# Regulated DC Power supply with Op-amp

Discussion in 'Homework Help' started by desecrate, Feb 9, 2011.

1. ### desecrate Thread Starter New Member

May 26, 2010
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0
This is the last part of some lab work I am undertaking. Attached is a photo of the lab script. In the attachment is the circuit image and the questions.

The last two questions are trivial, naturally the difficulty I am having is in the brunt of (a).

Prerequisites
We havn't had a formal introduction to transistors yet, so we have been told to ignore that part and were given a simple description of what it is doing. We have been given a formula for regulation and for calculating the average value of a rectified waveform and another one for calculating voltage ripple.

Attempt
(a) is basically asking us to find the values of R1, R2, and R3.
The op-amp is operating as a comparator, taking a fraction of the output in as negative feedback.

To find R1, I have:
$v_o=i_{z1}R_1$
$(5-3.3)=(5\times 10^{-3})R_1$
$\therefore R_1=340\Omega$

To find R1 and R2, I am using:
$v_o=v_i(1+\frac{R_2}{R_3})$

Sometimes I understand that it's ok to pic an arbitrary value for one resistor (and rearrange for the other) as long as it lies within a certain region. In this case I have no idea what that region is. Is it ok to pic an arbitary value for R2 and find R3, or should I consider something when deciding a value for R2?

- Jim

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,513
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Well, simply assume that voltage divider current is equal 5mA.
And you can look at this circuit as a non-inverting amplifier.
Where Vin is equal Vz=3.3V.

And have you built this circuit in real life?
Because I have some doubts, if this circuit can work properly.

3. ### desecrate Thread Starter New Member

May 26, 2010
10
0
So to get R2 I do something like?:
$R_2=\frac{5}{5\times 10^{-3}} = 1k\Omega$

It is meant to be a real circuit that we are going to build tomorrow. There is a part of the circuit on the left missing. Where it says V_rect, is a transformer and rectifier circuit, so v_rect should be a full wave rectified signal.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,513
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Hmm,
R2 = (5V - 3.3V)/5mA = 340 = 330Ω
But i think that you should start from R2 = 10K. You don't need to waste that much amount of current for no reason, because opamp input bias current is "very small" 200pA. So you could chose voltage divider current much larger then input bias current Idv >> 200pA.

5. ### desecrate Thread Starter New Member

May 26, 2010
10
0
How did you calculate 200pA? Is that from the TL071 input impedance, from it's datasheet? And why choose R2, why not R3=10K? The ratio will be the same won't it?

Last edited: Feb 9, 2011
6. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,513
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See in datasheet input bias current.

It is designer choice, and it's not so important.
The only think that matters is R2/R3 ratio:
Vout/Vz - 1 = R2/R3 = 5V/3.3V - 1 = R2/R3 = 0.515
R2 = 0.515*R3 or R3 = R2/0.515

7. ### Audioguru Expert

Dec 20, 2007
10,606
1,182
The zener diode is 3.3V so R3 will also have 3.3V across it. You want an output of 5V so R2 will have 5V-3.3V= 1.7V across it.

If R3 is 10k then R2 can be 5.1k. Simple arithmatic.

If the input is at least 7V (the minimum supply for a TL071 opamp) then the circuit will work.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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An I think that this circuit can also have problem with "start-up".

9. ### Audioguru Expert

Dec 20, 2007
10,606
1,182
I see what you mean.
R1 should be connected to the input supply.

10. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
674
The TL071 output has to go to around 6.0-6.2V (5V plus TIP110 Vbe). Since the TL071 output can only reach ≈(vcc-1.5V), vcc needs to be at least ≈7.7V.

Last edited: Feb 12, 2011