Discussion in 'Homework Help' started by mynamesalex, Oct 4, 2012.

1. ### mynamesalex Thread Starter Member

Oct 4, 2012
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Hey guys! First post on this forum. I have a feeling i'll be here quite a lot. I'm a computer engineering major and have a project I need some help on. Basically, I'm making an ECG as a design project. I was thrown into it with my group and we have no background knowledge on circuits/breadboard function. I have some general knowledge on circuit function but I'm not entirely sure how to translate that to a breadboard. I was given a schematic in which I have to find resistor values and calculate gain? I've been trying to teach myself how to do this but it hasn't been as easy a task as I hoped it would be. I will attach the circuit diagram which I must translate to the breadboard after finding resistor values as well as gain. I've partially managed to understand how to put the breadboard together to form a circuit but not the extent of what I need to do. I'm using a machine that outputs a signal to mimic a heartbeat instead of using electrodes attached to a human. Basically, the heartbeat should light a LED diode at the end of the circuit to mimic the heartbeat. I've managed to light the LED but not get it blinking with the beat. I've done a great deal of googling as to how circuit diagrams translate to breadboards but it hasn't helped me much with this. If anyone would be willing to help me out with this, it would be greatly appreciated! Thanks a lot.

Edit: Also attached is a jpg on equations to find the gain which I'm not sure about how to go about doing.

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2. ### Sensacell Well-Known Member

Jun 19, 2012
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The attached schematic shows a 'conventional' differential amplifier.
To recover ECG signals, you are going to need much higher CMMR or common mode rejection ratio than this circuit can provide- to get a usable output signal.

3. ### MrChips Moderator

Oct 2, 2009
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3,354
Sensacell is correct. If you are never going to connect your circuit to a patient then consider yourself lucky.

What is the amplitude of the signal from your ECG simulator? Is this a function generator? Are you given the output impedance of the generator?

4. ### mynamesalex Thread Starter Member

Oct 4, 2012
53
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For the simulated signal: 1 Hz, 250 mVpp, 0V offset, 250 ms signal length.

I believe the machine is a wave generator of a sort. It shows an oscillated signal on the screen. I will check later in the day when I go to the lab. Any ideas on how I would calculate the gains and resistances?

5. ### Audioguru New Member

Dec 20, 2007
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250mVpp is a signal level about 125 times higher than from a human so the amplifier design can use ordinary opamps and resistors and the breadboard will be fine.

6. ### mynamesalex Thread Starter Member

Oct 4, 2012
53
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Yep! I believe I am to use three op amps? Thanks for all the help already guys. It's appreciated. I'm still a little fuzzy on how to actually place the resistors and create the circuit on the breadboard. It hasn't been so easy to find material on how to place things on breadboards.

7. ### Audioguru New Member

Dec 20, 2007
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Only the first opamp as the differential amplifier is needed to blink the LED. You do not need two more opamps.

8. ### MrChips Moderator

Oct 2, 2009
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If someone instructed you to use three op amps they are most likely referring to an "instrumentation amplifier" configuration which is what is essential for ECG recording.

In your case, signal from a signal generator may not require an instrumentation amplifier.

9. ### mynamesalex Thread Starter Member

Oct 4, 2012
53
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I think I may have been confused as we were originally going to use electrodes attached to a person. Now we're using the simulated signal which would make sense of using the one op amp.

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10. ### mynamesalex Thread Starter Member

Oct 4, 2012
53
1
So I finally managed to get the circuit to light the LED! I'm stuck at one last thing and need a tiny bit of help from you guys. I'm currently using a DC Power Supply to supply the 5v to the circuit. The power supply has two outputs which allows me to ground my circuit. I'm required to connect the V+ and V- outputs to supply the power and then connect the V- to ground. The Waveform Generator only has one output thus not allowing me to ground the V- (unless there is another way to do this) The way I have it set up, one output from the power supply goes to the positive bus on the breadboard and the negative goes to ground while the other output goes to the negative bus and the positive goes to ground. The circuit will not light without grounding. Is there any way to ground the V- on one output? Thanks again for everything guys. You've been incredibly helpful.

11. ### MrChips Moderator

Oct 2, 2009
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I think you are confused about what V+, V- and GND means.

When you say the power supply has two outputs, I think you mean the power supply has two banana jacks labelled + and -. This is only one output.

Show us a circuit diagram of your hookup and we will set you straight.
Or post a photo of the power supply and the function generator.

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12. ### mynamesalex Thread Starter Member

Oct 4, 2012
53
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The power supply actually has two separate outputs. Meaning two positive and negative banana jacks. The way the circuit is set up I need to ground the inputs to get the LED to light. I'm uploading pictures now to hopefully make this easier for you to understand. I just don't know how I would ground the input with one output.

Edit: Pictures attached! On the breadboard, the first bus to the left is the positive input. The second blue bus on the other side is the negative. The red bus next to that is the ground.

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13. ### MrChips Moderator

Oct 2, 2009
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Sorry about my misunderstanding. So you are using a dual power supply, presumably to give +5V and -5V. Is that correct?

Bench power supplies are usually "floating" supplies. That means there is no reference to earth GND.

You have to connect the -ve terminal of the +5V supply to GND,
and connect the +ve terminal of the -5V supply to GND. Do you understand this?

The function generator has two leads on the cable, a signal and GND.
The GND lead is already at earth GND. You can connect this GND to the GND of your circuit.

Edit: I read that this is what you are already doing. Hence you have to clarify your question.

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Mar 24, 2008
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15. ### mynamesalex Thread Starter Member

Oct 4, 2012
53
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I don't know if it was the correct thing to do but I set the offset voltage on the function generator to about 1.175V and managed to get a blinking LED. Still a bit confused but should I be able to get a voltage through the circuit with a 0V offset?

16. ### Audioguru New Member

Dec 20, 2007
9,411
896
Your opamps need a dual-polarity power supply, +5V and -5V.
The 0V the +5V supply connects to ground of the circuit and the 0V of the -5V supply also connects to the ground of the circuit.