either i am tired or i am doing something wrong...
i have a common N-channel fet RFP50N06
i am wanting to use it as a switch.. it claims to handle 60v @ 50A
it will be used to switch one a 20w halogen lamp
when i breadboard this,, the lamp never lights fully... 12volts in..maybe 9.4v out..
i have it wired as such...
12 volts into the lamp...the lamp to the Drain , the source to ground.
the gate is connected to the output of a 741 op amp.
the op amp is being configured with a CDS fotocell to be used to turn on the fet at night..
where am i going wrong with this.. the op amp circuit is working fine.. i just cannot get full output of the fet..
PS i have even tried to use a BUZ11A and a 2SK2843 both are mosfets.
they are all what i had in my junk boxes...
if these are not good to use,,,what shall i use?
FUSTRATED...transistor question
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dacflyer
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Pffffff , never mind all.... i double checked my work,, i found my mistake,,but maybe someone can explain to me why it matters,,,
i had my lamp between the ground and source and the B+ going directly to the drain...
i reconfiged the set up so that the lamp is between B+ and drain and the source to ground... it works great now..
but i cannot understand what the difference is of where the load is...
i had my lamp between the ground and source and the B+ going directly to the drain...
i reconfiged the set up so that the lamp is between B+ and drain and the source to ground... it works great now..
but i cannot understand what the difference is of where the load is...
Vgs is the difference between the gate voltage and the source voltage. It needs to be at least your mosfets spec of 10V to turn it on fully. When the mosfet is on the high side, (between V+ and the load), you aren't seeing a high enough Vgs. think of your load and the mosfet as 2 resistors in a voltage divider, determine their resistances (when fully on) and calculate the voltage at the mosfet's source. Then consider the voltage you are applying to the gate. The difference is your Vgs which I'd guess is about 3 or 4 volts. You probably need to get the gate voltage to something like 20V or more to get it to work as a high side driver, assuming you have a V+ of 12V.
When the mosfet is a low side driver (i.e. between the load and ground), you can develop the 10V needed and thus fully turn the mosfet on.
If you want a high side driver, use a P-Channel MOSFET.
By the way, I think you are ok with your 20W load but pay attention to the safe operating area chart of the datasheet (fig 9). The manufacturers play very loose with max ratings. That mosfet can't come close to handling 60V at 50A continuous. it looks more like 1.5V at 50A or 2A at 60V. They are all like this. It was kind of a shock to me when I realized just how far off the specs were.
When the mosfet is a low side driver (i.e. between the load and ground), you can develop the 10V needed and thus fully turn the mosfet on.
If you want a high side driver, use a P-Channel MOSFET.
By the way, I think you are ok with your 20W load but pay attention to the safe operating area chart of the datasheet (fig 9). The manufacturers play very loose with max ratings. That mosfet can't come close to handling 60V at 50A continuous. it looks more like 1.5V at 50A or 2A at 60V. They are all like this. It was kind of a shock to me when I realized just how far off the specs were.
Semiconductor manufacturers have a long history of doing this. After awhile you learn what to look for on the data sheet.
The first trick is to rate the device based on its absolute maximum operating temperature at the junction. In this case, that is 175 degrees centigrade. This is unrealistic, and will probably be unreliable. If I were designing a piece of military equipment, I would be asked some very pointed questions if the estimated junction temperature gets much over 110 or 120 degrees centigrade. They have design reviews devoted to asking just this type of question. Higher than that would be considered unreliable unless you have a strong argument to the contrary.
The next trick is to assume that the case or heatsink is at 25 degrees centigrade (77 farenheit) no matter what the power dissipation is. This can be done by using a cold plate cooled by chilled water. But this usually isn't available in most equipment.
This reduces the thermal path to the thickness of the silicon die and a copper base. The data sheet gives the thermal conductance of this path as .877 watts/degree centigrade. If you divide the power rating (131 watts) by the thermal conductance (.877 watts/degree centigrade) you get 149.3 degrees centigrade, which is almost exactly the maximum junction temperature (175 centigrade) minus the case temperature (25 centigrade).
This is all totally misleading, but none of the figures are actually false.
Now look at that 50 amp current rating. Dissipating 131 watts at 50 amps implies a voltage drop of 2.62 volts, which would result from a channel resistance of 52.4 milliohms. But the data sheet says this is only 22 millohms maximum. But look at figure 9. It shows the on resistance increasing by a factor of over 2 when the temperature increases from 25 degrees centigrade to 175 degrees centigrade. And, sure enough, 22 milliohms is about half of 52 milliohms.
A more realistic rating might be junction temperature 125 centigrade, ambient temperature 50 centigrade, and a heatsink with a thermal resistance on the order of 2.5 centigrade/watt. This would give a maximum dissipation of 20.6 watts. The channel resistance would be less than 36 milliohms, so a reasonable maximum current would be around 24 amps. Even here the voltage drop would be nearly a volt, which might be excessive if you needed high efficiency with a low voltage power supply.
The manufacturers have been writing their data sheets in this fashion for nearly fifty years. After a few suprises, you learn to be skeptical about device claims until you understand how they were arrived at.
The first trick is to rate the device based on its absolute maximum operating temperature at the junction. In this case, that is 175 degrees centigrade. This is unrealistic, and will probably be unreliable. If I were designing a piece of military equipment, I would be asked some very pointed questions if the estimated junction temperature gets much over 110 or 120 degrees centigrade. They have design reviews devoted to asking just this type of question. Higher than that would be considered unreliable unless you have a strong argument to the contrary.
The next trick is to assume that the case or heatsink is at 25 degrees centigrade (77 farenheit) no matter what the power dissipation is. This can be done by using a cold plate cooled by chilled water. But this usually isn't available in most equipment.
This reduces the thermal path to the thickness of the silicon die and a copper base. The data sheet gives the thermal conductance of this path as .877 watts/degree centigrade. If you divide the power rating (131 watts) by the thermal conductance (.877 watts/degree centigrade) you get 149.3 degrees centigrade, which is almost exactly the maximum junction temperature (175 centigrade) minus the case temperature (25 centigrade).
This is all totally misleading, but none of the figures are actually false.
Now look at that 50 amp current rating. Dissipating 131 watts at 50 amps implies a voltage drop of 2.62 volts, which would result from a channel resistance of 52.4 milliohms. But the data sheet says this is only 22 millohms maximum. But look at figure 9. It shows the on resistance increasing by a factor of over 2 when the temperature increases from 25 degrees centigrade to 175 degrees centigrade. And, sure enough, 22 milliohms is about half of 52 milliohms.
A more realistic rating might be junction temperature 125 centigrade, ambient temperature 50 centigrade, and a heatsink with a thermal resistance on the order of 2.5 centigrade/watt. This would give a maximum dissipation of 20.6 watts. The channel resistance would be less than 36 milliohms, so a reasonable maximum current would be around 24 amps. Even here the voltage drop would be nearly a volt, which might be excessive if you needed high efficiency with a low voltage power supply.
The manufacturers have been writing their data sheets in this fashion for nearly fifty years. After a few suprises, you learn to be skeptical about device claims until you understand how they were arrived at.
Yeah. however, I see the audio output power ratings as completely different. Component makers market to an educated audience who should know the difference. However, consumers are routinely taken in by these lies. THD is another one.rshayes wrote:They have been doing it for so long that it has become a habit. Probably since the late 1950's when power transistors first came on the market.
A few years later the audio equipment manufacturers started doing the same thing with power output ratings.
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Robert Reed
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Dacflyer
To simplify your original problem- when load was connected source to ground, you in effect had a source follower configuration. This means the source voltage tries to follow the gate voltage within certain limlts. The 9.4 volts you were seeing across the load was probably just following the gate drive voltage minus some losses.
To simplify your original problem- when load was connected source to ground, you in effect had a source follower configuration. This means the source voltage tries to follow the gate voltage within certain limlts. The 9.4 volts you were seeing across the load was probably just following the gate drive voltage minus some losses.
Isn't this what we call experience?
It's never wise to only look at, and base your life on headlines, wherever you see it.
I've seen 'engineers' select critical components from shortforms alone, with interesting results like magic smoke and bad air.
Shortforms and max ratings are only used to sort out unusable components, and then you need to dig into the core of the data. The main fault newbies do, is to add together all max ratings to select a components. The learning curve tends to be steep if high power and/or voltage is involved. It's nothing like the magic smoke to hasten it.
TOK
It's never wise to only look at, and base your life on headlines, wherever you see it.
I've seen 'engineers' select critical components from shortforms alone, with interesting results like magic smoke and bad air.
Shortforms and max ratings are only used to sort out unusable components, and then you need to dig into the core of the data. The main fault newbies do, is to add together all max ratings to select a components. The learning curve tends to be steep if high power and/or voltage is involved. It's nothing like the magic smoke to hasten it.
TOK
Gorgon the Caretaker - Character in a childrens TV-show from 1968.
Personally, I do not find these data sheets misleading at all. You just have to know what you're doing. Any engineer who does not check pertinent data and smokes his parts should get a refund on his degree.
A MOSFET advertised with maximum ratings of 60V and 50A is fine, but you have to realise that it can't do both at the same time. That would cause it to dissipate 60*50= 3 kilowatts. Even the maximum power ratings are advertised with juction temperature limitations. You have to consider the thermal resistance of junction-to-case, case to heat sink thermal resistance using suitable thermal compound, heat sink to still or forced air, and ambient temperature. All of these are listed as Degrees C per Watt. With these you can calculate juction temperature rise with your required power device dissipation. Then you look at the temperature derating curve on the data sheet to find your realistic power level.
As for consumer audio, the power amplifier output power levels were grossly overstated in the late 1960s. That situation changed around 1970 and I thought it was due to legislation that "PIP" watts, "PEAK" watts, "IHF" watts and the like were either outlawed by the FCC or eliminated by industry standards of practise. Sometime around 1990, a new generation of marketers thought up a "new" gimmick and started repeating history because they were in diapers the first time around. Now we have 2 inch computer speakers capable of 200 watts. 300 watt clock radios! Yup, were back to Bullspit Watts.
Bob
A MOSFET advertised with maximum ratings of 60V and 50A is fine, but you have to realise that it can't do both at the same time. That would cause it to dissipate 60*50= 3 kilowatts. Even the maximum power ratings are advertised with juction temperature limitations. You have to consider the thermal resistance of junction-to-case, case to heat sink thermal resistance using suitable thermal compound, heat sink to still or forced air, and ambient temperature. All of these are listed as Degrees C per Watt. With these you can calculate juction temperature rise with your required power device dissipation. Then you look at the temperature derating curve on the data sheet to find your realistic power level.
As for consumer audio, the power amplifier output power levels were grossly overstated in the late 1960s. That situation changed around 1970 and I thought it was due to legislation that "PIP" watts, "PEAK" watts, "IHF" watts and the like were either outlawed by the FCC or eliminated by industry standards of practise. Sometime around 1990, a new generation of marketers thought up a "new" gimmick and started repeating history because they were in diapers the first time around. Now we have 2 inch computer speakers capable of 200 watts. 300 watt clock radios! Yup, were back to Bullspit Watts.
Bob
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