[SOLVED] Zenner diode not working. No Iz enough ?

Hey guys,

I made a circuit for conditioning square waves at the range 0-10V to LVTTL.
I've used an opto-coupler to help removing noise, and to drive the opto-coupler's LED I'm using a BJT as the input's signal is weak. I also desired to create some kind of bias, where you have a tolerance for what the circuit see as low level at the input of it. I tried to use a 2.4v zenner diode to create that range, in a way that only more than 2.4V will power ON the opto-coupler's LED.



Attached there is a Picture of my schematic, but the circuit that is now in a PCB didn't work.

In the Picture above I'm introducing a square wave 0-10v in the "IN" point and I'm expecting the BJT to switch the opto-coupler.
The opto-coupler's output is digital LVTTL so I get the translation from 0-10V to LVTTL with schmitt trigger, noise immunity and with the so WANTED hysteresis (provided by zenner)

I tested before in a proto-board (same concept, but with some differents componentes) and worked.
Can you help me in trying to understand why the zenner diode is not working as I want ?

Thanks in advance.

Hi,

What input signal? Is it push pull? What low voltage and what high voltage? What frequency?

Do you have a scope to check the voltages?

How do you know the circuit is not working? Is the output always on or off?

Klaus

KlausST said:

Hi,

What input signal? Is it push pull? What low voltage and what high voltage? What frequency?

Do you have a scope to check the voltages?

How do you know the circuit is not working? Is the output always on or off?

Klaus

Click to expand...

The input signal is a square wave of less than 1Khz.
Low = 0V
High = 10V
The input signal comes from an op-amp from inside a sensor (not designed by me).
The idea is to make the op-amp from that sensor drive my opto-coupler but I need to remove noise and other problems so I'm using that.
I've a scope. I'll make some measurements and post the reply.
I have a microcontroller Reading the pulses after the opto-coupler.
I know that is not working because if I short the leads of zenner diode, I get the pulse read by the MCU, but if I put the diode in the middle, it doesn't read anything. I'll look with a scope the voltage levels after the zenner and at the BJT's collector.

Just one thing before: Do you agree with me that this circuit makes sense or is it wrong ?

I would forget the diodes, capacitor and R6. Change R5 to 10K. This will allow the voltage to directly bias the transistor, there isn't much point in dropping the voltage then limiting the current again afterwards. If the 10V signal can supply more than 1mA or so you might be better using it to directly supply the optocoupler LED instead of the transistor.

The reason it misbehaves is the capacitor is having two undesirable effects, firstly it acts like a reservoir and charges to give a more continuous bias current to the transistor and secondly it skews the waveform slightly.

Optocouplers do not reduce noise, they are only useful in situations where the input and output circuits have to be electrically isolated.

Brian.

Hi,

Optocouplers do not reduce noise, they are only useful in situations where the input and output circuits have to be electrically isolated.

Click to expand...

Hi agree with that. If you don´t need the galvanic isolation, then there is no need for an optocoupler.


About the R and C.

With the zener you can adjust noise immunity levels.
With the C you can adujst tofilter out high frequency (spikes).

So if the desired frequency and voltage levels are known then the D and C can improve your circuit. If calculated wrong, then it can make signal quality worse.

Klaus

A Zener in series with a noisy signal will make the situation worse, it maintains the noise level while dropping some of the DC so overall, the SNR is worse. To 'clip' the noise it has to be across the signal.

Simply using the signal to bias the transistor should be OK, given enough current the transistor's non-linearity as it approaches saturation will supress the noise.

Brian.

The datasheet for the BZX84 zener diode and for all other zener diodes shows that with such a low voltage rating its voltage regulation is awful, not much better than a resistor.

Hi guys, yeah the opto-coupler is actually needed for isolation also so it should remain at the circuit.
The problem that I have to drive the transistor directly is that the input "IN" is a conector, that can be left open, or connected to a sensor through a long cable.

Thus the R6 is essential for the circuit as IN may be High Z
The C6 will damp spikes and high frequency noise.
The D2 will remove any possible negative voltage on that port.

The only problem is the BZX84 zenner 2.4v which is not playing its role.
For the zenner effect to happen, I would need Iz that in the circuit above would flow through IN -> D1 -> R5 -> R6 - > GND
But it seems that I'm wrong because it is not working.

You are completely ignoring that the B-E junction of the transistor will carry most of the current, not R6.

I still say forget the Zener, I can't see it serves any useful purpose. If negative voltages are a problem it would make more sense to move D2 to across R6 so they are dissipated in R5. you could also try moving the capacitor across R6 as well.

For protection against a floating input you can increase the value of R6 considerably as all it has to do is drain away any transistor leakage. I would suggest 100K. Then calculate the value of R5 so the 10V high level of the input signal drives the transistor to saturation. Making some assumptions, hfe = 100 and the optoLED is to pass 10mA, Ib will be 100uA. Using the 10 times Ib rule of thumb figure to ensure saturation, R5 has to pass 1mA while dropping (10-0.7)V so it's value should be 9.3K. The nearest standard value erring on the cautious side of things is 9.1K.

Brian.

An unclear point is the actual signal driving the circuit in your test. It should work with a 0/10V 1kHz square wave, although the circuit is probably far from a reasonable design.

betwixt said:

You are completely ignoring that the B-E junction of the transistor will carry most of the current, not R6.

I still say forget the Zener, I can't see it serves any useful purpose. If negative voltages are a problem it would make more sense to move D2 to across R6 so they are dissipated in R5. you could also try moving the capacitor across R6 as well.

For protection against a floating input you can increase the value of R6 considerably as all it has to do is drain away any transistor leakage. I would suggest 100K. Then calculate the value of R5 so the 10V high level of the input signal drives the transistor to saturation. Making some assumptions, hfe = 100 and the optoLED is to pass 10mA, Ib will be 100uA. Using the 10 times Ib rule of thumb figure to ensure saturation, R5 has to pass 1mA while dropping (10-0.7)V so it's value should be 9.3K. The nearest standard value erring on the cautious side of things is 9.1K.

Brian.

Click to expand...

I completly agree with you at these points, but have in mind that the hfe = 100 is just an assumption.
It should change in many ways and also the PCB is assembled in China where they will use any BJT for that circuit.
The one that I'm using seems to be hfe=400 then I would have a problem.
As I said, I agree with you, and I even considered your approach, but I want something more "fixed", i.e. that doesn't depend of performance of the componentes so drastically.
Just for example, instead of a zenner, I could put a common 1n4148 in series in forward mode, it would give me the drop of 0.6v. Adding 2 would give me about 1.2 and so on... but that is kind of a kludge...
I still prefer to rely in the zenner that is set from factory to conduct at that exact voltage.

I'll report the result of my measurements in few hours.

Thanks everyone so far.

The datasheet for the BZX84-A2V4 and any other low voltage zener diode shows that it begins to conduct when the voltage across it is only 1V, not 2.4V. That is why I said its voltage regulation is awful.
The datasheet for a forward biased signal diode like a 1N914 or 1N4148 shows that it begins to conduct when the voltage across it is only 0.47V, not 0.6V.

Well ok, this deeper analysis of the diodes is great, but doesn't explains why between D1 and R5 I have nothing when I input a signal of 200Hz 0-7volts on "IN" point. I just would like to understand why the zenner is not conducting when I apply 7v to it on that configuration displayed in my picture.

Hi,

With the circuit of post1 i expext a trip point at about 1.6 ... 1.9V.

Check if you accidentely used a standard diode instead of a zener. Or if it is a zener with wrong voltage.

With your circuit, if you give 7V at the input (cathode) then you should see at least 5V ia the anode. If this is not, then replace the zener.

Klaus

KlausST said:

Hi,

With the circuit of post1 i expext a trip point at about 1.6 ... 1.9V.

Check if you accidentely used a standard diode instead of a zener. Or if it is a zener with wrong voltage.

With your circuit, if you give 7V at the input (cathode) then you should see at least 5V ia the anode. If this is not, then replace the zener.

Klaus

Click to expand...

:idea:

Well... Yeah the zenner is damaged.
I just tested other board and it works...
D'oh! :roll:

I was almost freaking out...
Thank you all guys.