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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
 Posted - Sep 16 2011 : 12:50:52 PM
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Juan,
1). For 1.5kW output you may need around 2.0kW input. At a nominal 13.5V input that requires about 148A. That means a minimum of 2 x IRF3205 FETs each side – preferably 3, as in the circuit.
The FETs don’t consume any power to switch them, but the gate capacitance has to be continuously charged and discharged by the 555 driver as the gate voltage is varied. The capacitance of 3 gates is about 10nF and the 555 can source or sink 100mA, so they will charge/discharge in less than a microsecond. That’s quite adequate for the job.
2). The IRF3205 can switch 100A when the gate is driven to +6V, and with +9V it is switched on hard. So there is enough gate drive.
3). I have no experience of using FETs to drive loads that are partially inductive, so I cannot answer your question about the series resoistor. But I’m sure that Audioguru can advise you whether 10 ohms is OK.
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JUAN DELA CRUZ
Mad Scientist
Philippines
476 Posts |
Posted - Sep 16 2011 : 1:44:09 PM
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quote:
Originally posted by pebe
Juan,
1). For 1.5kW output you may need around 2.0kW input. At a nominal 13.5V input that requires about 148A. That means a minimum of 2 x IRF3205 FETs each side – preferably 3, as in the circuit.
The FETs don’t consume any power to switch them, but the gate capacitance has to be continuously charged and discharged by the 555 driver as the gate voltage is varied. The capacitance of 3 gates is about 10nF and the 555 can source or sink 100mA, so they will charge/discharge in less than a microsecond. That’s quite adequate for the job.
2). The IRF3205 can switch 100A when the gate is driven to +6V, and with +9V it is switched on hard. So there is enough gate drive.
3). I have no experience of using FETs to drive loads that are partially inductive, so I cannot answer your question about the series resistor. But I’m sure that Audioguru can advise you whether 10 ohms is OK.
Good Morning here! Good Afternoon there?
Very well said.. thanks a lot sir for the great info.
I think thats fair enough for the driver section trouble..
How about the "cut-off circuitry part" and the " output frequency issue" included here in the schematic...
1. Do you think this is right?
I can't simulate it.. coz it is very crucial part. It might damage the Power Mosfets if it will malfunction right?
(I need to turnoff the inverter when the battery is low)
2. Do you think can I use SG3525 instead of with 24 without any adjustment in the circuit?
3. How about the frequency output of the inverter? ( I need 60Hz for my appliances )
Thanks in advance. |
juan dela cruz
Penniless INVENTOR |
Edited by - JUAN DELA CRUZ on Sep 16 2011 1:52:03 PM |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 16 2011 : 4:57:38 PM
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1. What about the cutoff circuitry? Do you think there is a fault there? Build it and try it without the FETs connected. On low battery volts the drive output transistors should give no output.
2. I don't know without going through both specs. Why don't you print out the data sheets for them both and then compare them for differences?
3. The frequency is set by two components as detailed on Page5 of the ST data sheet. I'm sure you can work it out for yourself. |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Sep 16 2011 : 6:58:20 PM
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quote:
Originally posted by pebe
A better solution than complementary transistors would be to use this circuit. It uses the CMOS version of the 555 timer.
It can source and sink 100mA, and with a swing to 9V it will fully turn on the IRF3205 FETs.
No.
The Cmos LMC555 has a minimum output sink current of 50mA when its voltage loss is 2V. Its minimum output source current is only 10mA when its voltage loss is 1.5V.
My idea to use complementary emitter followers have a minimum output current of hundreds of mA with a voltage loss of 0.7V or 0.8V for each polarity.
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 17 2011 : 10:38:27 AM
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quote:
Originally posted by audioguru
quote:
Originally posted by pebe
A better solution than complementary transistors would be to use this circuit. It uses the CMOS version of the 555 timer.
It can source and sink 100mA, and with a swing to 9V it will fully turn on the IRF3205 FETs.
No.
The Cmos LMC555 has a minimum output sink current of 50mA when its voltage loss is 2V. Its minimum output source current is only 10mA when its voltage loss is 1.5V.
My idea to use complementary emitter followers have a minimum output current of hundreds of mA with a voltage loss of 0.7V or 0.8V for each polarity.
You're quite right. I didn't look up the data sheet. I was going from memory and remembered wrongly. Something to do with 'senior moments', I believe.
Juan,
You can use the bipolar LM555 instead and power it from the battery supply. It can supply up to 200mA and swing the FET gates up to over 10V to switch them hard on.
Download Attachment:  FET driver.GIF
3.52 KB
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JUAN DELA CRUZ
Mad Scientist
Philippines
476 Posts |
Posted - Sep 19 2011 : 08:39:41 AM
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quote:
Originally posted by audioguru
No.
The Cmos LMC555 has a minimum output sink current of 50mA when its voltage loss is 2V. Its minimum output source current is only 10mA when its voltage loss is 1.5V.
My idea to use complementary emitter followers have a minimum output current of hundreds of mA with a voltage loss of 0.7V or 0.8V for each polarity.
@audioguru
Regarding with your idea of using complementary emitter-follower BJTs.. I was wondering if 2N2222 & 2N2907 with a 100-ohms pull-ups will drive Power MOSFETSs very well?
quote:
Original posted by pebe
You're quite right. I didn't look up the data sheet. I was going from memory and remembered wrongly. Something to do with 'senior moments', I believe.
Juan,
You can use the bipolar LM555 instead and power it from the battery supply. It can supply up to 200mA and swing the FET gates up to over 10V to switch them hard on.
Download Attachment: FET driver.GIF
3.52Â KB
@pebe,
Thanks sir, I will use LM555 instead of the Cmos type. |
juan dela cruz
Penniless INVENTOR |
Edited by - JUAN DELA CRUZ on Sep 19 2011 8:48:20 PM |
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JUAN DELA CRUZ
Mad Scientist
Philippines
476 Posts |
Posted - Sep 19 2011 : 8:53:15 PM
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quote:
Originally posted by audioguru
No.
The Cmos LMC555 has a minimum output sink current of 50mA when its voltage loss is 2V. Its minimum output source current is only 10mA when its voltage loss is 1.5V.
My idea to use complementary emitter followers have a minimum output current of hundreds of mA with a voltage loss of 0.7V or 0.8V for each polarity.
@audioguru
Regarding with your idea of using complementary emitter-follower BJTs.. I was wondering if 2N2222 & 2N2907 with a 100-ohms pull-ups will drive Power MOSFETSs very well?
quote:
Original posted by pebe
You're quite right. I didn't look up the data sheet. I was going from memory and remembered wrongly. Something to do with 'senior moments', I believe.
Juan,
You can use the bipolar LM555 instead and power it from the battery supply. It can supply up to 200mA and swing the FET gates up to over 10V to switch them hard on.
Download Attachment: FET driver.GIF
3.52 KB
@pebe,
Thanks sir, I will use LM555 instead of the Cmos type. |
juan dela cruz
Penniless INVENTOR |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Sep 19 2011 : 9:09:19 PM
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quote:
Originally posted by JUAN DELA CRUZ
Regarding with your idea of using complementary emitter-follower BJTs.. I was wondering if 2N2222 & 2N2907 with a 100-ohms pull-ups will drive Power MOSFETSs very well?
If you use the SG3525A IC then it already has the high current output transistors and it doesn't need pullup resistors. Also the 555 is not needed.
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JUAN DELA CRUZ
Mad Scientist
Philippines
476 Posts |
Posted - Sep 19 2011 : 11:27:19 PM
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quote:
Originally posted by audioguru
quote:
Originally posted by JUAN DELA CRUZ
Regarding with your idea of using complementary emitter-follower BJTs.. I was wondering if 2N2222 & 2N2907 with a 100-ohms pull-ups will drive Power MOSFETSs very well?
If you use the SG3525A IC then it already has the high current output transistors and it doesn't need pullup resistors. Also the 555 is not needed.
So, the output of the SG3525A can drive the base "directly" (w/out series resistor?) of the complementary emitter-follower BJTs.. then drive the gates of the Power Mosfets with a 10-ohms series resistor each?? |
juan dela cruz
Penniless INVENTOR |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Sep 20 2011 : 09:44:31 AM
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I said that the SG3525A already has the high current output transistors. Then you don't need the complementary transistors.
Yes, use a 10 ohm resistor in series with each Mosfet gate mounted very close to the gate pin. |
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JUAN DELA CRUZ
Mad Scientist
Philippines
476 Posts |
Posted - Sep 26 2011 : 12:15:55 PM
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quote:
Originally posted by audioguru
I said that the SG3525A already has the high current output transistors. Then you don't need the complementary transistors.
Yes, use a 10 ohm resistor in series with each Mosfet gate mounted very close to the gate pin.
Thanks audioguru..
I have read some articles regarding power mosfet driver.
And I'd found out that a "fast swithing diode" is needed in the gate of the power mosfet parallel to the gate resistor 10 ohms.
Do you think 1N4148 schottky diode will do?
... and how about its polarity? the anode to the gate of the power mosfet? |
juan dela cruz
Penniless INVENTOR |
Edited by - JUAN DELA CRUZ on Sep 26 2011 12:17:00 PM |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Sep 26 2011 : 6:59:19 PM
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I have never seen a Mosfet driven with a diode parallel with the 10 ohms resistor at its gate.
A diode will not do anything since the 10 ohms will quickly charge the gate capacitance for a fast turn on and the 10 ohms will also quickly discharge the gate capacitance for a fast turn off.
A 1N4148 is an ordinary fairly fast low current silicon signal diode. Its forward voltage is about 0.7V.
A Schottky diode is very fast and has a forward voltage of about 0.3V.
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JUAN DELA CRUZ
Mad Scientist
Philippines
476 Posts |
Posted - Sep 27 2011 : 03:23:30 AM
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quote:
Originally posted by audioguru
I have never seen a Mosfet driven with a diode parallel with the 10 ohms resistor at its gate.
A diode will not do anything since the 10 ohms will quickly charge the gate capacitance for a fast turn on and the 10 ohms will also quickly discharge the gate capacitance for a fast turn off.
A 1N4148 is an ordinary fairly fast low current silicon signal diode. Its forward voltage is about 0.7V.
A Schottky diode is very fast and has a forward voltage of about 0.3V.
Ok, thanks.
How about the "snubber ckt needed" in the primary winding of the transformer to filter out the spikes that might damage the IC?
What will be the value of the resistor and capacitor needed if the power inverter is operating with the frequency of 60Hz? |
juan dela cruz
Penniless INVENTOR |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Sep 27 2011 : 09:01:14 AM
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| We don't know if your transformer will produce voltage spikes. If it does then you must test different snubber parts to reduce the spikes since your transformer has no detailed spec's when it is driven with a square-wave. |
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JUAN DELA CRUZ
Mad Scientist
Philippines
476 Posts |
Posted - Sep 27 2011 : 1:34:03 PM
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quote:
Originally posted by audioguru
We don't know if your transformer will produce voltage spikes. If it does then you must test different snubber parts to reduce the spikes since your transformer has no detailed spec's when it is driven with a square-wave.
Ok, so the snubber ckt isnt that essential for the PWM IC protection?
Do you think I need to use discrete Zener diode (15V) to power the SG3525A IC instead of using 7809 regulator? |
juan dela cruz
Penniless INVENTOR |
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12/120V inverter again
