Testing the TK Tar Baby

[Groundloop]

@GL: thanks for running that test.

What do you think would happen if you used 72 volts instead of 24 volts, leaving everything else the same?

TK,

The amount of heat generate in the MOSFET is also depended on the duty cycle.

So if I run at 72 Volt with a duty cycle of 12.2% on, then Amp =  Volt / Rtotal = 72 Volt / 11,85 Ohm = 6 Ampere * 12.2/100 = 0,732 Ampere.
So the Ptotal in the MOSFET will be (0,732 * 0,732) * 1,6 = 0,8573184 Watt.

So nothing bad will happen.

PS: I'm getting a cold with fever so my math may be wrong, can't think straight right now.

>>>leaving everything else the same?

With 100% ON time the heating in the MOSFET will exceed the SOA of the transistor.

GL.

[TinselKoala]

TK,

The amount of heat generate in the MOSFET is also depended on the duty cycle.

So if I run at 72 Volt with a duty cycle of 12.2% on, then Amp =  Volt / Rtotal = 72 Volt / 11,85 Ohm = 6 Ampere * 12.2/100 = 0,732 Ampere.
So the Ptotal in the MOSFET will be (0,732 * 0,732) * 1,6 = 0,8573184 Watt.

So nothing bad will happen.

PS: I'm getting a cold with fever so my math may be wrong, can't think straight right now.

>>>leaving everything else the same?

With 100% ON time the heating in the MOSFET will exceed the SOA of the transistor.

GL.

I'm sorry to hear you aren't feeling well.

I think the problem goes like this, though.

Your Rtotal is 11.85 Ohms, your voltage is 72 V and so your current during the "on" state is 6 amps. And the mosfet has Rdss of 1.6 Ohms. This means that your instantaneous power dissipating in the mosfet is I^2R, or (6x6)x1.6, or 57.6 Watts. Right? For as long as the current is flowing.

But this instantaneous power is only happening for 12.2 percent of the time ON AVERAGE, but it certainly is happening when it happens.

So NOW is when the duty cycle is applied to the figure: 57.6 Watts x 0.122 = a bit over 7 Watts average dissipation at the mosfet. It will get quite warm. And... if the _period_ of the pulsations is long, even with the short duty cycle, there will still be plenty of time at 57.6 watts for the mosfet to heat significantly. That this is where the duty cycle comes in should be clear if you imagine very long periods. Imagine, say, a 2.5 MINUTE period at 12.2 percent ON. That means that the mosfet is ON, carrying 6 amps and dissipating 57.6 Watts, for 12.2 percent of 150 seconds, or almost twenty seconds. (Grab hold of a 60 Watt light bulb and turn it on for 20 seconds and see if it gets hot or not.) Then it gets two minutes and ten seconds to cool.... then another 20 seconds at 57.6 Watts. Right? So you were applying the duty cycle at the wrong place in the calculation, I think.

If you have time, 72 volts, and a spare mosfet, perhaps you'd like to do the obvious experiment. I will be doing it myself, tomorrow, probably.

Take it easy, drink plenty of your favourite fluids and try to rest. I hope you feel better soon.

--TK

(Also... you only know your Rdss to two digits of precision... so really you don't know your answer to seven digits of precision, do you? What you are saying when you give a number like "0,8573184 Watt" is that the answer isn't 0,8573183 Watt, nor is it 0,8573185 Watt.... in other words, even if the math is right, the degree of precision is wrong so that makes the answer "wrongish". However, if the TRUE value is something like 0,8573184 and you give the answer as 0,86   -- since you only know the Rds as 1.6, not 1.600342 for example .... you will be more correct technically because the error in your Rds precision -- the least precise value in your calculation -- determines the error in your overall calculation.)

[TinselKoala]

@GL: so... if you are still up, why do you think that they removed one battery, leaving only 48 volts, during the second half of the video demo, the high heat mode?

I don't think this has ever been explained by the NERDs but I think it is very important in analyzing the behaviour of the circuit and the claims made for it.

[Groundloop]

I'm sorry to hear you aren't feeling well.

I think the problem goes like this, though.

Your Rtotal is 11.85 Ohms, your voltage is 72 V and so your current during the "on" state is 6 amps. And the mosfet has Rdss of 1.6 Ohms. This means that your instantaneous power dissipating in the mosfet is I^2R, or (6x6)x1.6, or 57.6 Watts. Right? For as long as the current is flowing.

But this instantaneous power is only happening for 12.2 percent of the time ON AVERAGE, but it certainly is happening when it happens.

So NOW is when the duty cycle is applied to the figure: 57.6 Watts x 0.122 = a bit over 7 Watts average dissipation at the mosfet. It will get quite warm. And... if the _period_ of the pulsations is long, even with the short duty cycle, there will still be plenty of time at 57.6 watts for the mosfet to heat significantly. That this is where the duty cycle comes in should be clear if you imagine very long periods. Imagine, say, a 2.5 MINUTE period at 12.2 percent ON. That means that the mosfet is ON, carrying 6 amps and dissipating 57.6 Watts, for 12.2 percent of 150 seconds, or almost twenty seconds. (Grab hold of a 60 Watt light bulb and turn it on for 20 seconds and see if it gets hot or not.) Then it gets two minutes and ten seconds to cool.... then another 20 seconds at 57.6 Watts. Right? So you were applying the duty cycle at the wrong place in the calculation, I think.

If you have time, 72 volts, and a spare mosfet, perhaps you'd like to do the obvious experiment. I will be doing it myself, tomorrow, probably.

Take it easy, drink plenty of your favourite fluids and try to rest. I hope you feel better soon.

--TK

(Also... you only know your Rdss to two digits of precision... so really you don't know your answer to seven digits of precision, do you? What you are saying when you give a number like "0,8573184 Watt" is that the answer isn't 0,8573183 Watt, nor is it 0,8573185 Watt.... in other words, even if the math is right, the degree of precision is wrong so that makes the answer "wrongish". However, if the TRUE value is something like 0,8573184 and you give the answer as 0,86   -- since you only know the Rds as 1.6, not 1.600342 for example .... you will be more correct technically because the error in your Rds precision -- the least precise value in your calculation -- determines the error in your overall calculation.)

TK,

I looked at you math and it looks right to me. So when the MOSFET reaches 50 degrees Celsius in my
case, then I'm starting to go outside the SOA of my transistors capability to transfer current.
The Maximum Power Dissipation (MPD) the MOSFET can handle is 190 Watt. You will need a really
good heat sink to get rid of all those Watts. If you allow the transistor to heat up just a little bit, then
you are in trouble. I know about precision, and I agree, all my digits is a copy paste from the PC calculator,
does not mean much in real life.

GL.

[MileHigh]

Rosemary:

When are you going to start your new round of testing?  I can tell you a 'wish list' item that I think a lot of people would like to see:

Do whatever you have to do to do the dim bulb testing first.  You know the drill.  Get small batteries, set up your preferred oscillation mode and confirm it with your DSO.  Then determine how long you want to run the setup starting with fresh batteries.  Run the setup as long as required and then do the dim bulb tests.

That should your first and foremost goal, the dim bulb testing.

After that feel free to record DSO captures to your heart's content.

We really do not want to wait months while you do whatever you want to do.  If you run the dim bulb test as a hasty afterthought after a few months worth of DSO captures we will all be extremely disappointed.

MileHigh