Replacing a 50uf cap for a 47uf - Page 1
Forum section: electronics / beginners

author: gkmaia
author_rank: Frequent Contributor
timestamp: April 16, 2019, 09:18:59 am
content: Looking at the schematic attached is it ok to replace the caps 87 and 86 for a 47uf. Reason being quite hard to find 50uf at acceptable prices here.

author: bd139
author_rank: Super Contributor
timestamp: April 16, 2019, 09:28:08 am
content: 47uF is fine. The tolerance of them is so wide (usually +/-20%) that it's not much of a concern. I would make sure there's good headroom on the voltage though (160v+)

author: lordvader88
author_rank: Frequent Contributor
timestamp: April 16, 2019, 09:32:19 am
content: Yeah those are just big storage tanks basically, to provide some current when the AC drops. The bigger the better in fact, provided u could afford them. 47u vs 50u won't matter any real amount. But as a rule it's better to go bigger, than smaller for tank cap's

author: Zero999
author_rank: Super Contributor
timestamp: April 16, 2019, 09:55:59 am
content: The values for electronic components have been standardised on the E series of preferred numbers, a good 50+ years ago, so values such as 50µF are no longer common. Electrolytic capacitors are typically made in E6 values, with E3 i.e. 10, 22 and 47 being most common. As mentioned above, the tolerance is so wide, it doesn't matter. Indeed the manufactures won't have changed their processes to make 47µF, rather than 50µF, they would only change the label.https://en.wikipedia.org/wiki/E_series_of_preferred_numbersOlder schematics will have odd values, before they were standardised, but it's easy to replace them with the nearest E6 value, so 50µF becomes 47µF, 300µF -> 330µF, 800µF -> 1000µF etc.

author: soldar
author_rank: Super Contributor
timestamp: April 16, 2019, 11:44:01 am
content: Not to mention that capacitors used in filtering rectified power supplies are not of critical value and you could replace it with a slightly higher value with no problem.

author: Dave
author_rank: Super Contributor
timestamp: April 16, 2019, 12:25:32 pm
content: Quote from: lordvader88 on April 16, 2019, 09:32:19 amSlightly higher value is not going to be a problem, but increasing the capacitance significantly (doubling, tripling, or more) will increase the RMS current through the transformer and could lead to several other problems.

author: Zero999
author_rank: Super Contributor
timestamp: April 16, 2019, 09:04:33 pm
content: You could go for two 100V 100μF capacitors in series, if you don't have >120V 47μF.

author: gkmaia
author_rank: Frequent Contributor
timestamp: April 16, 2019, 09:17:51 pm
content: Quote from: Zero999 on April 16, 2019, 09:04:33 pmIf you put a caps in series do you effectively adds up their voltage rating but retain the capacitance?If you out them in parallel would you add the capacitance but retain the voltage?Same as series resistors, while you add the resistance?

author: gkmaia
author_rank: Frequent Contributor
timestamp: April 16, 2019, 09:27:28 pm
content: Also, would a Nichicon 68μF 200V dc Aluminium Electrolytic Capacitor be fine?It is 40% more capacitance. Is it too high?

author: helius
author_rank: Super Contributor
timestamp: April 16, 2019, 09:29:58 pm
content: Quote from: gkmaia on April 16, 2019, 09:17:51 pmIt's more like, half the capacitance. And at higher voltages this can be unsafe unless you also use balancing or bleeder resistors.Ideally the series C formula is the same as the parallel R formula:\$ C_{series} = {1 \over \frac{1}{C}+\frac{1}{C}+\frac{1}{C} ...} \$But ESR and leakage deviate from the ideal.QuoteYes, exactly.Quote(Again ideally) combining capacitors in parallel has the same effect on the overall C, as combining resistors in series has on the overall R. Another way of saying it is that the capacitor behavior is dual to the resistor behavior. This is a consequence of the KCL and KVL formulations of the circuit being duals.

author: Zero999
author_rank: Super Contributor
timestamp: April 17, 2019, 09:41:39 am
content: The capacitance can't stay the same, because it would mean the energy being stored would double. The energy stored in a capacitor is proportional to the voltage squared.E =1∕2CV2Suppose the two 100µF capacitors are connected in parallel and charged to 100V. The total capacitance would be 200µF.V = 100VC = 200×10-6FE =1∕2CV2E = 0.5*200×10-6*1002= 100×10-6*10000 = 1JThe capacitors are connected in series and charged to 200V. The total capacitance is now 50µF, but the total amount of energy stored in the capacitors is still 1J.V = 200VC = 50×10-6FE =1∕2CV2E = 0.5*50×10-6*2002= 25×10-6*40000 = 1JYes, in practise voltage balancing resistors are a good idea, but when you've got plenty of headroom, it's less critical.A 68μF 200V capacitor would be fine.