# Capacitor circuit needed

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#### Tony4t3

##### New Member
I am looking for a circuit that can be charge with about 4 volts but out put 7-12 volts into the load. I would like it to run the load for at least 10 seconds before it starts to charge back up. I understand the charge time will be much larger then 10 seconds. The output amperage needs to be at least 1 amp and can not exceed 5 amps.

Thank You for any help in getting the circuit created.

#### MikeMl

##### Well-Known Member
Welcome Tony4t3

If you charge a capacitor to 12V and discharge it at a current of 1A so that after 10sec the voltage is (say) 7V, then

I*t = C*ΔV = Q
C = I*t/ΔV = 1*10/(12-7) = 10/5 = 2Farads (Have you heard of Supercapacitors?)

If you get a step-up switcher that converts 4V to 12V, then ask how long it will take to charge a 2Farad capacitor (with no load during the charging phase)?
I*t = Q = C*ΔV = 2*5 = 10Coulombs

Suppose that the 4V source is constrained to supplying 1A, and the switcher is 80% efficient. The switcher requires 10/0.8 = 12.5C to be transferred from the 4V source to the 2F capacitor.
I*t = Q
t = Q/I = 12.5/1 = 12.5seconds.

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#### Colin

##### Active Member
A super-capacitor is quite useless for this application.

#### MikeMl

##### Well-Known Member
A super-capacitor is quite useless for this application.
I was hoping to educate Tony about Supercapacitors; why don't you teach him why you think they are useless

#### Tony4t3

##### New Member
Welcome Tony4t3

If you charge a capacitor to 12V and discharge it at a current of 1A so that after 10sec the voltage is (say) 7V, then

I*t = C*ΔV = Q
C = I*t/ΔV = 1*10/(12-7) = 10/5 = 2Farads (Have you heard of Supercapacitors?)

If you get a step-up switcher that converts 4V to 12V, then ask how long it will take to charge a 2Farad capacitor (with no load during the charging phase)?
I*t = Q = C*ΔV = 2*5 = 10Coulombs

Suppose that the 4V source is constrained to supplying 1A, and the switcher is 80% efficient. The switcher requires 10/0.8 = 12.5C to be transferred from the 4V source to the 2F capacitor.
I*t = Q
t = Q/I = 12.5/1 = 12.5seconds.

Thank you for your response. Some of that is Greek to me. The source has about 4V @ 0.5A. If I step the voltage up to 12V then the amperage would go down to about 0.166A. The load can run off of 8V so I can double the voltage and I can double the source. What would the circuit look like and the values for the components? Thank you for your help.

#### Tony4t3

##### New Member

The source is a cheap TEG(Thermoelectric Generator) and the load is a small circuit that does not need to run continuously. The circuit is rated for 5-12V and <=10A. I am thinking of getting a better TEG. One that will output more power.

#### MikeMl

##### Well-Known Member
I think you need to use the TEG and a small step-up smps to charge an 8V 1Ah SLA battery, or possibly a NiMh pack. This complicates it bit, because the smps also has to be a charge regulator for whatever battery you choose. A simple shunt-regulator would work on a SLA battery.

What does the timing as to when the load is on and off? Is it voltage controlled, or time controlled?

I have done something similar, but the source was a 20Vopen circuit, 1A max solar panel. It runs a 12V pump (~5A), so I used a 12V SLA and a 555 chip to sense the battery voltage. It starts the pump whenever the battery voltage >14.4V, and shuts the pump off when the battery voltage <12.5V. It has been running for several years in my wife's green house...

In my case, the panel had more voltage than the motor requires, but no where near enough current, so the 555 chip/battery makes it such that the motor runs less than 1/5 of the time. Obviously, it only runs when the sun is shining.

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