ljcox

X4 is operated for the time that the B4 button is pressed, so I don't see why you need to increase its release delay.

__________________
Len

ljcox

The reconstruction was driven by Tony Sale. Some years ago I looked at their web site and obtained some circuits on a CD that Tony sent to me (there was a modest cost).

They produced 2 input AND gates by connecting the inputs to the grid and suppressor grid of a pentode valve. And XOR gates using 2 AND gates and a wired OR.

The German teleprinter cypher (code named Tunny) was transmitted via the international standard teleprinter code but the text was encoded using relay XOR gates and switching wheels.

The code to be decyphered was stored on a loop of paper teleprinter tape (5 bit words). It rotated very quickly around a series of pulleys and was read by photo cells.

Apparently, the tape broke occasionally - with spectacular results.

__________________
Len

Hayrettin Demir

If B4 button 0,1sn pressed...
Then X4 0,1sn operated and DS 0,1sn active.
(DS --> 1-3sn active to be).
...
So ;
If B4 button 10sn or more pressed...
Then X4 10sn or more operated and DS 10sn or more active...
This is a problem for DS...(DS is break down !).
...
.............................................

Series resistor (82 Ohm) is only for charge required.

__________________
I am sorry my friends...
For my English.

ljcox

[quote=Hayrettin Demir]If B4 button 0,1sn pressed...
Then X4 0,1sn operated and DS 0,1sn active.
(DS --> 1-3sn active to be). So you want to limit the time that DS is energised regardless of how long B4 is pressed. If so, then you will need to do more than just making X4 slow to release....
So ;
If B4 button 10sn or more pressed...
Then X4 10sn or more operated and DS 10sn or more active...
This is a problem for DS...(DS is break down !). Yes, DS is likely to over heat and burn....
.............................................

Series resistor (82 Ohm) is only for charge required. Yes, but it makes little difference to the release time. As I said previously, this is a standard technique used to increase relay release time.
I have never seen a diode across the resistor.
What you are missing is that the resistor increases the time constant and may in fact sligthy increase the release time compared to the case where the diode is included.

It is a complex situation. The initial voltage across the coil will be 12 Volt, not 11.3 as you have shown since, due to the coil inductance, the current does not change when the button is released.

In your second diagram, ie. where there is no diode, the voltage across the coil will be 12 Volt and the voltage across the resistor will be I * R where I is the coil current and R is the resistance of the resistor. For example, if the coil resistance is 600 Ohm, the current will be 12/0.6 = 20 mA.

If R = 82 Ohm, the voltage across it will be 20/1000 * 82 = 1.64 volt.

__________________
Len

ljcox

This is what I mean

Oops I made an error. See the revised diagram.

Attached Images

Relay release.gif (3.7 KB, 4 views)

__________________
Len

Hayrettin Demir

__________________
I am sorry my friends...
For my English.

ljcox

You did not look at my latest edit. The initial relay current is 20 mA.

My error was that I forgot to adjust the relay coil voltage.

It is 10.36 Volt which obeys Kirchoff.

I expect your next point will be - that is not 12 Volt, so the relay delay will be less.

But the magnetic force on the armature is proportional to the current, not the voltage.

My diagram shows that the current does not change (due to the nature of inductance) when the switch is opened. It decays from 20 mA.

So there will be little difference in the release delay. In fact it may be slightly longer since the capacitive time constant is a little greater due to the 82 Ohm resistor.

It is not possible to do this by intuition. You need to solve a second order differential equation and include the relevant initial conditions.

See the attachments. Start at the heading "RLC Transient" near the bottom of page 247.

This gives you an idea of what I'm trying to say. In your case, the initial conditions are:-
1. the capacitor has an initial charge of 12 Volt
2. the inductance of the relay coil has an initial current of 20 mA (assuming a 600 Ohm coil resistance).

So these need to be included in the mathematics.

Attached Images

	p_247.gif (74.6 KB, 4 views)
	p_248.gif (82.3 KB, 5 views)

__________________
Len

Hayrettin Demir

Please check again...

__________________
I am sorry my friends...
For my English.

ljcox

You have added the capacitor and resistor voltages.

The arrows show the polarity. Therefore the voltage across C1 is 10.36 + 1.64 = 12 Volt. As I said previously.

The point you're missing is that the back EMF of the coil is 1.64 Volt

I have simulated the circuit using Switcher CAD III.

The relay coil is simulated by R3 and L1. I have assumed that the inductance is 1 Henry. I have use a transistor to simulate the switch.

The Red curve is the current through C1. It is negative when the capacitor is charging.

The Green curve is the transistor base voltage.

The Blue curve is the current through the coil.

The Pink curve is the collector voltage.

Note that, while the transistor is on, the coil current is 20 mA and the C1 charges to 12 Volt.

When the transistor is turned off, the coil current decays from 20 mA.

Attached Images

	Simulation circuit.gif (31.9 KB, 7 views)
	scope.GIF (11.1 KB, 5 views)

__________________
Len

ljcox

Here is a diagram showing the relay coil voltages at the instant that the switch is opened.

For simplicity, I have not shown the capacitor and 82 Ohm resistor that are connected across the coil.

It shows that the back EMF is 1.64 Volt and this is driving the current. The back EMF is of such a value as to keep the same current flowing that was flowing just before the switch opens,

The coil can be considered to be an inductance in series with a resistance.

EDIT:
Here is a more accurate wave form chart. The transistor was not switching fast enough, so I had to change some parameters in the simulation.

Notr that the C1 current is equal to the L1 current immediately the transistor is turned off - as expected. They both decay from 20 mA.

Attached Images

	relay coil voltages.GIF (24.0 KB, 4 views)
	scope.GIF (10.1 KB, 2 views)

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Len

Hayrettin Demir

All Relays have coil parallel diode...
So , this relays coils have not back EMF...
...
...

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I am sorry my friends...
For my English.

ljcox

Yes, they do have back EMF. That is why the diode is required.

The diode limits the back EMF to about 0.7 Volt and therefore protects the MOSFET and the switch contacts.

If the diode was not connected, the back EMF would be very large and it would destroy the MOSFET and cause arcing across the switch contacts as they open - thus eroding the contacts.

The 82 Ohm and capacitor are not shown for simplicity. But they are assumed to be connected across the relay coil as in previous diagrams.

If a resistor of 1000 Ohm was connected across the coil, the voltage across the coil at the instant the contacts open would be 20 mA * 1000 Ohm = 20 Volt.

I have simulated the cases with and witout a diode across the 82 Ohm resistor. See attachments

As you can see, the release delay is slightly longer without the diode.

This is because the time constant is 600 * 1000 uF = 600 mS with the diode and (600 + 82) * 1000 = 682 mS without the diode.

Since I don't know what relay you are using, I have assumed that the relay release current is 4 mA.

Attached Images

	without diode.GIF (21.4 KB, 2 views)
	with diode.GIF (23.4 KB, 1 views)

__________________
Len

ljcox

I did not fully answer your question - Is this 20 mA FIXED?

Answer - YES.

You need to understand inductance.

When the circuit is opened, the inductance wants to keep the same current flowing.

So if the resistance across the coil was say 100k, the back EMF would be 100k * 20 mA = 2000 Volt.

__________________
Len

Hayrettin Demir

Please see 1 -> 2 -> 3 -> 4 -> and 5 .

Attached Images

final.png (63.9 KB, 5 views)

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I am sorry my friends...
For my English.

Hayrettin Demir

Also;
With parallel diode R8 (82 Ohm).
Or more circuit simulations (Proteus-ISIS).

Attached Images

final diyotlu.png (64.6 KB, 3 views)

__________________
I am sorry my friends...
For my English.