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Old 5th March 2011, 05:52 PM   #911
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Quote:
Originally Posted by NMOS View Post
Hello, as beginner please can you teach me what is diode trick ?
Diode trick means bypassing the internal body diode. Using an other, better diode to conduct, or some other way to decrease losses. Check this tread:
Body diode bypass, elegant way!
But if you are a beginner, DO NOT go over 200V. Better stay till 150V, start with low-power, or later use full-bridge (>1kW is possible with 150V FETs and full bridge). The problems exponentially increase over 200V.
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Old 7th March 2011, 10:44 PM   #912
NMOS is offline NMOS  Germany
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Have a question,....

could this amp work stable with +/- 90V using IRFP4332 (IRFB 4332) or similar ?

is over and undervoltage protection important ? using torodial transformer power supply
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Old 8th March 2011, 07:19 AM   #913
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Quote:
Originally Posted by NMOS View Post
Have a question,....

could this amp work stable with +/- 90V using IRFP4332 (IRFB 4332) or similar ?

is over and undervoltage protection important ? using torodial transformer power supply
If you pushing the limits so far, then an overvoltge protection should be necessary. At these high powers, bus pumping can easily push you over limit.
Since the IRFP4332 is a 250V it should handle +-90, but you would need a really good PCB layout for that. So as you said if you are a beginner, try with lower power first, better use full bridge.
(a 90V single-supply will do the same, with really cheap fets like IRF540Z, no bus pumping, so 90V for a 100V fet should be no problem).
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Old 8th March 2011, 12:58 PM   #914
stewin is offline stewin  Kenya
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does irf540n need a discharging circuit if driven in full bridge
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Old 8th March 2011, 03:33 PM   #915
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Quote:
Originally Posted by stewin View Post
does irf540n need a discharging circuit if driven in full bridge
SPafi quotes >100nC for IR2110 driver, since IRF540N is <71nC, it is not needed. For this small gate charges this will not enhance, since the driver needs some time to turn on its output transistors, so charging/discharging current to go up, and if you put an additional transstor it just adds delay.
In the datasheet it is said, what time is needed to get the 2A otuput, and its 25/17ns for IR2110. So the calculation

- without extra transistor (just diode): for 17nsec the average discharge current is 1A (linear rise is assumed), that means 17nC, another 54nC is supplied after 27ns, so total discharge time is 44ns (diode turn on is <4 ns so it is neglected)

- with extra transistor: the discharge current is first dictated by the resistor. I have no data for BC327, but 2n3906 is a simialr HF transistor, which says a turn on of >10ns. So for 17ns the discharge current is about 0,5A for average conditions (charge set 2x faster then discharge), that means 8,5nC. Then for 10nS the discharge current is 1A, that means 10nC. Total 18,5nC for 27nS, remaining 52,5nC charge is supplied with 3,5A (quessed current gain), takes 15ns. So total discharge time is 42ns.

The difference is totally in our calculations' and measurements' and datasheets' error tolerances, so no gain. This calcualtion is a quess, but it shows the way. Do the calculation for a FET with 150nC (our FET is good in every other property: low rdson, low Qcc, so we are not talking about old fets like IRFP260N, FORGET THEM). For that high charge it will shows the adventages of the extra transistor.
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Old 8th March 2011, 06:43 PM   #916
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Laci!

The specified charge and discharge current is measured on short circuit. In the miller region it is only the half or less. This because the BJT may help also for lower gate charge somewhat. But I never tried it with low gate charge.

I don't know if it is needed (or beneficial) for IRF540N in this circuit, the builder has to try it!
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Old 8th March 2011, 06:54 PM   #917
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Quote:
Originally Posted by Pafi View Post
Laci!

The specified charge and discharge current is measured on short circuit. In the miller region it is only the half or less. This because the BJT may help also for lower gate charge somewhat. But I never tried it with low gate charge.

I don't know if it is needed (or beneficial) for IRF540N in this circuit, the builder has to try it!
Even if it is half or less, the added BJT has turn on time, so with a BJT of infite current gain, but same turn on time the discharge is at least 27ns. So it may help, but not much. For high gate charge its necessary (its a really good idea), but for low gate charge: do the added complexity worth the improvement?
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Old 8th March 2011, 07:12 PM   #918
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Does turn on time means no base current? I don't think so. Maybe VBE can go up to several volts, but for a very short time. Then it acts like a diode (+resistor).

Quote:
so with a BJT of infite current gain, but same turn on time the discharge is at least 27ns.
While without BJT it's ~80...100 ns.

Quote:
but for low gate charge: do the added complexity worth the improvement?
For me it doesn't, but not because of complexity, since the difference is only 2*1 pin.
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Old 9th March 2011, 09:01 AM   #919
stewin is offline stewin  Kenya
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btw guys i don't love using the old fets e.g irfp250 and 260 but i have no choice maybe to try 45v-0-45v full bridge using irf540 ,540n or irf640.
out of topic a bit but just wondering in discreet circuit like the below can the gate discharge bjt transistor be added and will it be effective since i am using oldschool fets
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Old 10th March 2011, 12:06 AM   #920
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Hello

To operate the amplifier UCD With no load voltage of the MOSFET 90VDC IRFP4229 is recommended for the job?. Has an acceptable gate charge and switching time acceptable.
He had used the IRFP4232 But I've come to realize that they are too much for the IR2110. I made a mistake. Too lost in them (no-load heating amplifier)

Of the manufacturer International Rectifier There is not much selection of MOSFETs with a Vds voltage greater than 250V and efficient switching parameters. What do you think?

The tests I did with the amp with IRFP4232 were nothing unusual, in charge with +-85V @ 4Ohms power is very good, with acceptable audio quality (to ear).
Now I have to detail is the DC offset at the output. Involving a re-design of the PCB.

Greetings!

PS: I use a translator
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