increasing output current

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neilxavier

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generally, how can I increase the output current of a circuit. i.e. increase a 0.6A output current to about 1.5A?
 
Neil, to post your image, click on Post reply (Not quick Reply), from there go to manage attachments and click. You can upload from your PC
 
I'm sorry for that.

i attached the file since i cant upload it to other image hosting sites. like i've said, we're trying to increase the supply current to around 1.5A from 0.6A. maybe we can add additional circuitry.
 

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neilxavier said:
I'm sorry for that.

i attached the file since i cant upload it to other image hosting sites. like i've said, we're trying to increase the supply current to around 1.5A from 0.6A. maybe we can add additional circuitry.
Click to expand...

hi,
The LT1935 datasheet states that it can switch upto 2Amps.
Do you have any component values to post.?
 
R load is 100 OHM and you need 1.5 AMP??
wat is the V output you want to have??

Robert-Jan
 
here.

basically we just changed the values of components on the datasheet. The values on the figure were obtain based on the actual implementation (hardware). High input and output capacitors were to minimize the ripple on the output voltage.

Yes the switch current was up to 2A. But I'm just confused if the switch current would also be of the same range as the output current. btw, what is the switch current all about? is it same as the output current?

Thanks!
 

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According to the data sheet, the shutdown circuitry will reduce your output current. Try just setting it to above 2v.
 
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that's alright.

looking at the datasheet, it shows a sample circuit that converts a 5V input to a 12V out and has 600mA output. when we measure our circuit, the output current is (obviously) equal to the current given by ohm's law. i.e. V=IR.
 
rjvh said:
No the RC combination depend the max output but only in start up (that's my interpatation)

Robert-Jan
Click to expand...

but looking at figure 6, it can be seen that the main difference between them is the start up but the final output current seem to be the same. I am not totally contradicting your understanding but that's how I see it. Thanks!
 
I think that increasing the coil value has an positive effect on the output voltage and in a lesser extend for the curent as they make use of the energy stored in the coil and give a puls aditional on the output when the coil release its energy to boost the output the capacitor should do it more for the current

as you see the curent decrease with the increase of the duty cycle of the puls and will at a sertain moment not be in time for the total energy release time of the coil and capacitor

Robert-Jan
 
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the current limit that decreased with the duty cycle is the switch current, right? And as I understand the datasheet, the changing the inductor value will change the frequency (with constant voltages and switch current), am I right?

You said that changing the coil value has a positive effect on output voltage, what is that positive effect? I just prefer to maintain the same output voltage (12V) since it is in accordance with the specs of the beacon.
 
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