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frequency response

I'm trying to plot the frequency of a power supply I've designed. Here is the net list:

PowerSupply

v1 4 5 dc 0 sin(0 34 60)

D1 4 2 1N4007 D2 0 5 1N4007 D3 5 2 1N4007 D4 0 4 1N4007

c1 2 0 1000u c2 2 0 100n cout out 0 100n

radj adj 0 5k r1 out adj 240

x 2 adj out LM317

.include parts.lib .control set units=degree destroy all

tran 0.01ms 100ms plot v(4,5) v(out) vs (time*1000)

destroy all

ac dec 10 60Hz 7000Hz plot ac1.v(out) vs ac1.frequency

.endc .end

--------------------------------------------------------------

I'm not doing it correctly though. For the second plot (freq response), I get values in excess of 6000V ! This is so wrong as my calculations have shown just above 28V is possible. Please assist me in doing a frequency response.

Thanks.

Reply to
ranger
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"ranger" schrieb im Newsbeitrag news: snipped-for-privacy@y80g2000hsf.googlegroups.com...

Hello,

I tried a transient analysis of your netlist with LTspice and it worked in principle but there is a small oscillation at the output. The value of Cout is undersized. Just increase its value to 10u and you will be saved.

How can you run any .AC-analysis without any source having AC specified? It's impossible. Please explain where your AC is specified.

AC-analysis is a linear system analysis in SPICE! It has nothing to do with the AC-voltage applied to your rectifier diodes. Where do you want to apply the AC-voltage or current?

Best regards, Helmut

PS: Nobody can exactly reproduce your circuit wihout knowing your LM317 model.

Reply to
Helmut Sennewald

Thanks for the suggestion.

I thought having "v1 4 5 dc 0 sin(0 34 60)" would allow me to do an ac analysis because its an ac source. But from your reply, I see that this is not the case. The applied ac voltage should be applied to the rectifier as v1 is now. What changes should I make to the v1 to allow me to perform this ac analysis, while retaining my peak value of 34v and frequency of 60Hz.

I'm using the following model:

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Thanks again.

Reply to
DDDiiD

"DDDiiD" schrieb im Newsbeitrag news: snipped-for-privacy@h2g2000hsg.googlegroups.com...

Hello,

I still think you expect something different from the .AC simulation. The .AC simulation could be used to measure the small signal input "noise" rejection of the regulator versus frequency or it could be used to measure the dynamic output resistance of your regulator. All these .AC simulations will require to setup the appropriated DC condition for the LM317. This means you would apply a DC voltage of e.g. 32V at the input with an additional AC definition. It would be also necessary to add the correct load device, e,g, a resistor load or a current source load.

The normal 60Hz ripple rejection has to be simulated with .TRAN because this is a really nonlinear system.

Best regards, Helmut

Reply to
Helmut Sennewald

"Helmut Sennewald" schrieb im Newsbeitrag news:f2d6up$pdc$00$ snipped-for-privacy@news.t-online.com...

Sorry, it should be DC 34V in your cicuit.

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Reply to
Helmut Sennewald

I'm not sure I quite understand this. I am suppose to change "v1 4 5 dc 0 sin(0 34 60)" to "v1 4 5 dc 34 sin(0 34 60)" thereby giving it a dc defintion?

Reply to
ranger

"ranger" schrieb im Newsbeitrag news: snipped-for-privacy@p77g2000hsh.googlegroups.com...

Hello,

A possible Spice-line could be as shown below.

V1 1 0 DC 38 AC 1

Overall I rate an AC-simulation useless at this point in the circuit, because your diode bridge is acting like a switch which conducts only for a fraction of each period.

Who the hell told you to do a .AC-simulation at this point in the circuit and for what reason?

Best regards, Helmut

Reply to
Helmut Sennewald

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