Hi

I need to supply a circuit with 80 and 135 VDC. I have a Variac(240V max & 5A max) connected to a 1:1 transformer ( for isolating purpose ) connected to a bridge rectifier. Please help me understand what smoothing capacitor value can I use in this case.

Thanks

 

Based on not knowing whether you have 50 Hz or 60 Hz
1.141 C x Eripple(peak to peak) x frequency = 5

 

You CAN use any value rated to 340V. The value you actually choose depends on the load current and your max acceptable ripple.

You could use a lower voltage rating on the cap if you could limit the output of the variac but Murphy's law dictates that the knob will get turned up to the max eventually. Best not to have caps explode when that happens.

 

Based on not knowing whether you have 50 Hz or 60 Hz
1.141 C x Eripple(peak to peak) x frequency = 5

Working with 50hz, but I do not know how I can find the Eripple?

Also I mis-informed you about the 1:1 transformer, it is 240V 500VA. So I assume, current output at 80V would be 0.7A & at 135V would be 1.2A(correct me if I am wrong)

For eripple (peak-peak) do i need an oscilloscope? How can I find it out if I dont have one?

 

I do not know how I can find the Eripple?

You don't "find" Eripple. You decide what is your limit for Eripple and calculate for it. That will tell you the minimum size of capacitor.

For eripple (peak-peak) do i need an oscilloscope? How can I find it out if I dont have one?

Same answer. You don't find Eripple. You decide what your limit is and calculate the size of capacitor which accomplishes that.

 

You CAN use any value rated to 340V. The value you actually choose depends on the load current and your max acceptable ripple.

You could use a lower voltage rating on the cap if you could limit the output of the variac but Murphy's law dictates that the knob will get turned up to the max eventually. Best not to have caps explode when that happens.

Raise the value of the cap to the point where the RC time constant is say at least 10 times the input period, then try the equation again.

 

Working with 50hz, but I do not know how I can find the Eripple?

Also I mis-informed you about the 1:1 transformer, it is 240V 500VA. So I assume, current output at 80V would be 0.7A & at 135V would be 1.2A(correct me if I am wrong)

For eripple (peak-peak) do i need an oscilloscope? How can I find it out if I dont have one?

The value of the reservoir capacitor will be C = I Δt / ΔV, where:

C is the value of the capacitor, in farads,
I is the value of the load current, in amperes,
Δt is the period of the ripple waveform, in seconds, and
ΔV is the allowable ripple across the load, in volts.

For example,
Since your isolation transformer is rated to be able to transfer 500 VA into a load, with 240 V across its secondary, that's about 2 amperes into a 115 ohm load.

Then, if you needed to keep the ripple down to 1 volt across that load, with 50Hz mains supplying the power and a full-wave bridge topping up the reservoir, you could write: C = 2A X 0.01s / 1V = 0.02F = 20000μF

C = I Δt / ΔV = 2A X 0.01s

 

Mr. Fields used a slightly different equation than mine. I found my equation in an Audio Handbook by National Semiconductor and tested the results with my own parts, 1% DVM, and an oscilloscope. The resulting ripple was withing 1% of the calculated ripple.

I don't know why my equation works so well. Maybe it is the use of the peak-to-peak label that I use.

 

Your formula doesn't use the load current. I don't see how that can be left out – am I missing something? (And don't you mean 1.41 for the factor?)

I guess it's all moot until the TS explains what he needs to accomplish.

 

Your formula doesn't use the load current.

1.141 C x Eripple(peak to peak) x frequency = 5

Bad labeling. radical 2 C Er F = I
In this case, I = 5 amps

 

Thanks for all the replies but if you look at my earlier post which says

Working with 50hz, but I do not know how I can find the Eripple?

Also I mis-informed you about the 1:1 transformer, it is 240V 500VA. So I assume, current output at 80V would be 0.7A & at 135V would be 1.2A(correct me if I am wrong)

For eripple (peak-peak) do i need an oscilloscope? How can I find it out if I dont have one?

Which means I want to calculate smoothing capacitor for output 1)80V & 0.7A 2)135V & 1.2A.

Now using the calculation method by @EM Fields , I get 1)7000uF 150V 2) 2000uF 250V.

These capacitor values look unreal to me, correct me if I am wrong please

Also wanted to know if there any way of testing the output after using the smoothing capacitor without connecting to any load?

Thanks in advance for your replies.

 

Good grief. The ripple is defined BY YOU, and then calculations give you the estimated capacitance to achieve that level of ripple. The load determines the current draw, a necessary input to the equation, NOT the specifications of the transformer. No load, no ripple, no capacitor. Big load, big ripple OR big capacitor.

 

Thanks for all the replies but if you look at my earlier post which says

Working with 50hz, but I do not know how I can find the Eripple?

Also I mis-informed you about the 1:1 transformer, it is 240V 500VA. So I assume, current output at 80V would be 0.7A & at 135V would be 1.2A(correct me if I am wrong)

For eripple (peak-peak) do i need an oscilloscope? How can I find it out if I dont have one?

Which means I want to calculate smoothing capacitor for output 1)80V & 0.7A 2)135V & 1.2A.

Now using the calculation method by @EM Fields , I get 1)7000uF 150V 2) 2000uF 250V.

These capacitor values look unreal to me, correct me if I am wrong please

Also wanted to know if there any way of testing the output after using the smoothing capacitor without connecting to any load?

Thanks in advance for your replies.

I think the thing you are missing is that you get to decide how much ripple you have. If you allow more ripple you can use a smaller capacitor.

 

Ok.....been studying on this alot and I figured this out a bit. I understand that I can choose my ripple, all I dont get is why do I need to know the load? because I already know the max current draw. could you please specify a bit more. sorry I am noob here. @wayneh .

 

Well, you can set the ripple for the maximum current draw if you like. It will be less for loads below that, so your capacitor may be larger and more costly than you need.

Please confirm your maximum currents. I'm not familiar with your transformer but your estimates don't make sense to me. It seems to me the currents should go up as the voltage drops not vice versa, but as I said, I'm not sure about that.

 

I will explain my setup :

This is my connection:

Variac>1:1 Isolating Transformer>Bridge Rectifier>smoothing capacitor(in parallel).

Ratings:
Variac: 250V 5A 1.25kVA
1:1 Transformer: 240V 500VA

What outputs I want:

80VDC & 135VDC

I cant give you any load information.

According to my calculations my transformer outputs - 80V at 0.7A
135V at 1.2A

So for these two outputs I need 2 different smoothing capacitors to be used.

My calculations say I should use 2000uF 250V capacitor when my chosen Vripple is 2V.
Hope this is all making sense. If you guys could verify this, I will appreciate it very much. Spending all my day to get this issue sorted .

 

I will explain my setup :

This is my connection:

Variac>1:1 Isolating Transformer>Bridge Rectifier>smoothing capacitor(in parallel).

Ratings:
Variac: 250V 5A 1.25kVA
1:1 Transformer: 240V 500VA

What outputs I want:

80VDC & 135VDC

I cant give you any load information.

According to my calculations my transformer outputs - 80V at 0.7A
135V at 1.2A

So for these two outputs I need 2 different smoothing capacitors to be used.

My calculations say I should use 2000uF 250V capacitor when my chosen Vripple is 2V.
Hope this is all making sense. If you guys could verify this, I will appreciate it very much. Spending all my day to get this issue sorted .

I don't want to over complicate it but you need to derate your transformer because of the bridge rectifier. So your transformer can supply about 1.2 amps.
It can do this no matter what voltage you set the variac. The thing that limits the transformer is it's power. That is decided by formula P=I^2R.
Now, just because your transformer can put out 1.2 amps does not mean it will always put out 1.2 amps. The current depends on the load you put on the transformer supply. For example if you set it to 80 volts and put an 80 ohm resistor across it it will supply 1 amp and you can calculate the ripple for 1 amp. I=E/R
Or you might have 160 ohms across it so it would only supply 0.5 amps so you would have less ripple with the same capacitor.

http://www.hammondmfg.com/pdf/5c007.pdf

 

I don't want to over complicate it but you need to derate your transformer because of the bridge rectifier. So your transformer can supply about 1.2 amps.
It can do this no matter what voltage you set the variac. The thing that limits the transformer is it's power. That is decided by formula P=I^2R.
Now, just because your transformer can put out 1.2 amps does not mean it will always put out 1.2 amps. The current depends on the load you put on the transformer supply. For example if you set it to 80 volts and put an 80 ohm resistor across it it will supply 1 amp and you can calculate the ripple for 1 amp. I=E/R
Or you might have 160 ohms across it so it would only supply 0.5 amps so you would have less ripple with the same capacitor.

http://www.hammondmfg.com/pdf/5c007.pdf

I totally understand what you are saying, but all I am asking is if my choice of capacitor(for 2v ripple) correct or not because I blew up one capacitor once and would like not to do it again.

Thanks

 

C = I/F x Vrip
C = 1.2/ 100 x 2
C = 6000 Ufd

This will be very expensive.