Help in reduce circuit losses

GH Crash · 2024-09-15T23:14:02-0700

Any suggestions on how to reduce the resistance of the attached circuit.

A supercapacitor (EDLC) is used to power a small motor. There is an ATTiny85 chip in the circuit that is used as a timer and a PWM controller for the motor. The problem that I'm loosing about half a volt between the capacitor's voltage and the voltage across the motor's leads. I need to minimize losses in the circuit as much as possible.

Can you suggest means to reduce voltage loses to the motor?

Nigel Goodwin · 2024-09-16T08:47:48-0700

GH Crash said:
For those of you who question the use of a capacitor, the entire driving force behind this project is to make a device/circuit that will make a capacitor preform like a battery. Why? Because no one has done it.

There are good reasons for that, a capacitor makes a pretty poor battery.

GH Crash · 2024-09-16T10:05:43-0700

danadak said:
So where do you measure V loo\sr and how much is the loss in V ?

Ok, you lost the dummy again. "Measure V loo\sr" ??? Capacitor voltage is measured across the cap leads, motor voltage is across the motor leads. The difference between the two is about half a volt.

GH Crash · 2024-09-16T10:12:57-0700

Nigel Goodwin said:
There are good reasons for that, a capacitor makes a pretty poor battery.

Under some/most conditions, I agree with you, but in other cases, a battery makes a pretty poor power supply. As a free flight power source, batteries are heavy and take time to recharge. Capacitor are much lighter and re-charge within seconds.

AnalogKid · 2024-09-16T10:13:09-0700

GH Crash said:
For those of you who question the use of a capacitor, the entire driving force behind this project is to make a device/circuit that will make a capacitor preform like a battery. Why? Because no one has done it.

Actually, lotsa people have done it. Ultracaps are common on hybrid city busses, and Coleman (the camping lantern people) used to make a cordless screw driver that used caps instead of nicad cells.

ak

Nigel Goodwin · 2024-09-16T10:59:52-0700

GH Crash said:
Under some/most conditions, I agree with you, but in other cases, a battery makes a pretty poor power supply. As a free flight power source, batteries are heavy and take time to recharge. Capacitor are much lighter and re-charge within seconds.

They are lighter, and charge faster, mostly because they are much lower capacity - and to charge quickly requires large charging currents. It's simple physics - to charge faster needs higher current - to put hours of charge in a storage medium in seconds takes massive currents. For example, if a battery takes 5 hours to charge at 1A, a similar specification (if such existed) capacitor would need 60A to charge in 5 minutes, or 300A to charge in just one minute.

alec_t · 2024-09-16T11:41:38-0700

GH Crash said:
The difference between the two is about half a volt.

You shouldn't be losing 1/2 Volt if the current is 3A. There must be a poor connection or else the wire cross-section is too small.

danadak · 2024-09-16T14:13:57-0700

GH Crash said:
Ok, you lost the dummy again. "Measure V loo\sr" ??? Capacitor voltage is measured across the cap leads, motor voltage is across the motor leads. The difference between the two is about half a volt.

Meaure Voltage loss, where the drops occur, what path element creates the loss ?

Help in reduce circuit losses

GH Crash

Member

Attachments

Nigel Goodwin

Super Moderator

GH Crash

Member

GH Crash

Member

AnalogKid

Well-Known Member

Nigel Goodwin

Super Moderator

alec_t

Well-Known Member

danadak

Well-Known Member

Similar threads

Help in reduce circuit losses

Member

Attachments

Super Moderator

Member

Member

Well-Known Member

Super Moderator

Well-Known Member

Well-Known Member

Similar threads

Privacy & Transparency

Privacy & Transparency