Hi there,

the circuit below is from "The Art Of Electronics" and does the job of a
differentiator along with a signal rectifier. What role does R2
play, apart from modifying the RC filter for positive voltages?

 

Originally posted by skypher@Apr 16 2006, 07:46 PM
Hi there,

the circuit below is from "The Art Of Electronics" and does the job of a
differentiator along with a signal rectifier. What role does R2
play, apart from modifying the RC filter for positive voltages?

[post=16224]Quoted post[/post]​

On the resistor R2 one gets positive-edge triggered signal
On the resistor R3 one gets negative-edge triggered signal

 

Originally posted by dragan733@Apr 16 2006, 04:20 PM
On the resistor R2 one gets positive-edge triggered signal
On the resistor R3 one gets negative-edge triggered signal

[post=16227]Quoted post[/post]​

Could someone please expain how this circuit works in a bit more detail ?
thanks guys

 

Did you read the description about the circuit in AoE?

 

Originally posted by dragan733+Apr 16 2006, 11:20 PM--><div class='quotetop'>QUOTE(dragan733 @ Apr 16 2006, 11:20 PM)</div><div class='quotemain'>On the resistor R2 one gets positive-edge triggered signal
On the resistor R3 one gets negative-edge triggered signal

[post=16227]Quoted post[/post]​

[/b]

But I thought R3 forms a voltage divider with the diode, to compensate for the forward drop of the rectifier diode...?

<!--QuoteBegin-JoeJester@Apr 17 2006, 04:05 AM
Did you read the description about the circuit in AoE?

[post=16236]Quoted post[/post]​

[/quote]
Sure, but it does not say anything about the additional resistor, it just explains the subcircuit that provides diode forward drop voltage compensation.

 

Originally posted by skypher@Apr 18 2006, 04:25 AM
But I thought R3 forms a voltage divider with the diode, to compensate for the forward drop of the rectifier diode...?
Sure, but it does not say anything about the additional resistor, it just explains the subcircuit that provides diode forward drop voltage compensation.

[post=16249]Quoted post[/post]​

Just having a guess here seeing they don't tell you what it is there for. Probably just the load resistor.

 

Originally posted by Mazaag@Apr 17 2006, 02:24 AM
Could someone please expain how this circuit works in a bit more detail ?
thanks guys

[post=16235]Quoted post[/post]​

R1-C is a differentiator (in fact a high-pass filter with ~1,6 MHz) along with a signal rectifier D2 to get only the positive voltages. The R3-D1 circuit with the additional +5V supply is just for compensating the forward drop voltage of D2.

 

Originally posted by skypher@Apr 18 2006, 04:28 AM
R1-C is a differentiator (in fact a high-pass filter with ~1,6 MHz) along with a signal rectifier D2 to get only the positive voltages. The R3-D1 circuit with the additional +5V supply is just for compensating the forward drop voltage of D2.

[post=16282]Quoted post[/post]​

So basically we are compensating for the 0.7v loss across the D1 ? but how are we doing that with this circuit? I can't really see that.. are we adding an offset so to speak of 0.7v to the output ?

Thanks

 

Originally posted by Mazaag@Apr 18 2006, 06:28 PM
So basically we are compensating for the 0.7v loss across the D1 ? but how are we doing that with this circuit? I can't really see that.. are we adding an offset so to speak of 0.7v to the output ?

[post=16289]Quoted post[/post]​

R3 and D1 form a voltage divider. The voltage drop at R3 is 5V minus D1's voltage drop, which is very close to D2's (provided they are the same type). So the voltage drop that gets lost at D2 gets added.

 

Originally posted by skypher@Apr 18 2006, 12:27 PM
R3 and D1 form a voltage divider. The voltage drop at R3 is 5V minus D1's voltage drop, which is very close to D2's (provided they are the same type). So the voltage drop that gets lost at D2 gets added.

[post=16290]Quoted post[/post]​

skypher,
I'm kinda confused (sorry pleaes excuse my noobinism).

I understand that R3 and D1 form a voltage divider. So wouldn't the potential across R1 would be LESS by 0.7 ? I mean, if you were to consider the circuit without R3 and D1 ( say its ground ) , the voltage across R1 would be say 4volts. When we introduce R3 D1, the voltage ACROSS the resistor would now be decreased by 0.7 volts ( right ? ) , but then it would drop another 0.7 volts across D2 .... ? wouldn't it ? I'm kinda confused... I still dont' see how that divider circuits "adds" 0.7v to compensate for D2 forward votlage drop of 0.7v....

thanks for your patience

 

Originally posted by skypher@Apr 19 2006, 04:27 AM
R3 and D1 form a voltage divider. The voltage drop at R3 is 5V minus D1's voltage drop, which is very close to D2's (provided they are the same type). So the voltage drop that gets lost at D2 gets added.

[post=16290]Quoted post[/post]​

Weren't you trying to find out what R2 was for?

 

Originally posted by windoze killa@Apr 19 2006, 12:45 AM
Weren't you trying to find out what R2 was for?

[post=16296]Quoted post[/post]​

Yes, and I still do, but that doesn't keep me from explaining the other parts of the circuit, which I do understand

 

Originally posted by Mazaag@Apr 18 2006, 08:26 PM
(sorry pleaes excuse my noobinism).

Please keep on asking until you understand it all!

I understand that R3 and D1 form a voltage divider. So wouldn't the potential across R1 would be LESS by 0.7 ? I mean, if you were to consider the circuit without R3 and D1 ( say its ground )

If R3 and D1 were replaced by ground, all current from the 5V supply would be shorted to ground and no voltage whatsoever would be dropped across R1.

I'm kinda confused... I still dont' see how that divider circuits "adds" 0.7v to compensate for D2 forward votlage drop of 0.7v....

Try to look at R1 and D2 as a second voltage divider in line with the first (formed by R3 and D1).
Both work the same way.

 

Originally posted by windoze killa@Apr 18 2006, 02:03 AM
Just having a guess here seeing they don't tell you what it is there for. Probably just the load resistor.

[post=16266]Quoted post[/post]​

I don't think so. It's not consistent with the style of the book found elsewhere. Load resistors are usually dashed or marked specially otherwise.

 

Skypher started the thread with this question

the circuit below is from "The Art Of Electronics" and does the job of a
differentiator along with a signal rectifier. What role does R2
play, apart from modifying the RC filter for positive voltages?

The applicable section from Horowitz & Hill’s Art of Electronics, Chapter 1, Foundations, Section 1.30 Titled, Circuit applications of diodes, subsection Signal Rectifier, on page 48 and 49.


The part we didn’t get on the thread, as part of the original question was the circuit skypher posted was shown as an improvement over another circuit when dealing with an input less than the Vf of D1 in the circuit below.

Skypher posted the improved circuit, redrawn, but here it is from AoE.

R2 provided a different differentiated time for positive pulses compared to R1 on the negative pulses, if they are different values as per the improved drawing. R3 and D1 provide a bias to D2 to allow rectified differentiated waves less than Vf of D2 in figure 1.82.

Since only the positive pulses are of interest, no one cares for the differentiated times of the negative pulses.

 

Originally posted by skypher@Apr 19 2006, 12:54 PM
Please keep on asking until you understand it all!
If R3 and D1 were replaced by ground, all current from the 5V supply would be shorted to ground and no voltage whatsoever would be dropped across R1.
Try to look at R1 and D2 as a second voltage divider in line with the first (formed by R3 and D1).
Both work the same way.

[post=16328]Quoted post[/post]​

skypher thanks for you help...however I still don't get it !

when i said "say it was ground" , i meant "consider the circuit without D1 R3 and the 5V "

so basically we have a filter and a load resistor with a D2 in place ...

Now,
because D2 is forward biased, we get a voltage drop of 0.7v across it... right ?

first question, does that mean the AC signal superimposed will be offset by 0.7v downwards?

so we want to compensate for this 0.7 voltage drop by D2. So somehow , we need to ADD 0.7v to the output of the filter.. such that when we run the AC signal, it will start off 0.7 higher than it should be , drop by 0.7v because of D2 , and everything is fine... right ?

okay , so we need to add 0.7 volts to the node between the Capacitor and R1.. correct ?

so we have a circuit ( 5 volts with R3 and D1... ) this is a voltage divider, with 0.7v across the diode , and 4.3 across R3... so far so good ?

now.. we connect up the other end of R1 to the node between the D1 and R3.... ( which has a voltage of 0.7v to ground.. )

second question.. how do I analyze from this point, to show that the voltage at the node between the capacitor and R1 has increased by 0.7v......

I would appreciate a response with similar level of detail ... j/k.. just try your best to explain ..and I thank you for your patience..

 

Originally posted by JoeJester@Apr 20 2006, 03:51 AM
R2 provided a different differentiated time for positive pulses compared to R1 on the negative pulses,

Why would I want that?

 

Originally posted by Mazaag@Apr 20 2006, 04:39 AM
first question, does that mean the AC signal superimposed will be offset by 0.7v downwards?

Yes.

so we want to compensate for this 0.7 voltage drop by D2. So somehow , we need to ADD 0.7v to the output of the filter.. such that when we run the AC signal, it will start off 0.7 higher than it should be , drop by 0.7v because of D2 , and everything is fine... right ?

okay , so we need to add 0.7 volts to the node between the Capacitor and R1.. correct ?

so we have a circuit ( 5 volts with R3 and D1... ) this is a voltage divider, with 0.7v across the diode , and 4.3 across R3... so far so good ?

Yes to all this.

second question.. how do I analyze from this point, to show that the voltage at the node between the capacitor and R1 has increased by 0.7v......

Easy. Forget the capacitor and its input, we can do without them for explaining this. A voltage divider, formed by D2 and R1, remains. 0.7V has to fall off at the diode, so R1 does not do much to the voltage.

 

Originally posted by skypher@Apr 20 2006, 07:13 AM
Yes.
Yes to all this.
Easy. Forget the capacitor and its input, we can do without them for explaining this. A voltage divider, formed by D2 and R1, remains. 0.7V has to fall off at the diode, so R1 does not do much to the voltage.

[post=16357]Quoted post[/post]​

So because the voltages at the nodes between R1 are the same, "no current" flows and therefore we can consider the resistor shorted ? and therefore we get an addition of 0.7 ?

 

Yes. Ohm's law gives I = (U_r - 0.7V)/1k = 0V/1k = 0A.