Need help to understand simple circuit

[Nora]

HI-
I've attached a simple circuit to this post.
Can someone explain it to me?
Main question is how does the voltage change from 3.3V to affect the ADC?
Thanks!
N_N

 

[FvM]

The circuit is delivering 0.4 to 2.0 V to the ADC. 3.3V is the Z-diode nominal breakdown voltage, that should be never reached in normal operation. It's for overload protection only. There's no 3.3V present in the circuit.

As a severe disadvantage, a 3.3V Z-diode has considerable leakage current at 2 V, affecting the circuit linearity. It shouldn't be used as shown to my opinion. The ADC substrate diodes can achieve a usual overload protection. If high energy bursts shall be clamped, a Z-diode with a higher breakdown voltage or a varistor may be used additionally.

 

[Nora]

Thanks for the reply.
R2 and C3 is a low pass filter?

What is R1 and C1 used for?
N_N

Added after 4 hours 2 minutes:


So if Isource is 4 to 20mA and approaches the Z-diode as Iout, then Iout is actually 0.73 - 3.6mA (by using formula for current divider Iout = Isource*(R4/(R4+R3)) ).

Carrying the current over R2 gives an approximate Voltage into the ADC of 0.73 - 3.3 (3.6V but limited by the Z-diode).
This is an inherited design to me, but shouldn't there be a zero voltage option into the ADC?

Am I missing something?
Thanks!
N_N

 

[laktronics]

Hi,
Yes, you are missing a ground connection, D1 anode to be connected to ground.

Regards,
Laktronics

 

[FvM]

I had been complementing the missing ground connection on my own cause the circuit wouldn't make minimum sense without it.

 

[Nora]

Yes, that was a mistake...there should be a connection from D1 to signal ground.
New schematic attached- note also that C3 is now C2.

FvM- does the analysis I suggested make sense?

N_N

 

[FvM]

does the analysis I suggested make sense

No. Assuming a high impedance ADC input and ignoring zener leakage current, you have 1k+1k = 2k in parallel to 220R, a current divider factor of 0.10 and a 1k shunt to the ADC. As a result, you get the output voltages I previously mentioned.

 

[Nora]

Thanks FvM, that makes sense.
I simulated the circuit as well as following your calculations and got the same results.

When analyzing a circuit with a zener in it for voltages and currents, how do you analyze the zener- ? I thought of it as a high impedance- open circuit- non-existant..??

Also, if R2 is changed, say to 2k, then the output voltage to the ADC should double.
However, in the simulation, it remains the same.

:?:

Thanks in advance- this is really helpful
N_N

 

[Kral]

Nora, FvM is correct regarding the "soft" regulation characteristic of a 3.3V Zener. In general, the higher the Zener voltage, the sharper the knee. If you have a 3.3V supply available, then a better way to clip the input is to connect a diode between the point to be regulated (junction of R1, R3) and the 3.3V supply. The cathode goes to the 3.3V supply. Now you only have to be concerned with the reverse leakage of the diode. The voltage will be clipped at Vf +3.3V, where Vf is the forward drop across the diode (approx .65V). The 3.3V supply could be a 3.3V Zener with a high current (approx 20 mA) flowing through it, to ensure that the diode is reverse biased at voltages lower than 3.3V
Regards,
Kral

 

[FvM]

R2 is only determining the second filter time constant and - if the value is too high -causing offset errors due to ADC input leakage current. A sampling ADC can actually have a non-neglectable input current, so some caution is advisable here. Many ADC datsheets are specifying a maximum source resistance that shouldn't be exceeded to avoid input current induced errors.