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-   -   Ecg Heart Rate Monitor Design (http://forum.allaboutcircuits.com/showthread.php?t=19555)

keane2097 02-19-2009 03:10 PM

Ecg Heart Rate Monitor Design
 
I'm currently building a heart rate monitor based on the MSP430 from Texas Instruments. This is the schematic I'm basing my preamplifier design on, it's a TI design with some electrical isolation added:

http://forum.allaboutcircuits.com/ca...rontendcp3.jpg

I've built a test version of the circuit on breadboard and I want to check that I've done it properly without attaching it to myself...

Unfortunately I'm not sure what input signals I should apply to test it suitably.

Basically, can anyone tell me what kind of signal I should apply to the right-leg driven circuitry and to each of the inputs that will be connected to the patient's arms?

Thanks

keane2097 02-19-2009 07:30 PM

Could somebody possibly explain this section of the schematic to me specifically please?

http://forum.allaboutcircuits.com/ca...291/inaho3.jpg

I can't really understand what it's telling me to do here with regards to the connection of + and - Vcc, seemingly to the Vin+/- pins?

bertus 02-19-2009 07:35 PM

Hello,

The diodes to +Vcc and -Vcc and the inputs - and +, are protecting the inputs.
The inputs can not go higher than +Vcc + Vdiode (+5.7)
and lower than -Vcc - Vdiode (-5.7), in this schematic.

Greetings,
Bertus

keane2097 02-19-2009 07:50 PM

Quote:

Originally Posted by bertus (Post 120454)
Hello,

The diodes to +Vcc and -Vcc and the inputs - and +, are protecting the inputs.
The inputs can not go higher than +Vcc + Vdiode (+5.7)
and lower than -Vcc - Vdiode (-5.7), in this schematic.

Greetings,
Bertus

What I'm confused about is the fact that it seems to be telling me to attach both + and - VCC to the same pin... Is this correct?

If so, is it some sort of protection for the circuit?

bertus 02-19-2009 07:55 PM

Hello,

As I told you the diodes are placed there to protect the inputs of the instrumentation amplifier.
The way they are used is called clipping.
http://www.allaboutcircuits.com/vol_3/chpt_3/6.html

Greetings,
Bertus

keane2097 02-19-2009 07:58 PM

Quote:

Originally Posted by bertus (Post 120460)
Hello,

As I told you the diodes are placed there to protect the inputs of the instrumentation amplifier.
The way they are used is called clipping.
http://www.allaboutcircuits.com/vol_3/chpt_3/6.html

Greetings,
Bertus

So I connect both + and - Vcc to the positive and negative input pins of the amplifier, each through a diode?

Thanks for the link.

SgtWookie 02-19-2009 10:24 PM

No.
On the vertical edge of the triangular amplifier symbol, you will see the noninverting (+) input and the inverting (-) input. Those are signal inputs, not Vcc/Vee.

With instrumentation amplifiers, there are two additional lines on the vertical edge of the symbol; these are used to set the gain of the amplifier.

keane2097 02-20-2009 01:57 PM

Quote:

Originally Posted by SgtWookie (Post 120483)
No.
On the vertical edge of the triangular amplifier symbol, you will see the noninverting (+) input and the inverting (-) input. Those are signal inputs, not Vcc/Vee.

With instrumentation amplifiers, there are two additional lines on the vertical edge of the symbol; these are used to set the gain of the amplifier.

This is the pin-out diagram of the chip i'm using:-

http://forum.allaboutcircuits.com/ca...[/URL]

From what you've said I don't connect the diodes and Vcc+/- to the noninverting (+) input and the inverting (-) input, that there are two other input lines for this, but it doesn't look like this in the pin-out diagram.

From what I can see, I need to connect my two 40 ohm resistors up between pins 1 and 8, Connect pin 7 to a 5V source, and connect the diodes and Vcc+/- to pins two and three...

Is this correct?

beenthere 02-20-2009 02:55 PM

From the data sheet, you also need to have pins 4 and 5 to circuit ground. Pin 5 is the reference (always circuit ground, unless you need an offset in the output), and 4 is the negative supply pin (V-). The boxes that indicate "input protection circuitry" are explained in the data sheet. You don't need clipping diodes.

Operating the INA118 with a single supply means that it can't respond to a signal that drives the output negative with respect to ground. Unless you are certain that all signals will be positive-going with reference to ground, a -5 volt supply might be a good idea.

Do you have the data sheet for the device? It's always a good idea to have it, as it has example applications illustrated.

keane2097 02-21-2009 02:52 PM

Quote:

Originally Posted by beenthere (Post 120649)
From the data sheet, you also need to have pins 4 and 5 to circuit ground. Pin 5 is the reference (always circuit ground, unless you need an offset in the output), and 4 is the negative supply pin (V-). The boxes that indicate "input protection circuitry" are explained in the data sheet. You don't need clipping diodes.

Operating the INA118 with a single supply means that it can't respond to a signal that drives the output negative with respect to ground. Unless you are certain that all signals will be positive-going with reference to ground, a -5 volt supply might be a good idea.

Do you have the data sheet for the device? It's always a good idea to have it, as it has example applications illustrated.

Thanks for this post - it's made things somewhat clearer. I have the datasheet for the INA118 alright.

Having studied the schematic a little more I've come to the following conclusions:

Pin 1: 40 kOhm Resistor, sets amp gain along with pin 8
Pin 2: Inverting Input - connected through 390 kOhm resistor to left arm
Pin 3: Non-Inverting Input - connected through 390 kOhm to right arm
Pin 4: Negative Supply Voltage - Connected to ground
Pin 5: See below
Pin 6: Instrumentation Amp Output line
Pin 7: Positive supply voltage - 5V applied
Pin 8: 40 kOhm Resistor, sets amp gain along with pin 8

The final thing I'm unclear about is setting up the reference (i.e. Pin 5). In your post you say it should be connected to ground, however in my schematic it seems to be connected through a parallel resistor and capacitor to the integrator op-amp (A4).

Does this provide the offset voltage you spoke about, or am I again mislabelling the pins?

Thanks for all the help, I'm fairly sure I have the circuit constructed correctly now and just want to be sure everything's right...


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