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different zener diodes in parallel- what about resistors?

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kris1

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hi

In my circuit I had a 230/12. rectified, and filtered(using 470uF), supply which supplies an electrmagnetic relay of 12v 200 ohms. taking approximately 80 mA.

Now I need to tap 5.1 V, 9.1 V and 15V from the same source, for which i used Zener diodes. 5.1V zener is for charging 4700uF Capacitor. 9.1V is for 555timer and 15V is for IC4011. Accordingly what are the resistors to be used?
I used 10k resistor for each zener. Is it OK?

also, From pspice i found that filtering capacitor(470uF) power dissipation is having around 15 watts pulses. why is it so?
 
kris1 said:
hi

In my circuit I had a 230/12. rectified, and filtered(using 470uF), supply which supplies an electrmagnetic relay of 12v 200 ohms. taking approximately 80 mA.

Now I need to tap 5.1 V, 9.1 V and 15V from the same source, for which i used Zener diodes. 5.1V zener is for charging 4700uF Capacitor. 9.1V is for 555timer and 15V is for IC4011. Accordingly what are the resistors to be used?
I used 10k resistor for each zener. Is it OK?

also, From pspice i found that filtering capacitor(470uF) power dissipation is having around 15 watts pulses. why is it so?

hi kris,
Can you post a drawing on how you have wired the zener circuits etc.?
 
Is it ok

Now iam sending
the exact drawn circuit's jpg attachment. u may see it.
view this after saving.
 

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hi,
Just typed in a reply.. system crash so I'll try again.

The 10K and 5.1V zener will charge the capacitor slowly to about 5v.

The 10K with the 9.1V for 555 [is it TTL?] is much too high.

The 10K and 15V zener will NEVER give you 15V, you cant get +15V from +12V in this way.

If you tell us what the project is for, we could suggest other ideas..:)
 
12 V means from ac rectification

sir as ac when rectified gives the peak voltage as its output. i.e.,
12 *1.414=16.968 volts(approximately) as dc voltage. then the filtering capacitor taken is of 470uF 35v. will smooth it to about 15 v.
My doubt is whether we can connect these different zeners in parallel. And whether the remaining two zener resistors are ok or not .
 
While it is easy to purchase and use 7805, 7809(or their 78L counterparts with input &output pin reversal), 15 volts is rather difficult after loading the the rectfied supply.

Perhaps it is preferable to try out boost converter version of lm257x series. How ever, CMOS chips work well at 12V also. 15V is rather on the higher edge of its use.
 
The simplest solution would be a 78L05 and a 78L09. The 4011 will happily work with 9V.

Mike.
 
kris1 said:
sir as ac when rectified gives the peak voltage as its output. i.e.,
12 *1.414=16.968 volts(approximately) as dc voltage. then the filtering capacitor taken is of 470uF 35v. will smooth it to about 15 v.
My doubt is whether we can connect these different zeners in parallel. And whether the remaining two zener resistors are ok or not .
hi kris,

Lets take it one zener at a time.
Consider the 9.1V for the 555 timer.

Q1.. is the 555, CMOS or TTL.?
Q2. What is the 555 output driving at what current.?

Like to help.:)
 
Lets take it one zener at a time.
Consider the 9.1V for the 555 timer.

Q1.. is the 555, CMOS or TTL.?
Q2. What is the 555 output driving at what current.?
555 is CMOS
555 is driving the input of 4011 ic i.e., two common inputs of NAND gate.
The output of NAND gates, which is acting as nor gate, drives the transistor base(BC547)
 
kris1 said:
Lets take it one zener at a time.
Consider the 9.1V for the 555 timer.

Q1.. is the 555, CMOS or TTL.?
Q2. What is the 555 output driving at what current.?
555 is CMOS
555 is driving the input of 4011 ic i.e., two common inputs of NAND gate.
The output of NAND gates, which is acting as nor gate, drives the transistor base(BC547)

hi,
Assume the 555 and the logic and the base drive require 15mA and that the supply is +12v.
Allow an additional current of 5mA into the zener, equals 20mA total.

So Resistor = (12 -9.1)/.02 = 147R,, npv = 150R

If you had a supply of +Vs then the formula is

Res= (Vs -Vz)/ (Iz + Iload)
Set Iz= 5mA and calculate the current required by the rest of the circuit the zener output is driving.

Does this explain this part OK.?
 
The 4011 is driven by the C555 so they both must have the same supply voltage. If they both use 9.1V then the current in the 10k resistor is too low for the 4011 to drive anything. The current is only (15V - 9.1V)/10k= 0.59mA.
If the 10k resistor is changed to 1k then there is 6mA available for the 4011 to drive the transistor plus a few mA for the zener diode.
 
Hi Kris,
Downloaded your drawing, that you added to a previous post.

I have added some components marked in BLUE and voltages in RED.

Reposted your circuit to help other members locate easier.

Can you explain exactly what you are trying to do with this circuit.
 
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explaining

Q1 & Q2 for High voltage and low voltage cut respectively.
the nand gates provide drive to base of transistors.
The connection is such that the relay cannot energise by itself. the transistor Q3 provides the closed path if it is switched on by timer.
Initially the timer will be getting supply so its output will be high so it may immediately turn on Q3. But I mean to protect he equipment from sudden trip and resumption. so i should have a time delay.
timer output will be high for a certain time & then goes low. So i used one NAND gate here as inverter. i.e., providing low input to Q3 initially & then high after a TIME DELAY. This will happen only if Q1 &Q2 are in On state.
Immediately after switching on of relay, timer will not get any supply.
Q4 is pnp and will be in off state as long as the relay is in energized state.
so will be the buzzer. Hence it gives brief alarm during power failure; continuous beep for high/low voltages. Also beeps after power restoration until circuit is turned ON automatically.
The components and some part of composition was not totally mine. But the logic of this connection is my thought.
Thank you for your support. and thanks to electro-tech.
Kris
 
hi kris,
Saved your descriptive text OK.

I have gone thru part of the circuit already, there are many parts of the circuit that have problems.

Do you have the spec on the buzzer, volts/current.?

This design could take some time to sort out. :)
 
The 4011 has a 15V supply. Its minimum input-high voltage is 11V. Some might work with an input high voltage of 7.5V.

But the 555 has a supply voltage of much less than 9.1V. So its output high voltage is less than 7.8V and most 4011 ICs won't work.
 
Generally high and low voltage sensing is done before filter(C1). Both these signals could be ANDed and drive a single transistor. the so called delay created by the timer cold be also created by RC network (R in series and C in parallel after R).If timer is used, perhaps timer and cmos IC can both have same voltage, say 9V. you may also see the recent comments of Audioguru, in this context. Finally nothing should prevent operating the buzzer from operating from the same voltage even.
 
mvs sarma said:
Generally high and low voltage sensing is done before filter(C1). Both these signals could be ANDed and drive a single transistor. the so called delay created by the timer cold be also created by RC network (R in series and C in parallel after R).If timer is used, perhaps timer and cmos IC can both have same voltage, say 9V. you may also see the recent comments of Audioguru, in this context. Finally nothing should prevent operating the buzzer from operating from the same voltage even.

hi sarma,
As I have posted, the design needs a lot of reworking, do you want to volunteer..;)

Regards
 
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If we give the Sensed voltages to the same NAND gate or and gate. as two inputs, we get 1 0 ---->1
0 1 ----->1
0 0------>1
1 1------>0
That means it can sense only one abnormal condition but not both.

As you said same supply for 4011 and timer. The timer will not get supply as long as relay is energised.
As u said i also feel that we can have a simple RC instead of TIMER.
But it need to be discharge soon after switching on of relay.
Then only we can have protection from sudden tripping and resumption.
 
ericgibbs said:
hi sarma,
As I have posted, the design needs a lot of reworking, do you want to volunteer..;)

Regards
Eric, As you are already on the process i would like Kris1 to make it work and learn from the design.
I would, in the mean time try and come out with a so called v.2. Perhaps it is more important that Kris1 should apply himself and learn to improve the design.
 
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