What should expect from this kind of circuit???
A.
B.
A.
An externally hosted image should be here but it was not working when we last tested it.
B.
An externally hosted image should be here but it was not working when we last tested it.
Something for fast switching. Bipolars providing low z drive to the mosfet gates for v-fast on/ off.
At a guess it's part of a switch mode PSU, outputs a&b are probably driven in antiphase and the mosfets connect to opposite ends of a centre-tapped inductor where the tap connects to +V. Where is the circuit from. Only the BD139 can source 12V.
Infact that's a SMPS (push-pull topology).
And will i destroy the IRF if i bypass the transistor block and connect the gate resistor directly to sg3524?
And will i destroy the IRF if i bypass the transistor block and connect the gate resistor directly to sg3524?
I doubt bypassing the BDs will damage the mosfets but the psu will work less efficiently and run hotter (maybe overheat) because the lower current output directly from the SG3524 will not be able to switch the mosfets on and off as fast.
Just for ask! 🙂
Someone tell methat in the diode's bridge there's a 1.414 factor.. what's it exactly?
Thanxa lot.. is incredible to see how there's always someone that give an answer at all my stupid question..
Too if i don't reply to too mych question..
Too if i don't reply to too mych question..
If you bypass the BD139/140 you'll probably destroy the MOSFETs.
Fast switching is important, very important...
Fast switching is important, very important...
Hi MaXiZ.
that's a nice copy from ESP's switchmode power supply.
What do the red voltages mean?
You CAN'T leave out the BD139/BD140, since the SG3524 can't switch the IRF's fast enough.
A too low frequency won't destroy the IRF's, but if the IRF-set from output A is open while the IRF-set from output B is open, you will blow all the IRF's. (if you're lucky, these are the only components that blow).
If you test this circuit, it's important that you test the design this way, that means WITHOUT the transformer. Look at the signals at the output of the A-set of IRF's and the output of the B-set of IRF's at the same time, and make sure the "open"-times of both the sets are NEVER at the same time.
Grtz, Joris
that's a nice copy from ESP's switchmode power supply.
What do the red voltages mean?
You CAN'T leave out the BD139/BD140, since the SG3524 can't switch the IRF's fast enough.
A too low frequency won't destroy the IRF's, but if the IRF-set from output A is open while the IRF-set from output B is open, you will blow all the IRF's. (if you're lucky, these are the only components that blow).
If you test this circuit, it's important that you test the design this way, that means WITHOUT the transformer. Look at the signals at the output of the A-set of IRF's and the output of the B-set of IRF's at the same time, and make sure the "open"-times of both the sets are NEVER at the same time.
Grtz, Joris
Mean only what i should expect from that circuit.. if should have the case A or case B.
If the transformer isn't connected too???
HOW?
Hi MaXiZ,
if the transformer is not connected, then nothing will go wrong!
If you have built the schematic as far as this, then test it the way I described,WITHOUT the IRF's attached:
The measure-points are on the base of the points where the lines split for the 4.7 Ohms resistors.
If this is OK, then measure again with the IRF's attached, and is that's OK also, then attach the transformer and go on from there.
If the transformer is attached, then watch the signal over the primary windings, that should be a wave form (if you watch both windings at the same time, then you should see two square-waves with a dead moment between the rises of each square)
If that's OK, then attach a small load (that the output current is a few milli-amps), and watch the output. From then, just lower the resistance of the load and keep watching the output signal (but watch out for a TO low resistance!)
You can change the dead time by changing the resistance between pin 5 and 7 of the SG3524.
Grtz, Joris
if the transformer is not connected, then nothing will go wrong!
If you have built the schematic as far as this, then test it the way I described,WITHOUT the IRF's attached:
The measure-points are on the base of the points where the lines split for the 4.7 Ohms resistors.
If this is OK, then measure again with the IRF's attached, and is that's OK also, then attach the transformer and go on from there.
If the transformer is attached, then watch the signal over the primary windings, that should be a wave form (if you watch both windings at the same time, then you should see two square-waves with a dead moment between the rises of each square)
If that's OK, then attach a small load (that the output current is a few milli-amps), and watch the output. From then, just lower the resistance of the load and keep watching the output signal (but watch out for a TO low resistance!)
You can change the dead time by changing the resistance between pin 5 and 7 of the SG3524.
Grtz, Joris
If the transformer is not connected, you will also measure nothing! You need some kind of test load, like a resistor of a few 100 Ohms from the FET outputs to the supply, so you can measure voltage switching at the FETs.
Jan Didden
Jan Didden
Indeed Janneman,
if there's no load, there's no current, there's nothing to measure!
@MaXiZ, when the transformer is not connected, and you don't use any load for the IRF's at all, then nothing can go wrong.
If you do use a load for the IRF's, the IRF's blow only when 1) your load has a to low resistance 2) the resistance of the should-be-4.7-ohms-resistors is to low
Grtz, Joris
if there's no load, there's no current, there's nothing to measure!
@MaXiZ, when the transformer is not connected, and you don't use any load for the IRF's at all, then nothing can go wrong.
If you do use a load for the IRF's, the IRF's blow only when 1) your load has a to low resistance 2) the resistance of the should-be-4.7-ohms-resistors is to low
Grtz, Joris
There are some errors in these schematics :
1:
SG3524 IC has two independent output transistors with anti-saturation network and uncommited collectors and emitters
You can configure it to source current when ON and do nothing when OFF [emitter follower] or to sink current when ON and do nothing when OFF [common emitter]
With this IC you must provide adequate turn off means since the IC can´t turn off anything connected to their outputs, this can be done simply by configuring the IC as emitter follower and adding 1K or so resistors from outputs to ground
Current through these resistors will be amplified by PNP transistor ensuring fast turn-off and some inmunity to parasitistic gate turn-on during high dVds/dt pulses [usually when the other side turns on]
--------------------------------------------------------------------------
2:
Using two independent 6.8 ohm resistors causes the B-E junction of the PNP transistor to be reverse biased during gate turn-on and B-E junction of the NPN to be reverse biased during gate turn-off
With more than 6V of reverse biasing you can easily cause reverse breakdon and if excessive current flows in this mode B-E junction will be damaged causing loss of beta or transistor failure with time
If the purpose of using separate resistors [but of different value] was to match turn-on and turn-off times, this also can be done with a resistor in series with a diode, emitters must be connected together
------------------------------------------------------------------------------
3:
Voltage drop in the IC is about 1.4V when its outputs are in high state and configured as emitter followers and you have to add 0.6V of voltage drop from the external NPN so total voltage drop is about 2V
With 12V power supply MOSFET gates will be only charged to 10V making Rds-ON higher tan desirable
The best way to overcome this is to get an auxiliary regulated 20V supply from transformer output or an auxiliary winding and use it to supply control circuit [and MOSFET gates]
With 20V supply you will also have 2V of voltage drop so maximum Vgs will be 18V, 2V less than maximum Vgs rating of most MOSFETS [+-20V] so Rds-on will be the lowest possible
[During start-up a bootstrap diode is needed to get power to control circuit from 12V line]
-------------------------------------------------------------------------------
You can test the circuit with transformer connected as long as it doesn't saturate, and this will happen if : frecuency is too low, power supply to the transformer is too high, MOSFTETs dont switch properly or dead time is too short
A saturated transformer acts as if there were no core so windings are like connecting a piece of wire from + to ground, troublesome high currents will flow
ICs are not perfect and generate asymetric pulses on its outputs so some minimum 5..10% deadtime must be provided in order to help the transformer to balance and reset properly, capacitive coupling of the transformer is another alternative but required capacitor size is only reasonable for high voltage switching or very high frequencies
Oscilloscope is almost mandatory to experiment with this kind of circuits since you have to check for clear waveforms, expected switching frequencies, and to adjust rise and fall times and damp transformer ringing
PD: Sorry for my bad english
1:
SG3524 IC has two independent output transistors with anti-saturation network and uncommited collectors and emitters
You can configure it to source current when ON and do nothing when OFF [emitter follower] or to sink current when ON and do nothing when OFF [common emitter]
With this IC you must provide adequate turn off means since the IC can´t turn off anything connected to their outputs, this can be done simply by configuring the IC as emitter follower and adding 1K or so resistors from outputs to ground
Current through these resistors will be amplified by PNP transistor ensuring fast turn-off and some inmunity to parasitistic gate turn-on during high dVds/dt pulses [usually when the other side turns on]
--------------------------------------------------------------------------
2:
Using two independent 6.8 ohm resistors causes the B-E junction of the PNP transistor to be reverse biased during gate turn-on and B-E junction of the NPN to be reverse biased during gate turn-off
With more than 6V of reverse biasing you can easily cause reverse breakdon and if excessive current flows in this mode B-E junction will be damaged causing loss of beta or transistor failure with time
If the purpose of using separate resistors [but of different value] was to match turn-on and turn-off times, this also can be done with a resistor in series with a diode, emitters must be connected together
------------------------------------------------------------------------------
3:
Voltage drop in the IC is about 1.4V when its outputs are in high state and configured as emitter followers and you have to add 0.6V of voltage drop from the external NPN so total voltage drop is about 2V
With 12V power supply MOSFET gates will be only charged to 10V making Rds-ON higher tan desirable
The best way to overcome this is to get an auxiliary regulated 20V supply from transformer output or an auxiliary winding and use it to supply control circuit [and MOSFET gates]
With 20V supply you will also have 2V of voltage drop so maximum Vgs will be 18V, 2V less than maximum Vgs rating of most MOSFETS [+-20V] so Rds-on will be the lowest possible
[During start-up a bootstrap diode is needed to get power to control circuit from 12V line]
-------------------------------------------------------------------------------
You can test the circuit with transformer connected as long as it doesn't saturate, and this will happen if : frecuency is too low, power supply to the transformer is too high, MOSFTETs dont switch properly or dead time is too short
A saturated transformer acts as if there were no core so windings are like connecting a piece of wire from + to ground, troublesome high currents will flow
ICs are not perfect and generate asymetric pulses on its outputs so some minimum 5..10% deadtime must be provided in order to help the transformer to balance and reset properly, capacitive coupling of the transformer is another alternative but required capacitor size is only reasonable for high voltage switching or very high frequencies
Oscilloscope is almost mandatory to experiment with this kind of circuits since you have to check for clear waveforms, expected switching frequencies, and to adjust rise and fall times and damp transformer ringing
PD: Sorry for my bad english
- Status
- Not open for further replies.