Aaron's Homepage Forum
Aaron's Homepage Forum
Home | Profile | Register | Active Topics | Members | Search | FAQ
Username:
Password:
Save Password
Forgot your Password?

 All Forums
 Electronics
 Power Supply
 12/120V inverter again
 New Topic  Reply to Topic
 Printer Friendly
Previous Page | Next Page
Author Previous Topic Topic Next Topic
Page: of 63

pebe
Nobel Prize Winner

United Kingdom
1078 Posts

Posted - Sep 13 2011 :  02:22:21 AM  Show Profile  Reply with Quote
Juan,
A further problem with the circuit.

When Q8 conducts its collector will go negative. By auto-transformer action, the collector of Q7 will go positive, and D1 will fry ! Ditto for D2 when Q7 conducts.
Go to Top of Page

JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts

Posted - Sep 13 2011 :  06:07:54 AM  Show Profile  Reply with Quote
quote:
Originally posted by pebe

Juan,
A further problem with the circuit.

When Q8 conducts its collector will go negative. By auto-transformer action, the collector of Q7 will go positive, and D1 will fry ! Ditto for D2 when Q7 conducts.



How about in this circuit? (http://www.edaboard.com/thread192166.html)

I need to include complementary emitter-follower BJTs to drive Power Mosfets instead. Do you think 2N2222 & 2N2907 will do? ..with a 1k pull up resistor from the IC?


juan dela cruz
Penniless INVENTOR

Edited by - JUAN DELA CRUZ on Sep 13 2011 08:59:02 AM
Go to Top of Page

audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - Sep 13 2011 :  1:19:58 PM  Show Profile  Reply with Quote
Juan, the schematic you posted is too small to see.
It is the circuit at EDA board with the diodes that fry. Also it might have other problems.
It doesn't have pullup resistors since the outputs that are used from the IC are emitter-followers.

You can build the defective circuits that are shown on the internet and see them burn.
You need to learn about the basics of electronics so you can spot errors in circuits and so you can modify circuits.
Go to Top of Page

JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts

Posted - Sep 13 2011 :  7:32:48 PM  Show Profile  Reply with Quote
quote:
Originally posted by audioguru

Juan, the schematic you posted is too small to see.
[quote]Originally posted by audioguru

Juan, the schematic you posted is too small to see.
It is the circuit at EDA board with the diodes that fry. Also it might have other problems.
It doesn't have pullup resistors since the outputs that are used from the IC are emitter-followers.

You can build the defective circuits that are shown on the internet and see them burn.
You need to learn about the basics of electronics so you can spot errors in circuits and so you can modify circuits.



I do have knowledge in electronics, but not that much that's why Iam asking for some assistance.

Ok how about this circuit?

http://www.simplecircuitsandprojects.com/circuits/power-inverter-circuit-with-low-battery-shutdown1.html


juan dela cruz
Penniless INVENTOR

Edited by - JUAN DELA CRUZ on Sep 13 2011 8:01:00 PM
Go to Top of Page

audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - Sep 13 2011 :  11:19:15 PM  Show Profile  Reply with Quote
quote:
Originally posted by JUAN DELA CRUZ
Ok how about this circuit?

I discussed the errors of this circuit about 18 hours ago. The high resistor values in series with the gates of the Mosfets and the high values of the pullup and pull-down resistors cause the Mosfets to turn on and off very slowly (because their high gate capacitance takes a long time to charge and discharge by the high resistor values) causes severe heating of the Mosfets.

Mosfets need a 10 ohms to 47 ohms resistor in series with the gate to prevent high frequency oscillation. They need a high current drive to quickly charge and discharge their high gate capacitance.
Go to Top of Page

pebe
Nobel Prize Winner

United Kingdom
1078 Posts

Posted - Sep 15 2011 :  07:16:52 AM  Show Profile  Reply with Quote
quote:
What value of complementary emitter-follower BJTs can you suggest me to use that can be added to the Driver circuit in the link that I provided?
(do you think 2N22222A & 2N29072A will enable to drive a lot of Power mosfets to have a power output of approx. 1.5KW from a bank of deep cycle batteries ?

Juan,
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.

Use an identical circuit for pins 13 and 14 of the SG3524.



Download Attachment: FET driver.GIF
3.44 KB

Go to Top of Page

JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts

Posted - Sep 15 2011 :  7:01:11 PM  Show Profile  Reply with Quote
quote:
Originally posted by pebe

quote:
What value of complementary emitter-follower BJTs can you suggest me to use that can be added to the Driver circuit in the link that I provided?
(do you think 2N22222A & 2N29072A will enable to drive a lot of Power mosfets to have a power output of approx. 1.5KW from a bank of deep cycle batteries ?


Juan,

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.

Use an identical circuit for pins 13 and 14 of the SG3524.

Download Attachment: FET driver.GIF
3.44†KB






Great! Thanks sir for the circuit. . .

1. How many Power Mosfet do you think in can handle? ( I need approx. 1.5KW OUTPUT from bank of deep cycle batteries )

2. How about the "voltage loss issue" of the SG3524 accdg to audioguru?

3. *10 ohms series resistor will be enough to the gate of the Power Mosfet from PIN 3 of 555 Cmos chip?

Thanks.

juan dela cruz
Penniless INVENTOR

Edited by - JUAN DELA CRUZ on Sep 15 2011 7:08:33 PM
Go to Top of Page

pebe
Nobel Prize Winner

United Kingdom
1078 Posts

Posted - Sep 16 2011 :  12:50:52 PM  Show Profile  Reply with Quote
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 Ė preferably 3, as in the circuit.

The FETs donít 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. Thatís 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 Iím sure that Audioguru can advise you whether 10 ohms is OK.
Go to Top of Page

JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts

Posted - Sep 16 2011 :  1:44:09 PM  Show Profile  Reply with Quote
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 Ė preferably 3, as in the circuit.

The FETs donít 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. Thatís 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 Iím 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
Go to Top of Page

pebe
Nobel Prize Winner

United Kingdom
1078 Posts

Posted - Sep 16 2011 :  4:57:38 PM  Show Profile  Reply with Quote
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.
Go to Top of Page

audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - Sep 16 2011 :  6:58:20 PM  Show Profile  Reply with Quote
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.
Go to Top of Page

pebe
Nobel Prize Winner

United Kingdom
1078 Posts

Posted - Sep 17 2011 :  10:38:27 AM  Show Profile  Reply with Quote
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

Go to Top of Page

JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts

Posted - Sep 19 2011 :  08:39:41 AM  Show Profile  Reply with Quote


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

Edited by - JUAN DELA CRUZ on Sep 19 2011 8:48:20 PM
Go to Top of Page

JUAN DELA CRUZ
Mad Scientist

Philippines
476 Posts

Posted - Sep 19 2011 :  8:53:15 PM  Show Profile  Reply with Quote

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
Go to Top of Page

audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - Sep 19 2011 :  9:09:19 PM  Show Profile  Reply with Quote
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.
Go to Top of Page
Page: of 63 Previous Topic Topic Next Topic  
Previous Page | Next Page
 New Topic  Reply to Topic
 Printer Friendly
Jump To:
Aaron's Homepage Forum © 1995-2020 AARONCAKE.NET Go To Top Of Page
This page was generated in 0.19 seconds. Snitz Forums 2000