roineust
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Re: Reducing voltage from 5V to 3-4 volts.
Hey people!
Thanks for the replies!
I will check your advices and return with questions, in case they will pop up!
Peter -
Sorry!
My mistake!
Here are the servos that i must use, because of weight limits of my application, which causes all the trouble of finding how to drop down the voltage:
https://www.hobbyking.com/hobbyking...0_075kg.html?gclid=CLLh96aV87kCFYm_3godl3gAhw
**broken link removed**
Regarding the current of LiPo, sorry, again my mistake, yes now i recall that i should multiply the C rating by the mAh.
My question is: when connecting a diode this way, does the rule that says that current (in contrast to voltage), is something that is drawn from the servo side and not 'pushed' towards the servo, does this rule still apply in this diode case? so if a diode is rated 1A, the servo will not be able to draw more than 1A, but no damage will occur to the diode, even if it is a 1000mAh X 20C LiPo?
The question also comes up, in light of the fact, that i did check the 2X 1N4004 (1A,400V) diodes setup, that did reduce the voltage, from 5.0V to 4.06V, while using a 3S 11.1V 1000mAh X 20C LiPo - and nothing was 'fried', or even heated up...was that just a matter of time, before that would happen?
Thanks.
- - - Updated - - -
Hey ard,
A question has already popped up! Peter, you opinion would help as well!
For example, the following component:
https://www.ebay.com/itm/DC-5V-to-3...731?pt=LH_DefaultDomain_0&hash=item417060fd4b
It says: " output: 3.3 V, 800 mA (load current can't more than 800 mA) "
This means, that by NO means, i can connect on the other side of this component, a LiPo, that is, say, 1000mAh X 20C ??
Thanks.
Hey kam1787!
As mentioned above, this is not clear to me...i know very little about electronics, and from that little amount of knowledge gathered here, i concluded, that current is somthing that is 'drawn' by the servo and not 'pushed' towards it (like voltage is?) - so how come, you talk in terms of a current, that can 'fry' the component? This might be too complicated for me to understand...please try to make the explanation as simple as possible....
Thanks.
Hey people!
Thanks for the replies!
I will check your advices and return with questions, in case they will pop up!
Peter -
Sorry!
My mistake!
Here are the servos that i must use, because of weight limits of my application, which causes all the trouble of finding how to drop down the voltage:
https://www.hobbyking.com/hobbyking...0_075kg.html?gclid=CLLh96aV87kCFYm_3godl3gAhw
**broken link removed**
Regarding the current of LiPo, sorry, again my mistake, yes now i recall that i should multiply the C rating by the mAh.
My question is: when connecting a diode this way, does the rule that says that current (in contrast to voltage), is something that is drawn from the servo side and not 'pushed' towards the servo, does this rule still apply in this diode case? so if a diode is rated 1A, the servo will not be able to draw more than 1A, but no damage will occur to the diode, even if it is a 1000mAh X 20C LiPo?
The question also comes up, in light of the fact, that i did check the 2X 1N4004 (1A,400V) diodes setup, that did reduce the voltage, from 5.0V to 4.06V, while using a 3S 11.1V 1000mAh X 20C LiPo - and nothing was 'fried', or even heated up...was that just a matter of time, before that would happen?
Thanks.
- - - Updated - - -
Hey ard,
A question has already popped up! Peter, you opinion would help as well!
For example, the following component:
https://www.ebay.com/itm/DC-5V-to-3...731?pt=LH_DefaultDomain_0&hash=item417060fd4b
It says: " output: 3.3 V, 800 mA (load current can't more than 800 mA) "
This means, that by NO means, i can connect on the other side of this component, a LiPo, that is, say, 1000mAh X 20C ??
Thanks.
Hey kam1787!
As mentioned above, this is not clear to me...i know very little about electronics, and from that little amount of knowledge gathered here, i concluded, that current is somthing that is 'drawn' by the servo and not 'pushed' towards it (like voltage is?) - so how come, you talk in terms of a current, that can 'fry' the component? This might be too complicated for me to understand...please try to make the explanation as simple as possible....
Thanks.
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