006
13,522
It is all about the quiescent and maximum current demand of the chip. If you have a 3.3 VDC rail and the chip draws 50 μA, the voltage drop across the resistor will be 0.5VDC. That means the chip would be operating with a Vcc of 2.8 volts. If the chip needs more current, then the voltage drop across the 10K resistor will increase in proportion to the current. At some current draw from the 3.3V rail, the chip will simply stop working. Unless you can specify, with certainty, the current draw I would rate this idea a -5 on a scale from 0 to 10

Edit: The diagram is so small that the resistor looked like a 10K. For a 10Ω resistor the 0.5V drop would occur at a current draw of 50 mA. What I said above still applies but with a different scale factor. Considering the power dissipation you would have 50 mA squared times 10 = 25 mW, so a 1/10 watt resistor would do the job.

The corner frequency of that RC combination seems high to me.
 
Last edited:

ronsimpson

Joined Oct 7, 2019
610
In switching power supplies we go to great trouble to reduce RF-noise. You will find R1, C3 very close to the Vcc1 pin to keep noise off the printed circuit board.
 

Thread Starter

TechWise

Joined Aug 24, 2018
77
The thought of the power supply voltage being reduced during periods of high current draw seemed wrong to me. The quiescent current is 1.5mA. As @Papabravo points out, this would lead to a 15mV drop which I suppose is perfectly tolerable, given that it is specified for operation down to 3V.

I asked TI what their explanation for this RC filter was. In the reference design, a buck converter powers the whole board with 5V. This feeds a 3.3V LDO which powers the low voltage side of the gate driver. The 5V also feeds the input side of an isolated DC/DC converter which switches at 400kHz and supplies the high voltage side of the gate driver for high-side gate driving. There is some concern about the 400kHz switching noise propogating back onto the 5V rail, and the 3.3V LDO has a poor power supply rejection ratio at this frequency, hence the inclusion of an RC filter with cutoff set at half the target ripple frequency. This is the reason for the seemingly high cutoff frequency.
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Papabravo

Joined Feb 24, 2006
13,522
I see the reasoning, but I would ask if you know how much attenuation a single pole RC provides at 400 kHz, with a corner at 159+ kHz. If it doesn't do the the job you may have to raise the price of poker.
 
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