Help with circuit diagram

[axro]

In this schematic:

**broken link removed**

Can R2 and R3 be combined into 1 940 ohm resistor(if there was a resistor of that value). Or is there a reason they are split into 2?

 

[ericgibbs]

In this schematic:

**broken link removed**

Can R2 and R3 be combined into 1 940 ohm resistor(if there was a resistor of that value). Or is there a reason they are split into 2?

hi,
No.
The two 470R set the adjusting pot in centre of the Vcc/2

 

[MikeMl]

R2 and R3 do two things. They limit current into/out of the non-inverting input of the 741 in the event that the pot (P1) wiper is run all the way up or down, respectively.
They also decrease the sensitivity of P1 by better spreading it's adjustment range.

You could eliminate R2 and R3, and then put a 1K resistor between the wiper to the non-inverting input. This current limiting into the input is not really needed with most modern op-amps. The 741 is an archaic design which may need the extra protection.

 

[chuddleston]

Am I correct in thinking you could just get rid of R2 and R3?

edit: ah, that makes sense.

 

[MikeMl]

Am I correct in thinking you could just get rid of R2 and R3?

Already answered in my Post above yours.

 

[Chippie]

Am I correct in thinking you could just get rid of R2 and R3?

edit: ah, that makes sense.

No for the reasons previously stated.........

 

[ericgibbs]

Am I correct in thinking you could just get rid of R2 and R3?

You could providing on that 741 circuit you didnt turn the pot to the top [V+] end.

 

[bailey45]

With a supply voltage of less than 15 volts there is no limit to the input range of a LM741. R2 and R3 can be eliminated, however, the circuit will be very difficult to adjust/calibrate with using and expensive 10 turn potentiometer. Leave R2 and 3 in the circuit and you will have not problem using a cheap 1 turn pot.

 

[ericgibbs]

With a supply voltage of less than 15 volts there is no limit to the input range of a LM741. R2 and R3 can be eliminated, however, the circuit will be very difficult to adjust/calibrate with using and expensive 10 turn potentiometer. Leave R2 and 3 in the circuit and you will have not problem using a cheap 1 turn pot.

hi,
From my 741 data the input range should kept less than 2V of the supply rail.

 

[MikeMl]

... R2 and R3 can be eliminated, however, the circuit will be very difficult to adjust/calibrate with using and expensive 10 turn potentiometer. Leave R2 and 3 in the circuit and you will have not problem using a cheap 1 turn pot.

R2 and R3 would have to each be much bigger relative to the 10k of P1 in order to desensitize wiper movement along P1. As shown, R2+R3 is less than 10% of 10K, so the whole "expand the range of P1" discussion is moot.

 

[axro]

In the diagram the + voltage is 12volts. Does that mean the non-Inverting input of the op amp is seeing 6?

If so is that by design? Because some of you were saying you could basically get rid of R2/R3, but if so would it not then see 12volts? Or does that not really matter?

 

[Hero999]

I wouldn't recommend using the 741.

I think you use a comparator such as the LM311, it can directly drive a relay (no transistor required) and it completely disconnects the relay when it turns off.

You'll need to reverse the inverting and non-inverting pins.

I've also added a feedback resistor to give some hysteresis whcih will make it more stable.

 

[MikeMl]

In the diagram the + voltage is 12volts. Does that mean the non-Inverting input of the op amp is seeing 6?

As shown in the original circuit, the voltage at the non-inv input could be anywhere from ~1V to about ~11V, depending on where the wiper of P1 is.

If so is that by design? Because some of you were saying you could basically get rid of R2/R3, but if so would it not then see 12volts? Or does that not really matter?

If you left out R2,R3, the entire 12V would appear end-to-end across the 10K pot. The non-inv input could then be adjusted from 0V to 12V as the wiper is moved.

The discussion about current limiting into the non-inv input of the 741 has to do with what happens as the wiper is run all the way up, and the input goes to the same voltage as the 741's V+ pin?

 

[axro]

I'm not actually planning on building this. I'm just looking at schematics right now to try to understand how everything works. I'm just looking at it to say "Ok that's why that goes there...and that goes there". Thats why I was asking about those resistors.

 

[axro]

As shown in the original circuit, the voltage at the non-inv input could be anywhere from ~1V to about ~11V, depending on where the wiper of P1 is.



If you left out R2,R3, the entire 12V would appear end-to-end across the 10K pot. The non-inv input could then be adjusted from 0V to 12V as the wiper is moved.

The discussion about current limiting into the non-inv input of the 741 has to do with what happens as the wiper is run all the way up, and the input goes to the same voltage as the 741's V+ pin?

Sorry bear with me here. So if R2 and R3 are gone, the noninverting would see the full 12v? How can it see 0->12V then? Doesn't the pot only limit current? Not voltage?

 

[MikeMl]

Sorry bear with me here. So if R2 and R3 are gone, the noninverting would see the full 12v? How can it see 0->12V then? Doesn't the pot only limit current? Not voltage?

Read up on "Voltage Divider". The Pot is a voltage divider consisting of the resistance from the top end of the pot to the wiper, and the resistance from the wiper to the bottom end. The wiper can be moved all the way from the bottom to the top, so the wiper voltage "taps off" any voltage from 0 to 12V as the wiper is moved.

 

[axro]

Ok I didn't know that a Pot was a voltage divider. I though a Pot was the same as a single resistor.

I am correct in saying that you need at least 2 resistors in series to be a voltage divider though correct?

 

[BrownOut]

You can think of the pot as two different resistors, seperated in value at the wiper. You could, for example, measure the resistance from one terminal to the center tap, then form the other end to the tap, and substitute two resistors who's respective values are the same as the values you measured on the pot, thus create a more 'familiar' voltage divider.

That said, I would not recommend that you remove the two fixed resistors. Even though modern devices allow you to get away with doing so, it's not really good practive to do so. The resistors cost about a penny apiece, and don't take up much room. Leave them in.

EDIT: Sorry to repeat what other's have said, we were all posting at the same time.

 

[axro]

So with voltage dividers. Say you have a 9volt battery. Could you use 2 resistors to make it 5 volts and connect the voltage supply to a circuit in between those 2.

Basically a makeshift voltage regulator. Or does that not really work?

 

[BrownOut]

So with voltage dividers. Say you have a 9volt battery. Could you use 2 resistors to make it 5 volts and connect the voltage supply to a circuit in between those 2.

Basically a makeshift voltage regulator. Or does that not really work?

It will work as long as the device connected does not draw appreciable current. This scheme is sufficient for producing bias voltages. For any reasoable load, you'll need something more sophisticated.