Hi all,

I am using the MAX7400 lowpass filter IC, where I am inputting a signal at 50Hz, 1Vp and a 2.5V offset.
The external clock is set to 10kHz, which according to the datasheet should set the cutoff frequency to 100Hz.
Why is the output staying at 0V?

I also tried using a capacitor instead of an external clock, but it still did not work.

Datasheet: http://www.mouser.com/ds/2/256/MAX7400-MAX7407-90493.pdf

Below is the circuit used:

 

Hi all,

I am using the MAX7400 lowpass filter IC, where I am inputting a signal at 50Hz, 1Vp and a 2.5V offset.
The external clock is set to 10kHz, which according to the datasheet should set the cutoff frequency to 100Hz.
Why is the output staying at 0V?

I also tried using a capacitor instead of an external clock, but it still did not work.

With the limited information you have provided, my first guess is failure to read and understand the datasheet. How about a schematic and a link to the datasheet so we can be more helpful.

 

With the limited information you have provided, my first guess is failure to read and understand the datasheet. How about a schematic and a link to the datasheet so we can be more helpful.

Sorry forgot to upload the image. Please see updated post.

 

Are you doing this in a simulator, or on a real breadboard?

You might need a load on the output. Datasheet recommends 10 KΩ || 50 pf.

This is not uncommon with filters that they require a load, in the same way that a transmission line has different behavior when open, shorted, and terminated in it characteristic impedance.

 

Hello,

As you use a single supply, you will need to connect the OS input with a voltage of about 1/2 Vcc.
This circuit is given in the by you attached datasheet:



Bertus

 

Hello,

As you use a single supply, you will need to connect the OS input with a voltage of about 1/2 Vcc.
This circuit is given in the by you attached datasheet:

View attachment 104859

Bertus

Good catch. I failed to appreciate from the datasheet that requirement. I was looking for a valid input signal range and found only the OS input range.

 

Are you doing this in a simulator, or on a real breadboard?

You might need a load on the output. Datasheet recommends 10 KΩ || 50 pf.

This is not uncommon with filters that they require a load, in the same way that a transmission line has different behavior when open, shorted, and terminated in it characteristic impedance.

Thanks for the reply.
I don't have the IC yet. I am trying to simulate the circuit on simulation software first.
The circuit is still not working with a 10 KΩ || 50 pf connected at the output.

 

@bertus , Thanks for the reply. Isn't that pin adjusted to set the output offset? The input voltage already has an offset of 2.5V.
I tried it with a potential divider but the output ramained at 0V.

 

Thanks for the reply.
I don't have the IC yet. I am trying to simulate the circuit on simulation software first.
The circuit is still not working with a 10 KΩ || 50 pf connected at the output.

Then I would question the behavioral model in the simulator. Where did you get the model from? I hope the answer is Maxim.

 

Hello,

From the datasheet:

Offset and Common-Mode
Input Adjustment
The voltage at COM sets the common-mode input volt-
age and is internally biased at midsupply by a resistor-
divider. Bypass COM with a 0.1μF capacitor and
connect OS to COM. For applications requiring offset
adjustment or DC level shifting, apply an external bias
voltage through a resistor-divider network to OS, as
shown in Figure 4. (Note: Do not leave OS unconnect-
ed.) The output voltage is represented by the following
equation:
VOUT = (VIN - VCOM) + VOS
with VCOM = VDD / 2 (typical), and where (VIN - VCOM)
is lowpass filtered by the SCF, and VOS is added at the
output stage. See the Electrical Characteristics for
COM and OS input voltage ranges. Changing the volt-
age on COM or OS significantly from midsupply
reduces the filter’s dynamic range.

Try making the OS voltage adjustable as in fig 4 I posted.


Bertus

 

Hi again,

Can someone help me determine the maximum peak-to-peak voltage that can be applied to the input of the MAX7400?

I cannot really understand the paragraph on page 11. There is also a table showing a Vcc of 3V and an Fc of 1kHz. It indicates Vin(Vp-p) is 2V. Is that the maximum peak-to-peak when having the provided specifications?

I was planning to use the IC with a 3.3V single supply rail, with an input signal that should not exceed 2kHz and with a maximum peak-to-peak of around 3V (with a 1.65V offset). I am guessing that this is not possible to achieve, right?

Datasheet: http://www.mouser.com/ds/2/256/MAX7400-MAX7407-90493.pdf

 

Hello,

The MAX7400 is indended to be used at 5 Volts.
The input voltage will be 4 Vp-p.
The MAX7400 will not work correctly at 3.3 Volts.
For 3.3 Volts use the MAX7404/7407, with an input voltage of 2 Vp-p.

Bertus

 

Hello,

The MAX7400 is indended to be used at 5 Volts.
The input voltage will be 4 Vp-p.
The MAX7400 will not work correctly at 3.3 Volts.
For 3.3 Volts use the MAX7404/7407, with an input voltage of 2 Vp-p.

Bertus

Sorry I was referring to the MAX7404. So it would be best to first filter the signal ans then apply the amplification.

 

As you use a single supply, you will need to connect the OS input with a voltage of about 1/2 Vcc.
This circuit is given in the by you attached datasheet:

The datasheet states:
" For applications requiring offset adjustment, apply an external bias voltage through a resistor-divider network to OS such as shown in Figure 3."
Can I connect the OS pin to the REF3318 voltage reference IC instead of using a resistor-divider?
If yes, can I use one REF3318 to drive the OS pins of five filters? Also, regarding the 0.1uF capacitor recommended in the MAX7404 filter IC datasheet, should I connect one at the output of the REF3318, or one at every OS pin for all filters?

 

Does anyone know if the OS pin can be driven from the REF3318 reference voltage IC? or does it have to be a high input impedance? (since the schematic shows quite large values for the voltage divider)