what are this Diode doing here ? and dose it affects on R (sense) voltage value to be 0.7 allways ?

Thread Starter

Eng Mohamad Yahya

Joined May 28, 2016
4
Hello everybody, I'm trying to make a (resistor / ma) specially 4-20ma
you can find the circuit in this
site
, I understood a lot if details in this circuit except those things >>>



1- what is the purpose of D1 ?
2- the diode is in parallel with Rsense ,,, i think it will limit the Rsense Voltage to 0.7 ? Am I right ? and if that ,,, how could the resistance Rsence be a voltage sensing element in the cuircuit ?
3-what is the purpose of C1 ? ( dose it for absorb the sudden changing in U1A output (U1B input) values , and make the change in U1B output values be smoothly !! I don't know !!
please Help !
thanks in advanced :)

 

ian field

Joined Oct 27, 2012
6,536
Hello everybody, I'm trying to make a (resistor / ma) specially 4-20ma
you can find the circuit in this
site
, I understood a lot if details in this circuit except those things >>>



1- what is the purpose of D1 ?
2- the diode is in parallel with Rsense ,,, i think it will limit the Rsense Voltage to 0.7 ? Am I right ? and if that ,,, how could the resistance Rsence be a voltage sensing element in the cuircuit ?
The diode in parallel with Rsense wouldn't make much sense - I think that schematic is badly drawn with indistinct dots where crossing lines connect. These mistakes couldn't happen when schematics were drawn with crossing lines that did not connect; had a "hump back bridge" to jump over the line it isn't connected to.

IMO: the diode is simply in series with the regulator GND to lift its output by 0.7V. You could also return the regulator GND to a voltage divider from is output to set the voltage to any value above its rated output voltage.

Incidentally - the diode in your circuit should have a pull up resistor to develop its Vf.
 

AlbertHall

Joined Jun 4, 2014
12,346
I don't agree about the diode being to lift the regulator ground. Incidentally the regulator is incorrectly labelled - in and out are swapped.
I believe the diode is there to cope with what happens when the loop supply is switched on. Effectively. a discharged C3 is in series with Rsense across the supply, putting, potentially, 24V across Rsense. The diode prevents this.

However C1 confuses me because it applies positive feedback around U1B.
 

crutschow

Joined Mar 14, 2008
34,420
I don't agree about the diode being to lift the regulator ground. Incidentally the regulator is incorrectly labelled - in and out are swapped.
I believe the diode is there to cope with what happens when the loop supply is switched on. Effectively. a discharged C3 is in series with Rsense across the supply, putting, potentially, 24V across Rsense. The diode prevents this.

However C1 confuses me because it applies positive feedback around U1B.
I agree with your analysis.

Yes, it would appear that the diode is in parallel with Rsense and provides the start-up protection you state.
Under normal operation the diode should have little or no effect on circuit function.

I think the intention of C1 is to provide some frequency rolloff and stability to the circuit but it would need to be connected to provide negative feedback, such as from the base of Q1 to the (-) op amp input (with R4 and C1 providing rolloff).
 

ian field

Joined Oct 27, 2012
6,536
I don't agree about the diode being to lift the regulator ground. Incidentally the regulator is incorrectly labelled - in and out are swapped.
I believe the diode is there to cope with what happens when the loop supply is switched on. Effectively. a discharged C3 is in series with Rsense across the supply, putting, potentially, 24V across Rsense. The diode prevents this.

However C1 confuses me because it applies positive feedback around U1B.
If you're right about that - the bridge rectifier is also the wrong way round..........................
 

AlbertHall

Joined Jun 4, 2014
12,346
No, the 'AC' terminals of the bridge are connected to the loop so it doesn't matter which way you connect it. The positive output from the bridge feeds the, incorrectly marked, input of the regulator and the negative from the bridge goes to the ground terminal of the regulator via the sense resistor.
 
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