# Resistance Measurement

#### sonali_chawande

Joined Mar 20, 2007
8
i m having problem in designing a circuit for the AC source, a magnectic circuit is used to generate the AC i have to convert it to constant DC i have genarated the circuit for that, my problem is that the magnectic circut is giving the AC for about 150 Hz.And also to adjust the Rc time constsnt i want to measure the circuit resistance.
How to measure the circuit resistance?
what should i do to operate that circuit properly?
how to measure the RC time constsnt for that circuit?
What to be changed in the circuit capacitor/resistor?

Anyone can help, Thanks.

#### kender

Joined Jan 17, 2007
264
Sonali, it's hard to envision your circuit, it's resistance and RC time constant without a schematic. Could you, please, post it?

#### sonali_chawande

Joined Mar 20, 2007
8
sir actually i m just designing a voltage regulator for dynamo, whose circuit i m sending 2 u, its Ac input will be varying maximum up to 30Volts and at the output side current required is 30 Amps so i have used the 35Amps diodes for that and now my problem comes, that the dynamo can change the input frequency, it can go up to 150Hz. for that i require to check the time constant for my circuit...
sir i m sending u my circuit, so plz can u send me the formula to calculate that RC time constsnt, thanks.
can u tell me that if the frequency changes should i change the capacitor?
and if it changes how to calculate the capacitor value?

thanks for ur Reply and ur kind help.

Thanking u,

sonali

Sonali, it's hard to envision your circuit, it's resistance and RC time constant without a schematic. Could you, please, post it?

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#### beenthere

Joined Apr 20, 2004
15,819
Please do not hijack another's thread. I have moved these posts to a more appropriate location.

#### beenthere

Joined Apr 20, 2004
15,819
Your circuit cannot do what you want. The filter capacitor is too small by at least three orders of magnitude. Even at the higher frequency, it's way too small.

Your regulator circuit is not properly arranged. The darlington has to pass a regulated voltage, but its base is fed from the unregulated output from the bridge. I suggest you look up data sheets for three-terminal regulators and see how to arrange the circuit so the 7815 will control the darlington's conduction. In using the 7815, you are limiting regulation to 15 volts, by the way.

Another question - is the regulator supposed to be for the dynamo (alternator?) output, or is is intended to regulate the exciter current that holds the output steady under load? If that is the case, then you circuit will require more alteration.

#### sonali_chawande

Joined Mar 20, 2007
8
sir can u help me in the alteration for that circuit, i have to regulate the voltage up to 14 volts with the current of 30Amps to 35 Amps.

Thanks,

#### beenthere

Joined Apr 20, 2004
15,819
What is the range of voltage? You say up to 14 volts, but what is the lower limit?

#### sonali_chawande

Joined Mar 20, 2007
8
sir,
i have to regulate the voltage at constant 14 volts not more or less than that, and the load applied to my circuit will be in range 5Amps to 35 Amps,

thanks,

#### sonali_chawande

Joined Mar 20, 2007
8
sir,
i have seen that circuit u provided but what my problem is that, the dynamo will give the voltages from 1000RPM to 6000RPM starting fron 8.6Volts to 35Volts at 6000RPM, i have to regulate that, the circuit u provided is the best but i want the smaller circuit, i m having now only one problem in my circuit that is, the AC frequency will increase at output of the dynamo, i have to filter that AC after rectifying that, how can i do that?
which value of capacitor will be helpful for that?
or i will require to use a inductor also?

Thanks,

#### sonali_chawande

Joined Mar 20, 2007
8
Sonali, it's hard to envision your circuit, it's resistance and RC time constant without a schematic. Could you, please, post it?

i want to use a filter capacitor for my regulator circuit,
my rectifier will give output from 8.6Volts to 35 Volts,
with 30Amps of current, for such high current the AC frequency will be increasing up to 110Hzs,
so which value of capacitor i will have to use in that circuit ?
or i will have to use a inductor?

Thanks,

#### kender

Joined Jan 17, 2007
264

i want to use a filter capacitor for my regulator circuit,
my rectifier will give output from 8.6Volts to 35 Volts,
with 30Amps of current, for such high current the AC frequency will be increasing up to 110Hzs,
so which value of capacitor i will have to use in that circuit ?
or i will have to use a inductor?

Thanks,
Would you find it polite if I advise you to take a handful of capacitors and inductors and see what works and what doesn't? You also seem to have an LTSpice (?) model. What answers to your question does the model give you?

#### thingmaker3

Joined May 16, 2005
5,083
Here's a handy rule of thumb:

C = 0.7(I)/ΔE(f)

So, take your thirty amps and multiply by 0.7 for "21."

Divide this by your lowest expected frequency. (That's the frequency after rectification, so 2x input frequency.) At your lower speed of 1000RPM, 2x will be about 50Hz. 21/50=0.42

Divide 0.42 again by the highest allowable p-p ripple. Answer gives your minimum capacitor value. Example 1: If you can handle 1Vpp ripple, use a minimum 420000 Uf capacitor bank. Example 2: If ripple must be .05Vpp or less, use a minimum 8.4Farad capacitor bank.

It is the "35Amp" requirement which causes problems here...

#### beenthere

Joined Apr 20, 2004
15,819
The regulated voltage seems to be for the exciter. If this is the case, the excitation voltage can have some considerable ripple in it without problem. Lots of alternators use rectified but unfiltered DC to drive the field.

It would be nice to have the excitation drive requirement in hand.

#### sonali_chawande

Joined Mar 20, 2007
8
Here's a handy rule of thumb:

C = 0.7(I)/ΔE(f)

So, take your thirty amps and multiply by 0.7 for "21."

Divide this by your lowest expected frequency. (That's the frequency after rectification, so 2x input frequency.) At your lower speed of 1000RPM, 2x will be about 50Hz. 21/50=0.42

Divide 0.42 again by the highest allowable p-p ripple. Answer gives your minimum capacitor value. Example 1: If you can handle 1Vpp ripple, use a minimum 420000 Uf capacitor bank. Example 2: If ripple must be .05Vpp or less, use a minimum 8.4Farad capacitor bank.

It is the "35Amp" requirement which causes problems here...
thank u for ur kind help ur given formula helped me, but can i use the same formula for single phase and the three phase AC circuits?

Thanks and Regards,

#### thingmaker3

Joined May 16, 2005
5,083
Simply change the value of f as required by number of phases.

#### sonali_chawande

Joined Mar 20, 2007
8

as u have described in the formula that lowest expected frequency will be 2X Input frequency and u have given this lowest frequency as 50Hz, but if my input AC frequency is 166.667Hz, then will it come as 333.334Hz, is this the right calculation?
then what will be the capacity if the frequency is same and current is 25Amps and maximum allowable voltage is 40Volts (for which ripple will be 19.28 for bridge rectifier) ?

Thank u but i m not understsnding this calculation can u plz help me to calculate this capacitor value?

Here's a handy rule of thumb:

C = 0.7(I)/ΔE(f)

So, take your thirty amps and multiply by 0.7 for "21."

Divide this by your lowest expected frequency. (That's the frequency after rectification, so 2x input frequency.) At your lower speed of 1000RPM, 2x will be about 50Hz. 21/50=0.42

Divide 0.42 again by the highest allowable p-p ripple. Answer gives your minimum capacitor value. Example 1: If you can handle 1Vpp ripple, use a minimum 420000 Uf capacitor bank. Example 2: If ripple must be .05Vpp or less, use a minimum 8.4Farad capacitor bank.

It is the "35Amp" requirement which causes problems here...

#### thingmaker3

Joined May 16, 2005
5,083
C = 0.7(I)/ΔE(f)

C = 0.7(25A)/ΔE(333Hz)