OVER-/UNDER-VOLTAGE PROTECTION CIRCUIT ANALYSIS

[walid1]

Hi

The circuit attached below is OVER-/UNDER-VOLTAGE PROTECTION.
I have some questions:

1- R4 & R8 are 1K why not 10K?
2- Why D5?
3- the voltage divider consisting of the resistors R1, VR1 and R2, is to monitor the voltage, i can use the same voltage division with smaller or bigger values of resistors. my question is Why these particular values?
thank you very much

 

[AN920]

The 1k resistors are to make sure enough current is available into the base to saturate T1. The diode is added because the opamp is not a rail-to-rail opamp and may not switch low enough to completely turn off T1. The added drop over D5 gives more switch off margin.

 

[walid1]

Hi AN920

The diode is added because the opamp is not a rail-to-rail opamp and may not switch low enough to completely turn off T1. The added drop over D5 gives more switch off margin.

please can u explain it more using a numeric example.
thanks

 

[AN920]

You would expect the output of the opamp under ideal conditions to switch between supply and GND. Using a rail-to-rail opamp will bring you close to this. Older opamps can't switch like this. So if the low of your opamp switch to 0.4V the transistor may not be completely off. If the transistor operates in a high ambient temperature the b-e voltage threshold will be lowered at -2.5mV/degC

By adding the extra diode drop you make sure that the transistor is off with 0.4V coming from the opamp.

 

[walid1]

yes yes thank you very much, it is clear now.
still the question #3 can u tell me something about it?

 

[AN920]

Those values are not critical. As long as you get the same adjustment voltage range on the pot without too much current through the pot you will be ok. You should always pick your values to have very little dissipation in the pot by limiting the current flow through the slider.

I just noticed something; LED next to D6 should have a current limit resistor (~1k) in series otherwise it will blow when 12V is switched accross it. 

 

[audioguru2]

The circuit is not designed properly:
1) The LED will blow up! It doesn't have a current-limiting resistor. Maybe the transistor will also blow up when it tries to saturate.
2) An LM324 was used in the circuit but it is a quad opamp. An LM358 dual opamp is exactly the same opamp but is a dual and should have been used instead.
3) D5 is not necessary because the output of the LM324 typically goes to +10mVDC when it doesn't sink much current.

 

[walid1]

Hi

I put the scheme to make the discussion easier.

To guru

Maybe the transistor will also blow up when it tries to saturate.

why?

To AN920

The diode is added because the opamp is not a rail-to-rail opamp and may not switch low enough to completely turn off T1. The added drop over D5 gives more switch off margin.

Guru said:

D5 is not necessary because the output of the LM324 typically goes to +10mVDC when it doesn't sink much current.

what u say?

thank you very much

 

[audioguru2]

Hi Walid,
The LED doesn't have a series current-limiting resistor.
The transistor has a base current of 8mA and has a current gain of at least 80 when it has a collector current of 100mA.

So the current through the LED and transistor will be much more than 100mA.
The max allowed current for the LED is 30mA.
The max allowed current for the transistor is 100mA.
The max power dissipation allowed for the transistor is 625mW.

Will the LED or will the transistor blow up first?

 

[walid1]

Hi guru
It is a good answer and excellent, but it opened the way to further questions.
1-I believe that these calculations were in the case of D5 still in the circuit. If we remove this D5 entirely from the circuit, Are these calculations remain as they are?
2-is there a darlengton pair connected to the o/p of op-amp N1 makes a voltage drop of 1.2V (2* 0.6), that is the Vcc= 12 in o/p it 12- 1.2=10.8V?
3- I understand very well how IB to be equal 8mA,  But I do not assimilated  how VC = 10.2V, can you please explain it. If VCE = VC= 10.2, then T1 is nearly off.

I noted the name under the above picture and laughed it. That name is very insulting telling story honestly

Thank you guru.

 

[audioguru2]

The red LED is about 1.8V so the lowest voltage for the transistor's collector is +12V - 1.8v= +10.2V before one of them burns out.

With a base current of 8mA then the transistor is turned on as hard as it can with an extremely high current and it will be extremely hot.

Look at the datasheet for the old LM324 to see its darlington transistor at the output.

Nearly every schematic from that website is about 30 years old and has some errors in it.

 

[walid1]

Hi guru

Look at the datasheet for the old LM324 to see its darlington transistor at the output.

yes i see it.

questions:
1- what is the minimum value of  R4 to make T1 not smoking?
2-If I remove this LED, put the correct value of R4 and we know that the relay resistance is 200 ohm,  what  the value of VC at T1?
3- I have many relays, there is no resistance value printed on its case, how to know if this relay is 200 ohm or not, can i measure its resistance as if it a resistor?
thank you guru

 

[audioguru2]

1- what is the minimum value of  R4 to make T1 not smoking?

You need R4 to turn on T1 as hard as it can. The LED needs to have a current limiting resistor in series with it then the circuit will work fine.

2-If I remove this LED, put the correct value of R4 and we know that the relay resistance is 200 ohm,  what  the value of VC at T1?

Don't change R4.
A graph in the datasheet for the BC547 transistor shows that the collector saturation voltage is typically 0.2V when the collector current is 60mA (200 ohm relay with a 12V supply) and a base current of 6mA.

3- I have many relays, there is no resistance value printed on its case, how to know if this relay is 200 ohm or not, can i measure its resistance as if it a resistor?

Yes. Also the relay's coil must have the proper amount of voltage for it to activate.

 

[walid1]

Hi guru

I understand from your words that there is no problem in the value of the R4 (=1K) resistance. And all of the problems resulting from a lack of a current limiting resistor in series with the LED.
I am now comes over as I feel I did not understand anything. At the outset I thought I understood. But your last reply shows I did not come to the desired point yet. Therefore, I want to start again:
(1) If we assume that the LED and its limiting current resistor are removed at all from the circuit as shown below.
I noticed that IB is still = 8mA and this is a great value, if we make R4 more bigger (I don’t know to any level) we can reduce IB.
(2) please tell me about the importance of the presence of D5.
(3) Is it true that IC = (12-0.2)/200 = 59mA?

 

[audioguru2]

Hi Walid,
The transistor is not straining so hard without the LED. With the LED, its load was 60mA for the relay and about 25mA for the LED if it had a current-limiting resistor. The total collector current would have been 85mA so the 8mA base current is fine. Without the LED then you can increase the value of R4 a little for a base current of 6mA. But 2mA doesn't matter.

The transistor was poorly chosen as a BC547 with an absolute max allowed collector current of only 100mA. Its current gain drops above only 30mA. I would have used a relay that uses less current or a higher current BC327 or 2N4401 transistor. Their saturation voltage is much lower so the base current can also be much lower if you want.

D5 is needed if the opamps are ordinary ones that their output go down to only +1VDC.
The outputs of the LM324 and LM358 opamps go down to +10mV so the transistor will be off without D5.

 

[walid1]

Dear guru:

D5 is needed if the opamps are ordinary ones that their output go down to only +1VDC.

Give me an example (names) of that op-amps.

The outputs of the LM324 and LM358 opamps go down to +10mV

I Know the o/p high = Vcc – 1.2V, but don’t know the o/p Low = 10mV?

and about 25mA for the LED if it had a C.

I want to make a calculation,
Let  R = 330 ohm
12V – I * 330 – 1.8V – 0.2V =0
I = 10/330 = 30 mA
If we choose R = 470 ohm
I = 10/470 = 21 mA
If we choose R = 1K
I = 10/1000 = 10 mA
Ok
Can you please tell  Any value is the best? And why?

Now we go to a general questions:
(1) Is all of an op-amps suitable for use as a comparator or is that only certain types?
(2) How can know the value of the relay coil resistance?

Thank you guru for your patience

 

[audioguru2]

Give me an example (names) of that op-amps.

The 741, 1458 (dual 741), TL07x and many more ordinary opamps have outputs that go down to about +1V above their negative supply which could be ground. It is a little more than 1V if the current is higher than about 5 mA.

I Know the o/p high = Vcc

 

[indulis]

There a bit more difference between the two than just that...

Can compensation make a opamp switch slow??? Yes it can, but  it's not automatic, you you have to select a value (LARGE) on purpose for that to happen. Then, slow is a relative term, how slow is slow? One reason why opamps make lousy "fast" comparators is because you make the output saturate and it take time to come out of saturation, it's not exclusively because of compensation.

 

[audioguru2]

Hi Indulis,
The old LM358 low power dual opamp and the old low power LM393 dual comparator were made at about the same time, have the same amount of voltage gain and have very similar circuits.

The opamp slews 2.5V in 10us.
The comparator slews 2.5V in 0.38us positively and even quicker negatively.
The comparator is 26 times quicker than the opamp.

My example shows the output transistor coming out of saturation and the total response time of the comparator as about 0.26us. The response time of the opamp is not seen on its graph because its slew rate is so slow.

It is the frequency compensation capacitor in the opamp that causes its slew rate to be so slow.