# Not really a call for help, more than a confirmation/debunk of my work

#### Morten Andersen

Joined Jun 7, 2019
25
Hello all.
Circuit drawing / calculations attached.
I'm in the process of bettering myself for my schoolwork, and has therefore started doing some circuit solving assignment outside class.
I've meet an assignment which has me calculating a resistor size R3 to give a specific wattage over an IRF530 mosfet transistor.
My conclusion is that the Vref is similar to the voltage drop (not sure this is the correct term, as i'm not primary english speaking) over R2 due to the feedback from the amplifier, and as the wanted power in the IRF530 is 10W the desired voltage for Vref/VR2 should be 1V as there then will be a current of 10A due to R2 being 0.1 ohm - and from that i use the voltage division formula to calculate R3.
My question is: is my calculation/understanding of this correct?

#### Jony130

Joined Feb 17, 2009
5,089
So you want to dissipate a 10W via the MOSFET or via R2 resistor? Because the transistor power dissipation is P = Vds*Id = (20V - 1V)*10A = 190W

#### Morten Andersen

Joined Jun 7, 2019
25
i want to dissipate 10W in the MOSFET. i thought that the wattage at the resistor R2 would be the same as in the MOSFET, but seems like im missing something?

#### BobTPH

Joined Jun 5, 2013
2,108
Look at your equation for Vref again.

From your calculation of 50Ω for R3, I get 4V for Vref. Which means the opamp (at 5V) cannot turn on the MOSFET. Which means it does not blow up, like it would if the calculation was correct.

Bob

• Morten Andersen

#### sagor

Joined Mar 10, 2019
122
Basic design flaws:
1) IRF530 needs 10V on gate to be fully on. While 5V may turn it on partially, the RdsON will be high, producing more heat in the MOSFET
2) R2 will take the brunt of the current. When IRF530 is fully on with 10V on gate, you will get full 20V across the MOSFET (RdsON = 0.16 ohms) and the 0.1 ohm resistor. Total resistance = 0.26 ohms. That will give almost 77A current.
Using P= I^2xR, you get 1538W of power dissipation. Everything will go up in smoke most likely.

To get 10W in the MOSFET at full "ON" (10V on gate), you work backwards. 10W = I^2 x 0.16 ohms. Works to 7.9A
To get 7.9A total current, you need RdsOn + R2 = 2.53 ohms total. Thus, R2 = 2.53 - 0.16, meaning R2 = 2.37 ohms. R2 would have to be a minimum 150W resistor

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#### Jony130

Joined Feb 17, 2009
5,089
Basic design flaws:
But in this circuit, the MOSFET will never by working as a "fully open" switch (linear/triode region). In this circuit, the MOSFET is working in the saturation region as voltage controlled current source.

• Morten Andersen

#### WBahn

Joined Mar 31, 2012
24,971
Hello all.
Circuit drawing / calculations attached.
I'm in the process of bettering myself for my schoolwork, and has therefore started doing some circuit solving assignment outside class.
I've meet an assignment which has me calculating a resistor size R3 to give a specific wattage over an IRF530 mosfet transistor.
My conclusion is that the Vref is similar to the voltage drop (not sure this is the correct term, as i'm not primary english speaking) over R2 due to the feedback from the amplifier, and as the wanted power in the IRF530 is 10W the desired voltage for Vref/VR2 should be 1V as there then will be a current of 10A due to R2 being 0.1 ohm - and from that i use the voltage division formula to calculate R3.
My question is: is my calculation/understanding of this correct?
You need to break the problem down in pieces and go from what you know to determine what you need.

First, assume that the gate voltage is correct -- you don't know what that correct voltage is (yet), but just assume that whatever it needs to be, it is.

Now consider that the drain-source path of the MOSFET is in series with R2, meaning that they have the same current in them and you know that the sum of the voltage drops across them has to always add up to 20 V.

Can you determine what the voltage across the MOSFET is as a function of the current through it?

Can you determine what the power dissipated in the MOSFET is as a function of the current through it?

If so, then can you determine what current you need to have in the MOSFET in order to get it to dissipate 10 W and, from that, what the voltage across it needs to be?

Get that far and then take you best shot at figuring out what the gate-source voltage needs to be? Is that compatible with running your opamp from a 5 V supply?

• Morten Andersen

#### crutschow

Joined Mar 14, 2008
23,758
i want to dissipate 10W in the MOSFET. i thought that the wattage at the resistor R2 would be the same as in the MOSFET, but seems like im missing something?
Yes, you are missing something.
Why do you think that?
The wattage of R2 has nothing to do with the MOSFET wattage.
R2's dissipation is its resistance times the square of the current through it.
The MOSFETs dissipation is the voltage across it times the current through it.

• Morten Andersen

#### Morten Andersen

Joined Jun 7, 2019
25
I've made some simplifications to the circuit and just want you guys opinion on how the new setup looks, and if my new calculations is correct.
As it's my understanding that the MOSFET is fully open with 10V on gate, i've made a simple amplifier setup that converts 1V to 11V. And I found in the IRF530 datasheet that the MOSFETs resistance is 0.16ohm at fully open. So how does my new calculation look? is the 47,47W over the MOSFET correct?

#### djsfantasi

Joined Apr 11, 2010
5,837
How are you planning on getting 11V output with an op amp running with 5V?

From your diagram it appears that your creating voltage out of thin air.

#### crutschow

Joined Mar 14, 2008
23,758
I'm curious.
TS's on this forum often say a transistor is "open" when it is fully-on, which in not usual for electrical nomenclature.
Where does this designation come from?
For example, contacts are "closed" when they conduct.
Is the reference to something like a mechanical "gate" which is open when things pass through?

#### djsfantasi

Joined Apr 11, 2010
5,837
I'm curious.
TS's on this forum often say a transistor is "open" when it is fully-on, which in not usual for electrical nomenclature.
Where does this designation come from?
For example, contacts are "closed" when they conduct.
Is the reference to something like a mechanical "gate" which is open when things pass through?
I noticed that as well. My first interpretation was that the transistor was not conducting (C to E). Then it appeared in his context, that was the opposite of what he meant. I use conducting and open (or not conducting). And I’ll often ask for clarification.

#### BobTPH

Joined Jun 5, 2013
2,108
I think it comes from that damned water analogy.

Transistor is a valve which is open to allow charge to flow and closed to stop it.

Bob

#### BobTPH

Joined Jun 5, 2013
2,108
I think it is time to step back and restate the problem. Forget about your circuit, just tell us what you are trying to accomplish.

You seem to be saying you want to come up with a circuit that results in 10W of dissipation in a MOSFET.

This does not make a lot of sense unless you are trying to evaluate different heat sinks.

Bob

#### WBahn

Joined Mar 31, 2012
24,971
I think it is time to step back and restate the problem. Forget about your circuit, just tell us what you are trying to accomplish.

You seem to be saying you want to come up with a circuit that results in 10W of dissipation in a MOSFET.

This does not make a lot of sense unless you are trying to evaluate different heat sinks.

Bob
Like many academic problems, this one is contrived in order to force the student to apply analytical skills to a problem that brings into it different factors, such as power into a problem where they are trying to find a voltage.

#### WBahn

Joined Mar 31, 2012
24,971
I've made some simplifications to the circuit and just want you guys opinion on how the new setup looks, and if my new calculations is correct.
As it's my understanding that the MOSFET is fully open with 10V on gate, i've made a simple amplifier setup that converts 1V to 11V. And I found in the IRF530 datasheet that the MOSFETs resistance is 0.16ohm at fully open. So how does my new calculation look? is the 47,47W over the MOSFET correct?
It's better to say "fully on" or "fully conducting". In electronics, an "open" generally means an open circuit meaning that there is a break such that little to no current can flow, while "closed" generally means the opposite (such as a closed switch).

In your original post you said that the assignment was to find a value for R3 that results in 10 W being dissipated in the MOSFET. So, no, any solution that results in 47.47 W being dissipated in it clearly can't be correct.

Also, putting 10 V on the gate doesn't mean anything at all. It is the voltage DIFFERENCE between the gate and source, Vgs, that counts. If you put 10 V on the gate but the source is also at 10 V, then you have a Vgs of zero volts.

What, EXACTLY, is the problem statement that you are trying to solve?

#### Morten Andersen

Joined Jun 7, 2019
25
Alright. the original assignment is the picture below. The information given is that the voltage over Vref is equal to the voltage over R2.
The point of the assignment is to find the value of R3, so that you get the desired wattage over IRF530, and then by changing the size of R3 you should get different wattage use of the IRF530. i'm trying to determine how to set up the calculations, so it's understandable for myself.

#### BobTPH

Joined Jun 5, 2013
2,108
Is that the original statement, or are you paraphrasing or translating it?

Because, as stated, its solution must be a function of Vdd, not a number.

I am skeptical that this is stated correctly because one does not say
“the power across a MOSFET”. There is a voltage across the MOSFET and a current through it. One can talk about the power dissipated in the MOSFET, and that is the product of the voltage and current.

We cannot come up with a number for R3 that will dissipate 10W in the MOSFET because it would depend on Vdd, which can be 10 to 30V.

Bob

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#### The Electrician

Joined Oct 9, 2007
2,751
We cannot come up with a number for R3 that will dissipate 10W in the MOSFET because it would depend on Vdd, which can be 10 to 30V.
Bob
We may not be able to come up with a pure number, but we can come with an expression with Vdd left as a literal variable.

Morten Andersen, WBahn gave a viable path to a solution in post #7. Let me make it more explicit. Assume the current through the mosfet and R2 is (i) amps. What is the voltage across R2 if this current passes through it? Let's call that voltage Vr2. Since the voltage across the series combination of the mosfet and R2 is Vdd, what is the voltage across the mosfet? Let's call that voltage Vt1. Knowing i and Vt1, what is the power dissipation in T1?

#### BobTPH

Joined Jun 5, 2013
2,108
Which is exactly what I said earlier in the post:
Because, as stated, its solution must be a function of Vdd, not a number.
Bob