555 PWM circuit

Nov 18, 2007 23 Replies

I think that adding a current limiter is a good idea.

I searched in google and found several "home-made" circuits, most of them using a LM317, for instance:

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I want to limit the current to 20A

It's a good idea? Is any commercial current limiter available? (cheap :D )

Thanks.

Yes, limiting current is a good idea, and if you're willing to use a mongo bipolar transistor and buy a low-ohms precision resistor, you can do it with four^H^H^Hive parts:

+V | [LOAD] | / c 555-3 ------[1K]----+-----| NPN | \\> e | | [D] | k| | | [Rcl] [D] | k| | | | +---+---+ | [GND]

Where Rcl is "resistor, current limit", and its value will be how ever many ohms drop about .6~.7V. at your current limit value.

I'm not sure how much current a 555 will source - you could either buffer it with a PNP, or use a darlington, in which case you'd need 3 diodes in series.

Good Luck! Rich

Nope. At 20A, you really want to avoid putting any kind of resistance in the high-current path.

The way to perform current limiting for this kind of application is to cut the PWM duty cycle when the current exceeds a certain threshold.

You will need a sense resistor (typically a 4-terminal "Kelvin" type) to measure the current, and a current-sense amplifier (a differential amplifier designed for common-mode voltages close to the power rails). E.g.:

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The power dissipation in the sense resistor is given by I^2*R. For a 1W resistor, 20^2*R R R

Did you miss this bit:

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At 20A, limiting current by operating the "switch" in its linear region isn't the right solution, IMHO. Reducing the PWM duty cycle is the way to go.

I was going to suggest using a high-side sense resistor and a PNP transistor to bypass the charging resistor, but the fact that the

600-700mV Vbe would translate to 12-14W dissipation in the sense resistor put me off.

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