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pebe
Nobel Prize Winner

United Kingdom
1078 Posts
Posted - Aug 19 2008 :  04:33:47 AM  Show Profile  Reply with Quote
The simple rules for maths are:
1. Calculate functions inside the brackets first.
2. Then multiplication and/or division.
3. Then addition and/or subtraction.

So:
x = 5 + 1.25 x 26.66/31.66
x = 5 + 1.25 x .842
x = 5 + 1.0525
x = 6.0525
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JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts
Posted - Aug 19 2008 :  04:40:44 AM  Show Profile  Reply with Quote
quote:
Originally posted by pebe

The simple rules for maths are:
1. Calculate functions inside the brackets first.
2. Then multiplication and/or division.
3. Then addition and/or subtraction.

So:
x = 5 + 1.25 x 26.66/31.66
x = 5 + 1.25 x .842
x = 5 + 1.0525
x = 6.0525


Ooops.. sorry.

Ok.
For 11V cut-off.
Ref. Voltage = 7.7V
(1K resistor in the upper arm & 7V7 Zener in the lower arm voltage divider)

R3 = 27K
Re = 7.5K
..R1 & R2 = 15K

Is that right?


Thank you
juan dela cruz
Penniless INVENTOR
Edited by - JUAN DELA CRUZ on Aug 19 2008 04:50:09 AM
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pebe
Nobel Prize Winner

United Kingdom
1078 Posts
Posted - Aug 19 2008 :  04:53:10 AM  Show Profile  Reply with Quote
What input voltage range?
11V to 13.5?
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JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts
Posted - Aug 19 2008 :  04:59:06 AM  Show Profile  Reply with Quote
quote:
Originally posted by pebe

What input voltage range?
11V to 13.5?


Yes.
I need to cut-off the supply to the Inverter when the battery voltage reached 11V.

Thank you
juan dela cruz
Penniless INVENTOR
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pebe
Nobel Prize Winner

United Kingdom
1078 Posts
Posted - Aug 19 2008 :  05:46:12 AM  Show Profile  Reply with Quote
Opamp output range = 10V to 1V = 9V swing. Midpoint = 5.5V
Input voltage range = 13.5V - 11V = 2.5V swing. Midpoint = 12.25V

Source voltage of Re = input range/2
So effective input range = 1.25V. Midpoint = 6.125V

So R3/Re = 9/1.25 = ratio 7.2 to 1
If Re = 7.5K then R3 = 54K

So ref = 5.5 + (6.125 - 5.5) x (54/61.5)
ref = 5.5 + .625 x .878
ref = 5.5 + .548
ref = 6.048V

Use a 6.8V zener with a pot.
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JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts
Posted - Aug 19 2008 :  05:51:48 AM  Show Profile  Reply with Quote
quote:
Originally posted by pebe

Opamp output range = 10V to 1V = 9V swing. Midpoint = 5.5V
Input voltage range = 13.5V - 11V = 2.5V swing. Midpoint = 12.25V

Source voltage of Re = input range/2
So effective input range = 1.25V. Midpoint = 6.125V

So R3/Re = 9/1.25 = ratio 7.2 to 1
If Re = 7.5K then R3 = 54K

So ref = 5.5 + (6.125 - 5.5) x (54/61.5)
ref = 5.5 + .625 x .878
ref = 5.5 + .548
ref = 6.048V

Use a 6.8V zener with a pot.


Greetings

Why not using just a 6V Zener ?
ref = 6.048V.. almost 6V.

Thank you
juan dela cruz
Penniless INVENTOR
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pebe
Nobel Prize Winner

United Kingdom
1078 Posts
Posted - Aug 19 2008 :  05:58:38 AM  Show Profile  Reply with Quote
I didn't know anyone made a 6V zener. Can you buy one?
The nearest preferred value is 6.2V. A 6.8V type is probably easier to find.
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JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts
Posted - Aug 19 2008 :  06:16:08 AM  Show Profile  Reply with Quote
quote:
Originally posted by pebe

I didn't know anyone made a 6V zener. Can you buy one?
The nearest preferred value is 6.2V. A 6.8V type is probably easier to find.

No.. actually I didn't know the standard value of zener voltage available that suits in my ckt.. Now I know 6.2V or 6.8.

What if I need to use a fixed resistor instead of Pot.
What will be the formula in calculating the value of that resistor to decrease that 6.8V/ 6.2V (from the Zener) to 6.048V ?

*The resistance of a Zener diode use as the bottom element in a voltage divider is "0".
So, how can I calculate the value of the resistance needed using Thevenin's resistance ?


Thank you
juan dela cruz
Penniless INVENTOR
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pebe
Nobel Prize Winner

United Kingdom
1078 Posts
Posted - Aug 19 2008 :  06:23:19 AM  Show Profile  Reply with Quote
Thevenin doesn't come into it. It's just simply Ohms Law.
Take the voltage across the potential divider as 6.8V and calculate from there.
This is so simple - surely you don't want me to calculate it for you???
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JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts
Posted - Aug 19 2008 :  06:39:11 AM  Show Profile  Reply with Quote
quote:
Originally posted by pebe

Thevenin doesn't come into it. It's just simply Ohms Law.
Take the voltage across the potential divider as 6.8V and calculate from there.
This is so simple - surely you don't want me to calculate it for you???


Greetings...

Yes, of course Sir..
I want to calculate it by myself.
But I didn't know what is the current flowing in that Rx.

Using Ohm's Law...
Vrx = Irx x Rx (Rx is the needed resistor)
Rx = Vrx/ Irx
Rx = (6.8V-6.048V)/ Irx
Rx = 0.752V/ Irx
....Irx ?
(current will be 0.0135A (or 13.5mA) when I use 1K as the upper arm of the voltage divider that feeds the -ve input)

If so, Rx = 55.7 Ohms ?


ThanK you
juan dela cruz
Penniless INVENTOR
Edited by - JUAN DELA CRUZ on Aug 19 2008 06:50:43 AM
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pebe
Nobel Prize Winner

United Kingdom
1078 Posts
Posted - Aug 19 2008 :  06:59:31 AM  Show Profile  Reply with Quote
You have two resistors in series fed from 6.8V and you want to get 6V at their junction. Right?

There is no current flowing from the junction into the -ve opamp input. So the same current will flow through each resistor. So the ratio of the voltages dropped across each resistor will be proportional to each one's resistance.

The 6V that you require is dropped across the bottom resistor, leaving 0.8V dropped across the top one. The ratio of the top/bottom resistors is therefore 0.8/6. That's a ratio of 1:7.5

You could use a 10K for the top one and 2 x 150K in parallel for the bottom one.
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JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts
Posted - Aug 20 2008 :  04:26:47 AM  Show Profile  Reply with Quote

Greetings..

quote:
If you put a high on the 4047 reset (pin9) it will disable the Q drivers and you will get zero output to both FETs.

If you want to be even more sure after doing that, then disable the V+ supply to the 4047.


Sir, what components (except Logic gates) can be added in my ckt. (below) to put a "high" on the 4047 Pin9 (reset) to turn-off the MSW Inv. when the battery voltage reaches 11V ?


Thank you
juan dela cruz
Penniless INVENTOR
Edited by - JUAN DELA CRUZ on Aug 20 2008 04:30:25 AM
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pebe
Nobel Prize Winner

United Kingdom
1078 Posts
Posted - Aug 20 2008 :  04:33:14 AM  Show Profile  Reply with Quote
The top NOR gate will not turn off when the op-amp o/p goes low, so the flip-flop circuit will not work.

You need to reconfigure it using NAND gates.
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JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts
Posted - Aug 20 2008 :  04:47:31 AM  Show Profile  Reply with Quote
quote:
Originally posted by pebe

The top NOR gate will not turn off when the op-amp o/p goes low, so the flip-flop circuit will not work.

You need to reconfigure it using NAND gates.


Greetings..

I need to use "only" NOR gates.
Do you have an idea how to make a Set-rest latch using NOR gates ?

-- What if I remove the 100K from the bottom NOR gates leaving only the N.O. switch that can be activate to make the output "high"?


Thank you
juan dela cruz
Penniless INVENTOR
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pebe
Nobel Prize Winner

United Kingdom
1078 Posts
Posted - Aug 20 2008 :  05:01:18 AM  Show Profile  Reply with Quote
The o/p of the opamp is the wrong polarity. You need another NOR gate wired as an inverter placed in the feed to the top opamp.

Then apply the logic of using two NOR gates as a flip-flop and figure out how to wire up its four inputs.
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