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Mads Barnkob

Thu Oct 22 2009, 06:45AM

Registered Member #1403 Joined: Tue Mar 18 2008, 06:05PM
Location: Denmark, Odense C
Posts: 1968

I got a question form Angstrom

But i'm wondering if the 555 supplies enough current to charge the IRFP250 gate.

By my (estimated) calculations...
123nC gate charge, 48000Hz @ 83% duty cycle
0.000000123C * 48000 * 0.83 = about 5mA (easily supplied)
Now I don't think that can be right... what do you think?

So I calculated it this way.

I = (Capacitance * drive voltage) / ((1 / frequency) * Dutycycle)

I = ((123*10^-9) * 12) / ((1 / 48000) * 0.83) = 0.085A ~ 85 mA

Even here a 555 should be able to deliver this current, but reality shows something else.

So I looked into the 555 high output current vs. voltage drop and made the attached graph out from a TI datasheet.

So the voltage drop at the required current found in the equation above can be added to the supply voltage in order to achieve it, but it puzzles me abit with the duty cycle, im not sure if I should multiply with 0.83 or 0.17 for a 83% duty cycle, as it is in the above equation, current goes down with higher duty cycle which doesnt make sense to me, but the other way around with 0.17 the current seems rather high.

I lost myself last night to this and went to bed, hopefully someone can straight me about abit

Steve Conner

Thu Oct 22 2009, 08:36AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Your calculations give the average current, which is the relevant one for heating of the gate driver. (well, sort of) But to answer the question "can a 555 drive X MOSFET", the peak current is more relevant.

To answer this, you start by deciding what switching speed you need. Then look up the total gate charge from the MOSFET datasheet. Now use the equation for a capacitor, Q=I*t. You know t (the switching time) and Q (the gate charge) so you can solve for I, the required peak drive current. If it's more than the rated output current of the driver (250mA for a 555, IIRC?) then the MOSFET will not be able to switch as fast as you wanted.

If you don't know what switching time you wanted, my advice is just to try it and see if it works satisfactorily.

Mads Barnkob

Thu Oct 22 2009, 11:09AM

Registered Member #1403 Joined: Tue Mar 18 2008, 06:05PM
Location: Denmark, Odense C
Posts: 1968

I can see my trouble is I were looking for an answer to the problem of a heating MOSFET in a audio modulated flyback circuit.

Angstrom suggested that the 555 was not able to drive the IRFP250N and thus it were still somewhat in linear mode causing heating.

I have my circuit run at 43700 Hz and the gate capacitance of the MOSFET is 123 nF.

So it should be no problem for the 555 to deliver the (123*10^-9) / (1 / 43700) = ~6 mA peak needed.

I guess the heating problem is another issue.

Bjørn

Thu Oct 22 2009, 11:20AM

Registered Member #27 Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058

So it should be no problem for the 555 to deliver the (123*10^-9) / (1 / 43700) = ~6 mA peak needed.

This current causes the mosfet to be in the linear region all the time since when it reaches the top it is already time to go down again. In fact it will only get halfway there because you can not use more than half the wavelength before it is time to return.

Try somewhere between 1/20th and 1/200th the time to run cooler. Also double check the datasheet of your chip, they are not all the same.

klugesmith

Thu Oct 22 2009, 03:42PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

Steve and Bjorn put it very well.
6 mA is the current which will put 123 nC onto gate in about 20 microseconds, but that's the whole period.
If the 555 sources 123 mA, it would fully charge the gate in 1 microsecond (T=Q/I).
That means about 5% of the period is spent in the transition from OFF to ON. And probably a similar fraction in the transition between ON and OFF.

If you're uncertain about your "555" type, you can quickly measure its output characteristic (as shown in Mads's chart) with a voltmeter. Set up 555 with Vcc=12V and output continuously HIGH -- I think grounding pins 2&6 will do. Measure the voltage between Vcc and Vout -- should be near zero with no load (or just the MOSFET gate). Repeat measurement with loads of 1000 ohm, 100 ohm, then 50 ohms to ground. For example, if voltage drop is 1.8 volts with 100 ohm resistor then the current is about (12V-1.8V)/100R = 102 mA. Beware of overheating the resistor; you should be able to make the connection and take a reading in 1 second.

I have my circuit run at 43700 Hz and the gate capacitance of the MOSFET is 123 nF.

No, the gate charge is 123 nanocoulombs (at 12 volts) in that chart. About the same as a 10 nF capacitor at the same voltage, but you can see that the mosfet gate has a nonlinear charge-to-voltage curve.

Angstrom

Thu Oct 22 2009, 04:30PM

Registered Member #1900 Joined: Fri Jan 02 2009, 06:44PM
Location: Texas
Posts: 29

So we are using gate charge to calculate this... but that yields a very low current, which doesn't explain the heating. I looked at the datasheet for the IXYS 60N60 IGBTs used in DRSSTCs and the gate charge is almost the same... which makes no sense given how much more current is needed to drive them (unless i'm comparing apples and oranges). However there is a significant difference in the gate capacitance...

Mads Barnkob

Thu Oct 22 2009, 08:36PM

Registered Member #1403 Joined: Tue Mar 18 2008, 06:05PM
Location: Denmark, Odense C
Posts: 1968

Thanks for the explaining, I can now see I have had a wrong idea about how it worked.

It is somewhat embarrassing, but I was not even giving it a thought that I was mixing gate charge and gate capacitance.

To make sure I understand it, I made an example here, there is no deadtime so dutycycle is just what is left from rise/fall time of the period, gate voltage is just described as a logic value.

red is linear mode, green is on.

did I get it right?

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