Help with circuit diagram

[MikeMl]

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

The voltage at the tap of a voltage divider (or Pot, as you have learned) depends on two things: the ratio of the resistor values above and below the tap, and, the current being drawn out of the tap.

In the case of the opamp circuit you posted, the current into the non-inv input of the 741 (input bias current) is very small (uA), so the voltage at the tap (wiper) is proportional to the wiper position from the bottom. However, if you tried to draw more than a few hundred uA from the tap, then the voltage would sag, like a poorly regulated power supply.

The math is fairly simple. The principle is "Thevenin's Equivalent". Google it.

 

[Mark68]

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.

What program makes that circuit diagram?

 

[axro]

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.

What if you had a wallwart the put out 6v and 500mA. And you used really small resistors. I would then be able to draw the majority of the 500mA would I not?

Or does that not really work since and create a short?

 

[BrownOut]

If you connected a 12 Ohm 3 Watt resistor across the wall wart's output, you would draw 500mA from it.

Not sure if that answered your question or not.

 

[MikeMl]

What if you had a wallwart the put out 6v and 500mA. And you used really small resistors. I would then be able to draw the majority of the 500mA would I not?

Or does that not really work since and create a short?

Do the math: R = E/I = 6÷0.5 = 12Ω

Now, the power dissipated in the resistor would be P = IE = 0.5×6 = 3W,

so it would take a 12Ω 3W resistor.

Suppose all you had were 1/2W resistors. How many and what Value would you connect together to make 12Ω?

Right off the bat, we know that 6 1/2W resistors would dissipate the 3W.

So, what value would the individual resistors be if six identical ones were connected in parallel, and the resulting resistance is 12Ω?

The formula for six parallel resistors is 1/Rt=1/R1+1/R2+...+1/R6,

but R1-6 are all the same, so call that Rx. 1/Rt =6/Rx, or Rt = Rx/6, or Rx=6×Rt, where Rt = 12, so Rx = 72Ω

So, six 72Ω 1/2W resistors connected in parallel would load up your supply to full output, without burning up the supply, or any resistors

I did the hard one; now you do the easy one. Suppose again that all you had were 1/2W resistors. How many and what value would they have to be if you were to connect them in SERIES?

 

[Mark68]

**broken link removed**

Anybody know what program makes this schematic?

 

[mneary]

Anybody know what program makes this schematic?

I only see this on one website and mirrors and copies. I would gues they used Paint or something similar.

 

[Mark68]

The reason I ask because "hero" had changed it using the same diagram?

Actually on closer inspection, it looks like he could have just erased the section that he posted not really changed it.

 

[mneary]

At this stage, it's a 'gif' image and has to be manipulated as pixels and not as components. Inspecting the binary file, I don't see any 'signatures' or watermarks from the author or editing software.

 

[ericgibbs]

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?

hi,
You really should consider the circuit inputs as whole circuit.

Say the LDR resistance is equal in value to the 10K series resistor, ie: both 10K
that would make pin2 of the 741 approx 6V [Vs/2].

If you now adjust P1 the 10Kpot to its centre of rotation, then pin3 of the 741 would be at about 6V.

The output pin6 of the 741 could be either high or low. [12V-2V] or +1.2V.
Adjusting P1 about its central rotation would cause the output to switch high or low.
If pin3 voltage is higher than pin2 the 741 output will be high, if pin3 voltage is lower than pin2, the output will be low.

As the LDR resistance changes with the intensity of light falling on it, so will the voltage on pin2, this switches the 741 output high/low depending upon the light intensity.

Using a 470R at either end of the pot, means that voltage on pin3 cannot be set higher than +11.5V or lower than +0.5v.

So the voltage on pin3 is from +0.5v thru +11.5V, which IMO is too greater range, it makes the setting of P1 'coarse'.

A point to note, some comparator ic's will go into output inversion if the inputs exceed Vsupply -2V, ie: 10V.

A far better design would be based on knowing the resistance range of the LDR when used in this project, the resistors would be calculated to give the operating range of P1.

OK.

 

[axro]

Do the math: R = E/I = 6÷0.5 = 12Ω

Now, the power dissipated in the resistor would be P = IE = 0.5×6 = 3W,

so it would take a 12Ω 3W resistor.

Suppose all you had were 1/2W resistors. How many and what Value would you connect together to make 12Ω?

Right off the bat, we know that 6 1/2W resistors would dissipate the 3W.

So, what value would the individual resistors be if six identical ones were connected in parallel, and the resulting resistance is 12Ω?

The formula for six parallel resistors is 1/Rt=1/R1+1/R2+...+1/R6,

but R1-6 are all the same, so call that Rx. 1/Rt =6/Rx, or Rt = Rx/6, or Rx=6×Rt, where Rt = 12, so Rx = 72Ω

So, six 72Ω 1/2W resistors connected in parallel would load up your supply to full output, without burning up the supply, or any resistors

I did the hard one; now you do the easy one. Suppose again that all you had were 1/2W resistors. How many and what value would they have to be if you were to connect them in SERIES?

6, 2 Ohm resistors?

 

[MikeMl]

6, 2 Ohm resistors?

Right on !

 

[axro]

hi,
You really should consider the circuit inputs as whole circuit.

Say the LDR resistance is equal in value to the 10K series resistor, ie: both 10K
that would make pin2 of the 741 approx 6V [Vs/2].

If you now adjust P1 the 10Kpot to its centre of rotation, then pin3 of the 741 would be at about 6V.

The output pin6 of the 741 could be either high or low. [12V-2V] or +1.2V.
Adjusting P1 about its central rotation would cause the output to switch high or low.
If pin3 voltage is higher than pin2 the 741 output will be high, if pin3 voltage is lower than pin2, the output will be low.

As the LDR resistance changes with the intensity of light falling on it, so will the voltage on pin2, this switches the 741 output high/low depending upon the light intensity.

Using a 470R at either end of the pot, means that voltage on pin3 cannot be set higher than +11.5V or lower than +0.5v.

So the voltage on pin3 is from +0.5v thru +11.5V, which IMO is too greater range, it makes the setting of P1 'coarse'.

A point to note, some comparator ic's will go into output inversion if the inputs exceed Vsupply -2V, ie: 10V.

A far better design would be based on knowing the resistance range of the LDR when used in this project, the resistors would be calculated to give the operating range of P1.

OK.

You are going to have to help me with this.

I see how pin 2 would get 6V if the LDR is 10k. Both resistors are equal so 12V/2 = 6.

The part I'm a little confused on is Pin3. Correct me where I am wrong. If P1 is at the center of it's rotation then it would be at 5000 ohms correct? The total Resistance between V+ and ground would then be 5940 correct? so If you take 12/5940 you get .002 Amps. So the voltage across R2 is 0.94 volts. R3 is also 0.94 volts. Votage across P1 would then be 10volts.

Is all of the right or am I off?

If I am right I don't know where you get the +0.5v thru +11.5V for voltage on pin3. Am I missing something?

 

[ericgibbs]

You are going to have to help me with this.

I see how pin 2 would get 6V if the LDR is 10k. Both resistors are equal so 12V/2 = 6.

The part I'm a little confused on is Pin3. Correct me where I am wrong. If P1 is at the center of it's rotation then it would be at 5000 ohms correct? The total Resistance between V+ and ground would then be 5940 correct?
No, its 10K + 940R, total resistance.
So 12V/10940R = 10.9mA, hence 470R * 10.9mA = 0.515V

so If you take 12/5940 you get .002 Amps. So the voltage across R2 is 0.94 volts. R3 is also 0.94 volts. Votage across P1 would then be 10volts.

Is all of the right or am I off?

If I am right I don't know where you get the +0.5v thru +11.5V for voltage on pin3. Am I missing something?

You must include all the resistance in the chain. The 10K dosnt change value, only the position of the pots wiper.

 

[MikeMl]

You are going to have to help me with this....
The part I'm a little confused on is Pin3. Correct me where I am wrong. If P1 is at the center of it's rotation then it would be at 5000 ohms correct? The total Resistance between V+ and ground would then be 5940 correct? so If you take 12/5940 you get .002 Amps. So the voltage across R2 is 0.94 volts. R3 is also 0.94 volts. Votage across P1 would then be 10volts.

Is all of the right or am I off?

If I am right I don't know where you get the +0.5v thru +11.5V for voltage on pin3. Am I missing something?

Stare at this:

 

[axro]

You must include all the resistance in the chain. The 10K dosnt change value, only the position of the pots wiper.

Wait..why does the 10k not change value? Isn't that the purpose of a pot is that it can change from anywhere from 0-10k? Isn't that what moving the wiper does?

 

[ericgibbs]

Wait..why does the 10k not change value? Isn't that the purpose of a pot is that it can change from anywhere from 0-10k? Isn't that what moving the wiper does?

If one end of the pot was connected the wiper, it would change value from 0ohms to 10Kohms.

As the wiper is NOT connected to any end of the pot, the value measured at the two ends of the pot remains 10k.

OK.?

 

[axro]

If one end of the pot was connected the wiper, it would change value from 0ohms to 10Kohms.

As the wiper is NOT connected to any end of the pot, the value measured at the two ends of the pot remains 10k.

OK.?

So you are saying only 2 of the legs are used on the pot? And if the wiper is not connected to anything then it won't do anything when you turn it then right?

So why even have a pot and not just use a 10k resistor?

 

[MikeMl]

Wait..why does the 10k not change value? Isn't that the purpose of a pot is that it can change from anywhere from 0-10k? Isn't that what moving the wiper does?

Look up the difference between a "rehostat" and a "potentiometer". Hint: one has two connections to the outside world; the other has three.

 

[axro]

Look up the difference between a "rehostat" and a "potentiometer". Hint: one has two connections to the outside world; the other has three.

Ok so let me see if I have this right. R2 is connected to one end and R3 is connected to the other. There is always 10k ohms through the pot in this configuration. The wiper terminal is connected to the op amp, and moving it ONLY adjusts voltage beween max and min.

If this is correct, does the op amp see the 10kohms from the pot?