Op-amp voltage regulator

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  • #1
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Homework Statement


FIGURE 3 (on page 7) shows a PIR (passive infra-red) detector and its
associated amplifier, as used in burglar alarm systems1.

(a) The detector is powered from a 12 V unregulated supply that needs
to be stepped down to 5 V. Design a suitable 12-to-5 V voltage
converter using an op-amp and a diode that has the forward
characteristics given in FIGURE 3(b).

(b) Estimate the quiescent voltages at the inputs and outputs of the two
op-amps and the overall voltage gain (in decibels) of the circuit at the
frequency of operation. State any assumptions made.

Homework Equations


[itex]V_-=V_o\frac{R_2}{R_1+R_2}[/itex]
Vo = GV (V+ – V–)
Vd = V+

The Attempt at a Solution


I think pic 2 is the circuit they are asking for.
as for the rest.


Any help would be appreciated.
Thanks
 

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Answers and Replies

  • #2
donpacino
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What is your actual question?

Why don't you give the problem your best shot...
 
  • #3
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1
(a)
If we work with a forward voltage VD of 320mV this means the diode would require a forward current If of 1mA. The resister R3 would be requred to drop the remanig voltage and limit the current through the diode
[tex]R_3 = \frac{5-0.32}{0.001}=4.68K[/tex]
The voltage at V_ and V+ must be equal when the system is balance at the required output voltage.
[tex]V_D = V_+[/tex]
The inverting input is worked out as R1 and R2 form a potential divider.
[tex]V_o=V_D\frac{R_1+R_2}{R_2}=5[/tex]
[tex]\frac{V_D}{V_O}=\frac{R_2}{R_1+R_2}=\frac{0.32}{5}=0.064[/tex]
So if we give R2 a value of “1” we can find the ratio of R2 to R1.
[tex]\frac{1}{0.064}-1=14.625[/tex]
So if we give R2 3.3k then R1 will be 48k

As for part (b) I don't know where to star
 
  • #4
rude man
Homework Helper
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(a)
If we work with a forward voltage VD of 320mV this means the diode would require a forward current If of 1mA. The resister R3 would be requred to drop the remanig voltage and limit the current through the diode
[tex]R_3 = \frac{5-0.32}{0.001}=4.68K[/tex]
The voltage at V_ and V+ must be equal when the system is balance at the required output voltage.
[tex]V_D = V_+[/tex]
The inverting input is worked out as R1 and R2 form a potential divider.
[tex]V_o=V_D\frac{R_1+R_2}{R_2}=5[/tex]
[tex]\frac{V_D}{V_O}=\frac{R_2}{R_1+R_2}=\frac{0.32}{5}=0.064[/tex]
So if we give R2 a value of “1” we can find the ratio of R2 to R1.
[tex]\frac{1}{0.064}-1=14.625[/tex]
So if we give R2 3.3k then R1 will be 48k
All very well done for part (a).
As for part (b) I don't know where to star
As for part (b) you haven't given us the frequency of operation so can't do part (b).
Can you flip the image 90 deg for us?
Since they're asking for voltage gain you also have to define the input point. The overall circuit gain is volts/watt or volts/lumen or similar with infrared input quantity.
 
  • #5
donpacino
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As for part (b) you haven't given us the frequency of operation so can't do part (b).
Can you flip the image 90 deg for us?
Since they're asking for voltage gain you also have to define the input point. The overall circuit gain is volts/watt or volts/lumen or similar with infrared input quantity.
assuming they don't give you a frequency, you might need to make it generic (the answer will simply have f as a parameter).
 
  • #6
78
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No frequencies are given.
That all the info I have been given.
Although it says state any assumptions.
I assume the RC feedback is some kind of high or low pass filter. I really don't know where to start.
Thanks for you help.
 
  • #7
donpacino
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No frequencies are given.
That all the info I have been given.
Although it says state any assumptions.
I assume the RC feedback is some kind of high or low pass filter. I really don't know where to start.
Thanks for you help.
I would first calculate the operating points at DC.

Information I found on the web is saying PIR sensors operate in the 1 Hz range. You could calculate it there.
Is this for a homework problem. you can ask your professor for clarification. Or like I said earlier calculate as a function of frequency.
 
  • #8
donpacino
Gold Member
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282
I would first calculate the operating points at DC.

Information I found on the web is saying PIR sensors operate in the 1 Hz range. You could calculate it there.
Is this for a homework problem. you can ask your professor for clarification. Or like I said earlier calculate as a function of frequency.
if you really dont know where to start, maybe do nodal anlysis or some other system at DC (at dc capacitors are open circuit). It should be easy to find the voltages at the IOs of the op amps.
 
  • #9
78
1
I really don't have a clue.

so R1and R2 form a potential divider giving 2V at the non-inverting inputs of X3 and X1

but I am not sure about the inverting side this looks like it is held low at 0V
 
  • #10
donpacino
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I really don't have a clue.

so R1and R2 form a potential divider giving 2V at the non-inverting inputs of X3 and X1

but I am not sure about the inverting side this looks like it is held low at 0V
there is a property of op-amps that states with negative feedback the inverting and non inverting terminals will be equal
 
  • #11
LvW
836
222
I would first calculate the operating points at DC.
Information I found on the web is saying PIR sensors operate in the 1 Hz range. You could calculate it there.
Is this for a homework problem. you can ask your professor for clarification. Or like I said earlier calculate as a function of frequency.
I think, the whole circuit resembles a kind of bandpass amplifier. Why not using the mid frequency Fo as operating frequency?
A circuit simulation can give the value of Fo.
 
  • #12
donpacino
Gold Member
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I think, the whole circuit resembles a kind of bandpass amplifier. Why not using the mid frequency Fo as operating frequency?
A circuit simulation can give the value of Fo.
that could work (assuming the circuit is properly designed/tuned). I assume op is supposed to do this analytically.
 
  • #13
Nidum
Science Advisor
Gold Member
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(a) The detector is powered from a 12 V unregulated supply that needs
to be stepped down to 5 V. Design a suitable 12-to-5 V voltage
converter using an op-amp and a diode that has the forward
characteristics given in FIGURE 3(b).
There is no requirement to use that specific circuit is there ?

There are other ways of doing the same thing and some of these other ways will certainly be easier to analyse .
 
  • #14
donpacino
Gold Member
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282
There is no requirement to use that specific circuit is there ?

There are other ways of doing the same thing and some of these other ways will certainly be easier to analyse .
This looks like a homework problem. I doubt a "solution" would be resdesign the problem.
 

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