Calculating capacitur size / specifications? - Page 1
Forum section: electronics / beginners

author: FriedMule
author_rank: Frequent Contributor
timestamp: May 03, 2020, 02:03:33 am
content: HiI have tried to find information how to calculate what values the two caps needs to be in my power supply. It's only a diode bridge and two caps. As long the voltage never goes below 5V, the constant current should work fine.I have made this small diagram to show how my constant current is madehttp://tinyurl.com/y7j8555fThe LED is just to make the circuit work. The real load is using 145mA continue and max 2.4A peak in 10uS.I have tried to look at formulas but did not understand them.

author: TimFox
author_rank: Super Contributor
timestamp: May 03, 2020, 03:01:13 am
content: Any textbook will discuss the peak voltage out of the rectifier into the capacitor. If you know the load current, then the pk-pk ripple voltage depends on the capacitance, current, and ripple frequency (100 or 120 Hz).dV = Q / C = (I T) / C = I / (C f)Your high current pulse is an additional charge draw over the DC value in this equation.

author: FriedMule
author_rank: Frequent Contributor
timestamp: May 03, 2020, 04:54:18 am
content: Okay I think I may understand it. If I did do I not think that I did explain my question right.I am trying to calculate the micro Farad I need to keep 9V + ripple to stay above 5V or maybe 6V, just to make sure.My load look like this 145mA continuous and max 2.4A peak in 10uS. There are build a min 5V current source between the caps and the load.

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 03, 2020, 09:18:19 am
content: There are quite exact formulas but they need some data.So I want to suggest one of the easy ways: ripple will be lower then 10% if capacitor capacitive impedance is 10 times less then load impedance.For example: voltage is 5VDC, current is .15A so load impedance is 5/0.15 = 33.3 R.So cap impedance have to be about 3.3 Ohms or less.That is about 1000 uF or more.10% ripple means 5V × 0.1 = 0.5 V ripple amplitude.

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 03, 2020, 09:22:50 am
content: A ripple must be 10% or less - it is an electrolitic cap restriction.So if load voltsge is 9V then ripple must be no more then 0.9V to save el. caps safe.

author: FriedMule
author_rank: Frequent Contributor
timestamp: May 03, 2020, 09:53:43 am
content: Okay but what about 2.5A in 10uS without ripple goes below 6V from 9V?How do I calculate how many micro Farad it takes?

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 03, 2020, 10:00:24 am
content: 10 us is 1/100000 s so additional average current is 2.5A/100 000 = 2.5 uA, so I think it won't change average ripple too much.But for peak deep, we need to calc it...Lets use the same method: 9V/2.5 A is a 3.6R load. Lets take a ripple amplitude is 20% or 0.2.3.6R×0.2 is 0.72R - needed filter impedance. That is about 4700 uF cap.

author: FriedMule
author_rank: Frequent Contributor
timestamp: May 03, 2020, 10:05:29 am
content: Thank you a lot, I'll try to calculate, what you have allready done, just to learn and se what I do get:-)

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 03, 2020, 10:13:05 am
content: These my short calculations go from that TimFox formulae:dV = Q / C = (I T) / C = I / (C f)I only wrote it in a little more convenient way.

author: FriedMule
author_rank: Frequent Contributor
timestamp: May 03, 2020, 10:22:29 am
content: So 9V = 6.2 Ohm or less, and 6V = 4.1 Ohm.How did you get 1000uF, when looking at Mouser and selecting 3.6 Ohm ESR, do it show a 8.2uF. That is clearly the wrong way to do it:-)Also there are a specification, called ripple current?

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 03, 2020, 10:32:50 am
content: Quote from: FriedMule on May 03, 2020, 10:22:29 amNo, you don't get it (or I?)I mean capacitive impedance which is 1/(2×pi×f×C). Don't take ESR into consideration - it does't help much.

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 03, 2020, 10:36:42 am
content: 9V and 0.15A load current is 9/0.15 = 60 Ohm load impedance.So filter must have 60×0.1 = 6 Ohm reactive impedance (or less). Which cap have such reactive (capacitive) impedance, if we know that capacitive impedance is 1/(2×pi×f×C)?- 510 uF cap - we can use lower 470 uF cap so ripple will be higher. But it is better to use higher capacity cap which is 1000 uF. There is no need to use 680 uF cap.You mentioned 2.5 A pulses - they need 4700 uF cap.

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 03, 2020, 10:54:32 am
content: ESR and riple current are meaningful for high-frequency converters.If you go with <10% ripple rule for low-frequency (50/60 Hz) rectifiers - real ripple current will be lower than cap rated.

author: FriedMule
author_rank: Frequent Contributor
timestamp: May 03, 2020, 12:47:53 pm
content: Thank you so much! It's a great explanation, I hope to be able to give back one time!LOL if anyone has questions about Magic tricks, I may be able to help (wont revile trick unless they are no longer performed:-)

author: FriedMule
author_rank: Frequent Contributor
timestamp: May 05, 2020, 01:11:26 am
content: Trying to determine the size of the transformer.As written above do the real load use 145mA continues and max 2.4A peak in 10uS.How large do the transformer have to be, I know that it can only get to small but I do hope on using about the smallest transformer that will work fine.

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 05, 2020, 09:05:13 am
content: Low-frequency transformer can withstand high-current peaks very well (if they are short enough to not increase an average current). So you have to take into consideration only an average load current.If the load current is 0,15 A DC, then the transformer current (for a bridge rectifier and similar) has to be somewhere about 0,15*1,41 = 0,21 ACA (the secondary winding current form is distorted so we are talking here about RMS value). Secondary winding voltage is less then rectified and filtered and the winding current is respectively larger.

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 05, 2020, 09:13:52 am
content: 9V DC rectifier output voltage needs about 7-8 V AC secondary winding. Of cause that output voltage will be a bit higher without load.A transformer secondary voltage value depends on diode type too - Shottky or not. Shottky diodes have about twice as less voltage drop which is significant for rectifiers up to 20-30 V DC. Of cause it is much significant for the 5-10 V circuit.

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 05, 2020, 09:34:31 am
content: There are more accurate calculations but I'll try to tell a simpler one (but it may be accurate enough):9V plus the diode voltage drop, for a bridge, is 9V + 2V = 11 V DC (here 2 V is 2 * 1V - one volt for one diode. Two diodes in bridge open simultaneously).11V / 1,41 = 7,8 V AC. But it is for an unloaded transformer. If we load the transformer then the voltage will drop. Let's take that it drops for about 10% under load (that is accurate only if a transformer isn't too small otherwise it will have larger internal resistance and AC voltage drop).So (for loaded transf.) we can use coefficient 1.25-1,35 instead of 1,41. Then needed rated AC secondary voltage is about:11V / 1.30 = 8,5 V AC.If the load current is 0,15 A then peak capacitor charging current is about 0,15*3 = 0,45 A. That current is diode current too. You may look for your diode datasheet to find a diode voltage drop under this current. It may be less than 1 V (about 0,8V for 1N4007), so a transformer secondary voltage has to be respectively smaller (about 8V AC).It is much convenient to use Shottky diodes for low-voltage rectifiers (<10..15V), so we will have a lower output voltage variation.If we use Shottky diodes (let's take SB140 - 1A 40V diode for example) then voltage drop with 0,45 A charging current is about 0,45 V. Then:9V + 2*0,45V = 9,9V DC.9,9V / 1,3 = 7,6 V AC.

author: FriedMule
author_rank: Frequent Contributor
timestamp: May 05, 2020, 03:29:40 pm
content: Thank you a lot, I'll read trough your answers and try to come with my result, to get checked:-)