Some testing with 30VAC 300VA transformer...
I have made some changes to the device: I put transistor and diode into heatsinks, and added a trimpot to adjust the voltage.
The 150R base resistor heats up very quickly! I changed it to a 11W 150R power resistor and it got up to 70
View attachment 35500
I have made some changes to the device: I put transistor and diode into heatsinks, and added a trimpot to adjust the voltage.
The 150R base resistor heats up very quickly! I changed it to a 11W 150R power resistor and it got up to 70
View attachment 35500
vainajala,
Is the PNP transistor used as a switch? Normally in a PNP use of a transistor as a switch, the value of R2 = Supply Voltage / ( Maximum Current Required / Minimum HFE * 1.3 ) You can find the HFE from your data sheet on the transistor.
Resistor R3 is not essential to the circuit but is generally used for stability and to insure that the transistor switch is completely turned off. This resistor insures that the base of the transistor
does not go slightly negative which would cause a very small amount of collector current to flow. The value of this resistor is not critical but a value about 10 times R2 is normally chosen.
I also noticed that you have connected the collectors (C1 and C2) of the TL494 chip to the base of the PNP transistor instead of connecting the TL494 emitters (E1 and E2) to the PNP. Is this what you meant to do? The data sheet shows C1 and C2 connected to VCC.
Hope this is helpful.
MP
Is the PNP transistor used as a switch? Normally in a PNP use of a transistor as a switch, the value of R2 = Supply Voltage / ( Maximum Current Required / Minimum HFE * 1.3 ) You can find the HFE from your data sheet on the transistor.
Resistor R3 is not essential to the circuit but is generally used for stability and to insure that the transistor switch is completely turned off. This resistor insures that the base of the transistor
does not go slightly negative which would cause a very small amount of collector current to flow. The value of this resistor is not critical but a value about 10 times R2 is normally chosen.
I also noticed that you have connected the collectors (C1 and C2) of the TL494 chip to the base of the PNP transistor instead of connecting the TL494 emitters (E1 and E2) to the PNP. Is this what you meant to do? The data sheet shows C1 and C2 connected to VCC.
Hope this is helpful.
MP
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vainajala,
Try to use a 2.2k for R3. As far as I can see the emitters and collectors of the TL 494 is correctly connected. With the change of R3 the efficiency should improve a bit and the heat in R2 will be reduced. Have you access to a scope? I would be useful to see how Q1 performs and possibly make some minor changes to improve the efficiency.
Ante :
Try to use a 2.2k for R3. As far as I can see the emitters and collectors of the TL 494 is correctly connected. With the change of R3 the efficiency should improve a bit and the heat in R2 will be reduced. Have you access to a scope? I would be useful to see how Q1 performs and possibly make some minor changes to improve the efficiency.
Ante :
audioguru2
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Vainajala,
The 150 ohm resistor heats because it has pulses of nearly 40V across it. ON Semi's application circuit shows it being used with an input voltage of only 10V, where 150 ohms will dissipate an average of only about 1/4W. You have a very high supply voltage.
The data sheet for your MJ15004 does not spec hFE beyond a collector current of only 5A, where its hFE is guaranteed to be only 25, and will be much less at higher currents. Even a 150 ohm resistor won't supply it with enough base current for only 5A collector current. I think that a lower value (and higher power rating) base resistor is needed, or else a darlington arrangement. A power MOSFET would be perfect here.
TI (who invented the TL494), have a teriffic application note showing a complimentary darlington output stage that is operating safely from a 32V supply and outputting 10A. Its base resistor is reasonable. This note is here:
http://www-s.ti.com/sc/psheets/slva001b/slva001b.pdf
The 150 ohm resistor heats because it has pulses of nearly 40V across it. ON Semi's application circuit shows it being used with an input voltage of only 10V, where 150 ohms will dissipate an average of only about 1/4W. You have a very high supply voltage.
The data sheet for your MJ15004 does not spec hFE beyond a collector current of only 5A, where its hFE is guaranteed to be only 25, and will be much less at higher currents. Even a 150 ohm resistor won't supply it with enough base current for only 5A collector current. I think that a lower value (and higher power rating) base resistor is needed, or else a darlington arrangement. A power MOSFET would be perfect here.
TI (who invented the TL494), have a teriffic application note showing a complimentary darlington output stage that is operating safely from a 32V supply and outputting 10A. Its base resistor is reasonable. This note is here:
http://www-s.ti.com/sc/psheets/slva001b/slva001b.pdf
audioguru2
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Ante,
All the big ordinary transistors have a guaranteed gain of only 5 at 10A collector current. Expensive ones (they select them) have only a little more gain, 10 if you're lucky.
Darlingtons don't fully saturate and therefore will need a big heatsink.
So I believe that the best thing to do is put 3 or 4 ordinary (MJ2955) transistors in parallel and benefit from their much improved gain (maybe 40) at only a few amps of collector current each. They will saturate well and won't need much heatsink. Their base resistor will be reasonable because of their high gain. ;D ;D
All the big ordinary transistors have a guaranteed gain of only 5 at 10A collector current. Expensive ones (they select them) have only a little more gain, 10 if you're lucky.
Darlingtons don't fully saturate and therefore will need a big heatsink.
So I believe that the best thing to do is put 3 or 4 ordinary (MJ2955) transistors in parallel and benefit from their much improved gain (maybe 40) at only a few amps of collector current each. They will saturate well and won't need much heatsink. Their base resistor will be reasonable because of their high gain. ;D ;D
First of all, thanks for your help, it's surely needed
Audioguru,
What do you think about using three MJE2955's in parallel, with 0,1R emitter resistor in each? By the way, are the "equalizing resistors" needed in switching applications?
As far as I can understand in this transistor's datasheet, the HFE=~40 at IC of 3A. So with three transistors, max. 9A could be drawn with HFE ~40.
So base resistors: 45V / (3A / 40 * 1,3) =460 ohms, and wattage: P = 0,075A2 * 460 ohms = 2,58W.
Am I correct?
9A would be actually more than enough, the circuit's current limit start at ~7,5A...
MOSFET(s) might be good idea. Could P-channel fets be easily used in this circuit? IRF9540 or IRF5305 maybe? ???
MJE2955: http://www.onsemi.com/site/products/summary/0,4450,MJE2955T,00.html
View attachment 35502
Audioguru,
What do you think about using three MJE2955's in parallel, with 0,1R emitter resistor in each? By the way, are the "equalizing resistors" needed in switching applications?
As far as I can understand in this transistor's datasheet, the HFE=~40 at IC of 3A. So with three transistors, max. 9A could be drawn with HFE ~40.
So base resistors: 45V / (3A / 40 * 1,3) =460 ohms, and wattage: P = 0,075A2 * 460 ohms = 2,58W.
Am I correct?
9A would be actually more than enough, the circuit's current limit start at ~7,5A...
MOSFET(s) might be good idea. Could P-channel fets be easily used in this circuit? IRF9540 or IRF5305 maybe? ???
MJE2955: http://www.onsemi.com/site/products/summary/0,4450,MJE2955T,00.html
View attachment 35502
audioguru2
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Vainajala,
I just now noticed that your very high supply voltage will be putting more than 50V on the TL496's output transistors (42V absolute max. rated) when your SMPS has no load!
When paralleling the output transistors, connect their bases together, connect the TL494's output transistors together and use a single base resistor. Also use a single 15 ohm resistor from the bases to the + supply. The emitter resistors may not be important if you don't match the transistors, read on.
Your calculation for the base resistor is close. The voltage across the base resistor is 45V - 1.5V (max. Vbe of MJE2955) - 1.3V (max. saturation V of TL494 outputs = 42.2V.
Therefore the base resistor is 42.2V divided by [9A divided by 25 (min. hFE of MJE2955 at Ic 3A)] = 117 ohms. I don't use the 1.3 correction term because TI also doesn't. Using 120 ohms, its 352mA is within the max. recommended 400mA of the TL494's paralleled outputs. Its power dissipation will be 42.2V X 352mA = 14.9W full-time, or less than 7W using the SMPS's PWM. Note that many power resistors get extremely hot at their power rating (what temp. does nichrome melt?).
If the output current is reduced to 5A and a 4th MJE2955 is used, then the base resistor can be about 270 ohms, with much less power dissipation.
Ante,
A darlington transistor's output transistor cannot have a saturation voltage that is less than its Vbe, which for an MJ2955 could be as high as 2.5V. By itself, its saturation voltage is half that. So a darlington dissipates twice the heat.
The argument about emitter resistors with paralleled transistors continues. In this circuit, they are not necessary since the transistor with the most gain can hog the whole current without damage. Also, as its collector current increases, its gain decreases down to the gain of the least-gain transistor. Also, the least-gain transistor will conduct less current, so its gain will be more. Therefore they will share the current, but not equally.
I just now noticed that your very high supply voltage will be putting more than 50V on the TL496's output transistors (42V absolute max. rated) when your SMPS has no load!
When paralleling the output transistors, connect their bases together, connect the TL494's output transistors together and use a single base resistor. Also use a single 15 ohm resistor from the bases to the + supply. The emitter resistors may not be important if you don't match the transistors, read on.
Your calculation for the base resistor is close. The voltage across the base resistor is 45V - 1.5V (max. Vbe of MJE2955) - 1.3V (max. saturation V of TL494 outputs = 42.2V.
Therefore the base resistor is 42.2V divided by [9A divided by 25 (min. hFE of MJE2955 at Ic 3A)] = 117 ohms. I don't use the 1.3 correction term because TI also doesn't. Using 120 ohms, its 352mA is within the max. recommended 400mA of the TL494's paralleled outputs. Its power dissipation will be 42.2V X 352mA = 14.9W full-time, or less than 7W using the SMPS's PWM. Note that many power resistors get extremely hot at their power rating (what temp. does nichrome melt?).
If the output current is reduced to 5A and a 4th MJE2955 is used, then the base resistor can be about 270 ohms, with much less power dissipation.
Ante,
A darlington transistor's output transistor cannot have a saturation voltage that is less than its Vbe, which for an MJ2955 could be as high as 2.5V. By itself, its saturation voltage is half that. So a darlington dissipates twice the heat.
The argument about emitter resistors with paralleled transistors continues. In this circuit, they are not necessary since the transistor with the most gain can hog the whole current without damage. Also, as its collector current increases, its gain decreases down to the gain of the least-gain transistor. Also, the least-gain transistor will conduct less current, so its gain will be more. Therefore they will share the current, but not equally.
audioguru2
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Ante,
The darlington transistor that you posted has a max. REVERSE voltage rating of 8V, when its eb junctions have avalanche breakdown like a zener, with damaging results if the current is not limited. Its forward Vbe for BOTH transistors is 2.5V max. at 10A Ic.
I was talking about the Vbe for just the output transistor of a darlington, since when the input transistor conducts, it shorts the output transistor's base to its collector. So the output transistor's saturation voltage cannot be less than its Vbe.
In this darlington, its max. output saturation voltage is 2.0V at 10A Ic, with a whopping 0.5A of base drive. A single high-current transistor can saturate with only 1.0V or less (but not a 2N3055, it is lousy at high currents, I just looked at its 3.0V max. saturation voltage at 10A Ic). Also, look at this darlington's saturation voltage at low currents, it is always more than a base-emitter voltage drop of about 0.75V. A single transistor saturates with less than 0.1V at low currents.
Transistor manufacturers have a wide spread for hFE values, so that they can still sell the lousy ones. If they could make them all the same, then they would advertise it on their data sheet.
Look at the 40 to 400 hFE spread on your posted darlington, and even a 2N3055 has a spread of hFE from 20 to 70. The spreads could be much more because they don't spec. a max. hFE. If they can't sell the higher hFE ones at a higher price, then they throw them in with these "ordinary" ones. Since you don't know how much hFE that you are getting, then design the base drive for the worst one and hope that it conducts at least some of the load current. Or use emitter resistors to equallize the gains of paralleled transistors.
The darlington transistor that you posted has a max. REVERSE voltage rating of 8V, when its eb junctions have avalanche breakdown like a zener, with damaging results if the current is not limited. Its forward Vbe for BOTH transistors is 2.5V max. at 10A Ic.
I was talking about the Vbe for just the output transistor of a darlington, since when the input transistor conducts, it shorts the output transistor's base to its collector. So the output transistor's saturation voltage cannot be less than its Vbe.
In this darlington, its max. output saturation voltage is 2.0V at 10A Ic, with a whopping 0.5A of base drive. A single high-current transistor can saturate with only 1.0V or less (but not a 2N3055, it is lousy at high currents, I just looked at its 3.0V max. saturation voltage at 10A Ic). Also, look at this darlington's saturation voltage at low currents, it is always more than a base-emitter voltage drop of about 0.75V. A single transistor saturates with less than 0.1V at low currents.
Transistor manufacturers have a wide spread for hFE values, so that they can still sell the lousy ones. If they could make them all the same, then they would advertise it on their data sheet.
Look at the 40 to 400 hFE spread on your posted darlington, and even a 2N3055 has a spread of hFE from 20 to 70. The spreads could be much more because they don't spec. a max. hFE. If they can't sell the higher hFE ones at a higher price, then they throw them in with these "ordinary" ones. Since you don't know how much hFE that you are getting, then design the base drive for the worst one and hope that it conducts at least some of the load current. Or use emitter resistors to equallize the gains of paralleled transistors.
About darlingtons...
I tried a darlington transistor configuration with dual MJ15004 transistors. 55W halogen served as a load, again. Unfortunately I didn't measure the performance of that circuit (Iwas in bit of a hurry...), but the output-transistor got very hot (i almost burn my finger when touched it :)
I experimented some different base resistors from 330R to 10K with little or no impact on "heating-effect".
I see you're discussing about darlingtons, well how about Sziklai configuration (also known as the complementary Darlington)? I suppose that the Sziklai-type has lower saturation voltage when compared with ordinary Darlington... I found this information from some finnish electronics website. Text was in finnish, so I put a translation in a form of a picture
Some more Sziklai info here:
http://www.ampslab.com/c200cfp.htm
View attachment 35508
I tried a darlington transistor configuration with dual MJ15004 transistors. 55W halogen served as a load, again. Unfortunately I didn't measure the performance of that circuit (Iwas in bit of a hurry...), but the output-transistor got very hot (i almost burn my finger when touched it :
)I experimented some different base resistors from 330R to 10K with little or no impact on "heating-effect".
I see you're discussing about darlingtons, well how about Sziklai configuration (also known as the complementary Darlington)? I suppose that the Sziklai-type has lower saturation voltage when compared with ordinary Darlington... I found this information from some finnish electronics website. Text was in finnish, so I put a translation in a form of a picture
Some more Sziklai info here:
http://www.ampslab.com/c200cfp.htm
View attachment 35508
audioguru2
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Vainajala, Ante,
In order to keep the base resistor cool, try a PNP darlington transistor, like a TIP145T, which is rated at 60V. But the 100V rated TIP147T is cheaper, with the same specs. Their data sheet is here:
http://www.fairchildsemi.com/ds/TI/TIP147T.pdf
With a 10A collector current, and only 40mA of base drive, its max. saturation voltage is 3.0V. Its base resistor voltage is calculated to be 45V (supply) - 3.5V (max. Vbe saturated) - 0.2V (TL494 saturated) = 41.3V. The base resistor is calculated to be 41.3V divided by 40mA = 1033 ohms. Using 1000 ohms, its power dissipation is 1.7W continuous, or less than 0.85W with SMPS PWM.
If you are unlucky and get a TIP147T that doesn't saturate well, then it will dissipate about 15W, so will need a medium-sized heatsink.
In order to keep the base resistor cool, try a PNP darlington transistor, like a TIP145T, which is rated at 60V. But the 100V rated TIP147T is cheaper, with the same specs. Their data sheet is here:
http://www.fairchildsemi.com/ds/TI/TIP147T.pdf
With a 10A collector current, and only 40mA of base drive, its max. saturation voltage is 3.0V. Its base resistor voltage is calculated to be 45V (supply) - 3.5V (max. Vbe saturated) - 0.2V (TL494 saturated) = 41.3V. The base resistor is calculated to be 41.3V divided by 40mA = 1033 ohms. Using 1000 ohms, its power dissipation is 1.7W continuous, or less than 0.85W with SMPS PWM.
If you are unlucky and get a TIP147T that doesn't saturate well, then it will dissipate about 15W, so will need a medium-sized heatsink.
I decided to test N-channel MOSFET as a switch transistor, and it worked well! Typical efffiency was 77% with 21W and ~74% with 55W.
The heatsink was so small that the FET heated pretty quick with 55W and higher loads, so effiency in these loads is more "approximate".
I had to make modification to circuit to drive the FET: I tied the emitters to the FET's gate (instead of collectors) and collectors to Vcc.
This is very promising result, and at least using FETs, the base resistors can be eliminated. So I'm considering to use two or three N-ch FETs in parallel instead of BJTs...
NTP45N06L N-ch FET
http://www.onsemi.com/site/products/summary/0,4450,NTP45N06L,00.html
The heatsink was so small that the FET heated pretty quick with 55W and higher loads, so effiency in these loads is more "approximate".
I had to make modification to circuit to drive the FET: I tied the emitters to the FET's gate (instead of collectors) and collectors to Vcc.
This is very promising result, and at least using FETs, the base resistors can be eliminated. So I'm considering to use two or three N-ch FETs in parallel instead of BJTs...
NTP45N06L N-ch FET
http://www.onsemi.com/site/products/summary/0,4450,NTP45N06L,00.html
audioguru2
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Vainajala,
A darlington transistor does NOT saturate with 2 Vbe voltage drops, when used in a common-emitter configuration that your project needs. Your article refers to the darlington being used in a common-collector (emitter-follower) configuration, where it does have 2 Vbe voltage drops. The Sziklai connection has a single Vbe voltage drop in either configuration, like a darlington does as a common-emitter.
The output transistor of your home-made darlington heated too much because maybe it didn't have a base-emitter resistor to turn it off quickly.
That's good that you got a MOSFET to work well. It sounds like you are using the TL494's transistors as emitter-followers which drive the MOSFET as a source-follower. That configuration has an efficiency loss that is due to the MOSFET's gate-source turned-on voltage.
A P-channel MOSFET will be much more efficient when used in a common-source configuration, with its gate driven though a voltage-divider (so that its max. gate-source voltage rating is not exceeded) from the collectors of the TL494 grounded-emitters transistors. If its Rds is low enough, then maybe it won't even need a heatsink. The voltage-divider could have a zener in series with the collectors to protect them from over-voltage. The voltage-divider resistors must have a fairly low resistance to charge/discharge the high gate capacitance quickly.
A darlington transistor does NOT saturate with 2 Vbe voltage drops, when used in a common-emitter configuration that your project needs. Your article refers to the darlington being used in a common-collector (emitter-follower) configuration, where it does have 2 Vbe voltage drops. The Sziklai connection has a single Vbe voltage drop in either configuration, like a darlington does as a common-emitter.
The output transistor of your home-made darlington heated too much because maybe it didn't have a base-emitter resistor to turn it off quickly.
That's good that you got a MOSFET to work well. It sounds like you are using the TL494's transistors as emitter-followers which drive the MOSFET as a source-follower. That configuration has an efficiency loss that is due to the MOSFET's gate-source turned-on voltage.
A P-channel MOSFET will be much more efficient when used in a common-source configuration, with its gate driven though a voltage-divider (so that its max. gate-source voltage rating is not exceeded) from the collectors of the TL494 grounded-emitters transistors. If its Rds is low enough, then maybe it won't even need a heatsink. The voltage-divider could have a zener in series with the collectors to protect them from over-voltage. The voltage-divider resistors must have a fairly low resistance to charge/discharge the high gate capacitance quickly.
Audioguru, thanks for information! I must try this when I have more time for this project...
Ante, maybe the MAX626/627 driver could be used to drive the dual P-channel FETs then? And, should I get inverting (MAX626) or non-inverting (MAX627) driver?
Also the max. 18VDC supply must be arranged for the MAX62*, that's one concern too.
I'll maybe use 2-3 of IRF9540N FETs...
View attachment 35518
Ante, maybe the MAX626/627 driver could be used to drive the dual P-channel FETs then? And, should I get inverting (MAX626) or non-inverting (MAX627) driver?
Also the max. 18VDC supply must be arranged for the MAX62*, that's one concern too.
I'll maybe use 2-3 of IRF9540N FETs...
View attachment 35518
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