Ecg Heart Rate Monitor Design

[keane2097]

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:

**broken link removed**

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

 

[stevez]

I did read a book (library or my wife's collection, as a nurse, cardiac care was one of her areas of expertise) that did provide information on the potential you might expect to find at various points on the body. That might help to determine how to test the device.

 

[duffy]

Just connect it to yourself, what's the big deal? It would take as much or more to simulate it.

Was the NMR100C your idea or TI's?

 

[keane2097]

Just connect it to yourself, what's the big deal? It would take as much or more to simulate it.

Was the NMR100C your idea or TI's?

That was my idea. The design has to contain suitable electrical isolation to protect the circuit and the patient.

I'd be happy enough to hook it up to myself to test it, but I don't currently have suitable electrodes.

I'd also like to carry out somewhat more professional testing involving test waveforms and the outputs they excite as I want to be able to document the design properly.

 

[duffy]

Yeah, I figured that might be your idea. Getting a lot of noise now, are you?

 

[keane2097]

Yeah, I figured that might be your idea. Getting a lot of noise now, are you?

I'm not sure since I haven't managed to carry out any proper testing yet.

Do you think the transformer is a bad idea?

 

[keane2097]

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

**broken link removed**

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?

 

[duffy]

Last time I put a DC-DC converter module in a design with an instrumentation amp, I got quite a bit of noise in the circuit.

You don't really need the peel-and-stick electrodes for this initial work, just solder a wire to a couple of pennies and tape them on. Clean and shave the area. I've never done ECG, but I've done EMG.

Those diodes are hooked up to +5v and ground. They are for voltage protection.

 

[keane2097]

Last time I put a DC-DC converter module in a design with an instrumentation amp, I got quite a bit of noise in the circuit.

You don't really need the peel-and-stick electrodes for this initial work, just solder a wire to a couple of pennies and tape them on. Clean and shave the area. I've never done ECG, but I've done EMG.

Those diodes are hooked up to +5v and ground. They are for voltage protection.

So I connect both + and - Vcc to the positive and negative input pins of the amplifier, each through diodes? How does this voltage protection work?

Thanks for the electrode suggestion - I'll give it a try...

 

[duffy]

Not the input pins, the SUPPLY pins. The protection is very simple, if the input is above or below the supply rails, the diodes conduct.

Simple, but IMPORTANT. Those float voltages can be anything, easily high enough to damage this sensitive device.

 

[keane2097]

Not the input pins, the SUPPLY pins. The protection is very simple, if the input is above or below the supply rails, the diodes conduct.

Simple, but IMPORTANT. Those float voltages can be anything, easily high enough to damage this sensitive device.

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

**broken link removed**

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?

 

[duffy]

40k, and

diodes and Vcc+/- to pins two and three

"the diodes to Vcc+/- and pins two and three", which is probably what you meant, but obviously you don't connect Vcc+/- (the positive supply voltage and ground, respectively) up to the input of the amp.

 

[keane2097]

40k, and

"the diodes to Vcc+/- and pins two and three", which is probably what you meant, but obviously you don't connect Vcc+/- (the positive supply voltage and ground, respectively) up to the input of the amp.

Yeah I've had another study of the circuit (and read back on notes froma few years back!) and this is what I've now come up with:

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 voltage (i.e. Pin 5). I thought the reference 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 an offset to the output or something, 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...

 

[duffy]

The final thing I'm unclear about is setting up the reference voltage (i.e. Pin 5). I thought the reference 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 an offset to the output or something,

Yes, it essentially is connected to "ground". Yes, it is there to provide an offset adjustment.

Your "ground" is really the +2.5V "pedestal voltage" (by the way, where IS that being generated in your circuit? I don't see its source point anywhere). Opamps and instrumentation amps in particular want a bipolar supply. By giving it that +2.5V reference as ground (0V), the voltages on pins 7 and 4 look like a + and - 2.5V supply.

It can also be used as an offset adjust, it is part of an internal divider circuit and by raising it very slightly above or below the +2.5V reference, it can be used to cancel out offset in the final stage of the instrumentation amp.

In your circuit, it is used in conjunction with a high-pass filter. This is to reduce DC offset problems typically caused by electrode galvanic voltages.

 

[keane2097]

Yes, it essentially is connected to "ground". Yes, it is there to provide an offset adjustment.

Your "ground" is really the +2.5V "pedestal voltage" (by the way, where IS that being generated in your circuit? I don't see its source point anywhere). Opamps and instrumentation amps in particular want a bipolar supply. By giving it that +2.5V reference as ground (0V), the voltages on pins 7 and 4 look like a + and - 2.5V supply.

It can also be used as an offset adjust, it is part of an internal divider circuit and by raising it very slightly above or below the +2.5V reference, it can be used to cancel out offset in the final stage of the instrumentation amp.

In your circuit, it is used in conjunction with a high-pass filter. This is to reduce DC offset problems typically caused by electrode galvanic voltages.

Thanks for this post - extremely helpful. i think I understand a lot better now. As far as the pedestal voltage is concerned, it comes from the Voltage Reference for the ADC used to sample the signal for processing. I cut some parts out of the first schematic I posted unintentionally, here it is in its original form:-

**broken link removed**

The ADC shown is on board an experimenter's kit for the MSP430 MCU from TI. The 2.5V reference is generated internally on the board. Since I'm not ready to connect the board up to the analog circuit should I connect a battery to the positive input pins of A2, A3 & A4 to provide the 2.5V ?

 

[duffy]

If you have a battery that's exactly half the supply voltage, sure. Otherwise use a couple of precision 10k resistors in series (or a potentiometer so you can adjust it) across +5V and ground, the center tap will give you the 2.5V neatly divided output.

The values of the resistors aren't critical, if you have a 50k pot or a pair of 100k precision resistors, that will work just as well. Add a .1µf cap across the lower resistor to help eliminate noise.

 

[keane2097]

So I've gotten the circuit in the schematic built on breadboard and I'm now happy enough that all the connections are correct.



I'm having a problem with the supply voltages though.



I'm using a power supply to generate a +5 to -5 voltage. The +5 is being used to provide the instrumentation amplifier with its positive supply. I'm then using two voltage dividers from the +5 and -5 signals to get the 2.5V supplies for the OPA2335 op-amps.



Unfortunately when I connect the op amps to the voltage divider the voltage drops considerably and the amps aren't therefore supplied with the desired +/-2.5V.



I've tried buffering the two signals with voltage followers. Again the output from the voltage followers are the desired +/-2.5V but when I connect to the op-amps the voltage drops again and I don't get the supply I want.



Clearly I'm absolutely hopeless with analog circuitry. Can someone please give me an idea of how I can drive the desired voltages into my op-amps so i can get back to the land of 1s and 0s asap!?

 

[duffy]

That circuit is designed to run on a single-ended supply: +5V and ground. Hook it up that way and see how it responds.

 

[keane2097]

That circuit is designed to run on a single-ended supply: +5V and ground. Hook it up that way and see how it responds.

So drive the Instrumentation amplifier and the 4 OPA2335s on +5 and zero?

 

[duffy]

Yes - the reason you have that +2.5V reference is that it is there to provide a signal ground; it makes the 0V look like -2.5V, the +2.5V ref look like 0V, and the +5V supply look like +2.5V.

On the spec sheet for the OPA2335 it even boasts of its "Single supply" operation, right at the top of the page. Most opamps can run this way, but it's an indication that was what the designer intended for this circuit.