1. Dec 1, 2013

### mishima

I was looking for a simple circuit to breadboard that demonstrates how a buffer prevents "loading" effects. I would like to do something like this:

2. Add buffer to circuit and observe the elimination of this effect

I don't really understand what is meant by "loading effects" when it is mentioned in textbooks or here for example.

2. Dec 1, 2013

### phinds

for 1. use a wire
for 2. use the circuit you referenced

When you use a wire, the load on the output will load the input. The op-amp buffers the input and isolates it from the output so that the output load has no effect on the input.

3. Dec 1, 2013

### mishima

Thanks, I understand the last sentence above but I have never really seen this isolation for myself. I understand why having a high output impedance with a low input impedance would do this, I just need a basic setup with something I can measure.

Are you saying to short circuit a battery with 1) a wire, then instead to 2) use a buffer? How could I observe the loading effect?

Could I see this loading effect with a multimeter or the brightness of an LED?

4. Dec 1, 2013

### phinds

No, I wouldn't go straight from the battery to the load but rather put a resistor in series with the battery, then feeding a load resistor. This will act like a voltage divider and the output is that divided voltage between the two resistors. When you put an op-amp in-between the two, the output becomes the battery voltage, not a divided voltage, and the current out the battery is trivial rather than supplying the two resistors. It will be supplying one of them but only to feed a small current into the op-amp --- these comments on the battery current assume you are feeding the op-amp power from a separate source and are just using the battery to create an input to the op-amp via the resistor.

5. Dec 1, 2013

### Okefenokee

A source and load can be modeled like this: (The model of the source is called the Thevanin equivalent circuit fyi.)

Do you notice the output resistance of the source (Rs)? When current flows there is going to be a voltage drop across Rs. If your load is 100Ohm then by the voltage divider rule that drop won't be significant.

Now imagine what would happen if the resistors were flipped so that the output resistance of the source (Rs) is 100 and the load input (Rload) is 1 Ohm. Now most of the voltage drop will be across the output resistance and you won't get much voltage on the load.

Some sources are going to have high output resistance. Sensors are a good example. Some loads are going to have low input resistance.

The simple op amp follower allows you to connect the source to something with high input resistance. It could be in the MOhms range. That way the majority of the voltage is delivered to the amp by the voltage divider rule. The ouput of the amp will match the input and also have a low ouput resitance so you can deliver more current to the load without a voltage drop.

6. Dec 1, 2013

### Jony130

Suppose we have a 1K resistor voltage divider supply from 10V battery.
Without any load connect to the output terminal of our circuit output voltage is equal 5V.
Now we connect a 100Ω load resistor across the output terminal.
And now our voltage divider output voltage drops from 5V to 0.83V. So we ruin our circuit.
To fix this issue we add a buffer (emitter follower) see the diagram

7. Dec 1, 2013

### mishima

I tried this with a 741 and some 9V batteries. First I just made a voltage divider using 2 1000 ohm resistors. Then I used a multimeter to get the voltage at each node. The node connecting battery to 1st resistor was 7.19 V, between the 2 resistors was 3.63 V. Seemed about right. Current was everywhere the same as expected (3.93 mA).

Then I put a 741 (with +/-9 supply separate from the resistor battery) in between the resistors and remeasured everything. Between battery and 1st resistor was 8.3 V, between 1st resistor and 741 (pin 3) was 8V. Between 741 (pin 6) and 2nd resistor was 4 V. Current on the 1st resistor side was 0.36 mA, but on the 2nd resistor side current was 4.0 mA.

For some reason I was expecting the voltage at pin 3 and at pin 6 to be the same, but other than that it seems ok: very low power on source side, very high on load side. Is that right that the voltage still gets halved with the follower in place?