# Full-Wave Bridge Rectifier Filter + Regulator DC Power supply

Discussion in 'Homework Help' started by Petrucciowns, Jul 5, 2009.

1. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
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http://img104.imageshack.us/img104/3557/1234uyq.jpg

Ok, there are a few calculations on this problem that I am having difficulty with. This is a past homework assignment which is why the answers are filled in. They are all correct.

Anyways I approached the problem in this manner:

V secondary peak = V2/ .707

The voltage drop across RL is ZZT

IRL= ZZT/RL

VC1P= V2P - Both forward biased diode drops.

VC1 P-P (Ripple) = IDC / (2f * C)

VRS= Vc1 dc - Vzt

IRS= VRS /RS

The three that I am having trouble with are: Vc1-dc , IZ, and VRL p-p (Ripple)

Can anyone help me out?

Thanks.

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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1,097
If you calculate the
Vc_pp (ripple) then
Vc1_dc=( 32.5V + ( 32.5V-Vc_pp (ropple) ) / 2
So for Vc_(ripple)=9.4Vpp (in my calculation is 4.4Vpp)
Then
Vc_dc=(32.5V+23.1)/2=27.8V

Iz=IRS-IRL
VRL_(ripple)= [ rz / (rz+Rs) ] *Vcpp(ripple)= 20/520*9.4V=0.361V

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I tried a realistic simulation of the circuit - the ripple on Vc1 was about 3.1 Volts p-p. Seems curious there would be such a significant discrepancy. Anyone else tried a simulation?

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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If we calculate the Vc1 ripple we get:
$Vpp_{ripple}=(\frac{1 }{F*R*C})*Vin_p=4.4V$
And in my simulation that the ripple on Vc1 was about 3.1V

5. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
62
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See I was told the last formula you posted was used for rectifier filters. For the rectifier filter+ regulator I was given the following equation: VC1 P-P (Ripple) = IDC / (2f * C)

It may be different because it is asking for the ripple across the capacitor?

6. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
62
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For the Vc_dc where does the value of 23.1 come from?

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But this equation gives the same answer
Vc1_ripple=25mA/5.64m=4.4
Where 25mA is a current at the beginning of a discharge of a capacitor from I=(32.5V-20V)/500Ω=25mA

32.5 - Vc1_ripple=32.5-9.4V=23.1V

8. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
62
0
Well the equation I used : VC1 P-P (Ripple) = IDC / (2f * C)
The I DC is found by taking the Capacitor Dc voltage of 27.8 / Zzt+Rs which is further / (2f * C) this will give you the answer I came up with.

I can see that this works but is there a way to find the Dc voltage across the cap without having already found Vc1_ripple or is there a way to find Vc1_ripple without the the Dc voltage across the cap. Because at first at least one of them is going to be missing and I have no equation that I can use to find their values without the missing one.

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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And that is the conundrum.

Any formulas you plug into lead to approximations.

The heart of the problem is that you need to know the current to find the ripple and then the mean capacitor voltage - but don't you need the mean capacitor voltage to find the mean current in the first place...Hmmm?

My approach (without a careful numerical solution) would be something like this.

1. Estimate the mean capacitor voltage is going to be somewhere around 30V in this case.
2. The current I drawn from the rectifier + cap filter will be approximately I=(30-Vz)/500 = (30-20)/500 = 20mA
3. The (worst case estimate) capacitor ripple voltage will then be approx Vr = I/(2*F*C) = 3.6V
4. Hence an estimate for mean Vcap = 32.5 - 3.6/2 = 30.7 V
5. You can either leave it at that or replug a new estimate for mean Vcap = 30.7 V and redo the above steps.

In any case the 9.4V ripple value shown on the original JPEG file doesn't look correct. Hence my comment about the simulation showing the ripple is more like 3.1 V p-p.

It's worthwhile to try to get an appreciation of what's going on in the circuit physically so that the validity of any formulas used can be given a reality check.

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
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The capacitor ripple voltage can be find only by knowing the basic circuit theory equation.
C=Q/U and Ohms low
Capacitor will be charge to Vp=32.5V
And will be discharge by a current I=(32.5V-20V)/500Ω=25mA at t=0s.
Capacitor will be discharge by half of the period 1/60Hz=16.7ms---->8.3ms
And finally
C=Q/U
U=Q/C=(I*t)/C=(25mA*8.3ms)/47uF=4.41V
So the minimum voltage on the capacitor Vc_min=32.5V-4.41V=28V
Vdc on a capacitor will be equal the average voltage between 32.5V and minimum voltage on a capacitor 28V
Vdc=(32.5V+28V)/2=30.2V

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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The method works fine as you've explained but the result is really an approximation in that certain assumptions (non-obvious) are made - I'm being a pedant I know.

The current I can't be a constant (25mA) because the capacitor voltage is continually decreasing from a max of 32.5 V in the discharge phase.

Also the discharge interval isn't half the AC period of 8.33ms - it's something less than that - determined by the point at which the increasing half cycle of the rectified AC voltage just equals the instantaneous capacitor voltage.

The problem can't be solved with simple circuit theory - only approximated. I guess my point is that we forget about these matters sometimes and plug values into a formula thinking the result exactly reflects the physical situation.

My real issue is with the clearly wrong value of ripple voltage in the OP's submitted circuit.

12. ### millwood Guest

ripple is independent of the diode forward drop.

it is simply how quickly the capacitor will be discharged over half a cycle (this is a full wave rectification circuitry). how fast that capacitor will be discharged depends on C, and the equivalent "resistance" of the circuit, as determined by the resistor, zener and the load.

the average voltage on the capacitor is 28v, and the zener drops 20v, and the voltage drop over Rs = 8v, and the current going through Rs is 8v/500=16ma.

so now the question becomes how much voltage drop you get if you bleed a 47uf capacitor over 1/120th of a second.

and the answer is 16ma/47u * 1/120 = 2.8v.

13. ### millwood Guest

the last equation comes from the definition of C:

C=I/(dv/dt).

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Hi Millwood,

While this method works as a good approximation, the discharge of the capacitor is more closely represented by a 1st order exponential decay - since the current I is not really constant. This will be more pronounced as the ripple voltage increases - with ripple being dependent on the the choice of capacitor value and load current.

Further, you have stated the average DC capacitor voltage as 28V. But in reality you don't know this average value until you work out what the ripple is - and you have already assumed an average value to find the ripple. It's a chicken and egg argument.

In my analysis I get the average capacitor voltage closer to 30.9V. The actual discharge interval for the capacitor is about 7.3 milliseconds - rather than 1/120th of second.

I believe one must include the conducting diode voltage drops in the analysis - at least to determine the peak capacitor voltage - around 32.55 Volts with the 24V RMS supply - and assuming a forward diode drop of 0.7V.

I get a capacitor ripple voltage of about 3.13 V p-p.

15. ### millwood Guest

tnk, you are right on multiple front.

1) you can indeed use an exponential function to calculate the discharge. using a linear approach is just easier.

2) to get the average, you will have to make an assumption, do the calculation and see if it turns out to be right. in my case, the decline is 2.8v, so the average is more like 1.4v lower than the peak, making the average more like 30v, rather than 28v. then you recalculate the discharge current = (30-20)/500=20ma, and the decline is more like 20ma/47u*1/120=3.5v.

then the average is more like 1.8v lower than peak, or 31.4-1.8=29.6v, and the current is ......

with a few iterations, you should be able to get there.

my spreadsheet says that the ripple is 3.43v, assuming linear discharge. factoring the non-linearity, it is likely less.

as you can tell, the only place where the diode's forward voltage drop matters is the discharge current: high Vfwd will cause the peak voltage to drop thus causing the discharge current to drop thus reducing ripple. a .1v increase in Vfwd will cause the ripple to increase by about 0.03v, per my model.

16. ### millwood Guest

the real and more accurate answer is actually much more complicated than this. the diode will be conducting for a very short period of time. the duration during which the diode is conducting depends on many factors and is hard to quantify by hand.

a spice model will give a far more accurate answer than we are calculating but as you can tell, we aren't that far off.

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Yep - agreed on all that. There tends to be a short but significant spike of charging current into the capacitor and then things get really fuzzy.

I guess the OP has probably lost interest in this one (I wouldn't blame them!) - although it is a 'nasty' but tantalizing puzzle when you dig a bit deeper.

18. ### millwood Guest

if i were to put the exponential discharging into my model, I get a ripple of 3.404v, vs. 3.439v for linear discharging.

the rest doesn't change.

19. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
For what it's worth - this is how I solved it.

This gives a ripple of 3.18V - and an average capacitor voltage of 30.96V.

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20. ### millwood Guest

tnk, after reading your analysis, I realized that I didn't factor in the internal resistance of the zener: that addition will considerably complicate the analysis but may not add much to the end results: 20ma flowing through 20ohm yield 0.4v.

in your additional of Vdischarge, you should also consider the voltage drop over that resistor. so rather than 20v, it should be something slightly higher.