WBahn

Joined Mar 31, 2012
26,398
It seems that everyone is still guessing.

Vbe and Vbc are both 0.7V. Vce is infinite when the transistor is not conducting, but Vce is very close to 0 when the transistor is saturated. The controller uses PWM to maintain the voltage.

On what basis do you claim that Vbe and Vbc are both 0.7V. This is only true if Vce happens to be 1.4V. (Vce = Vcb+Vbe)

Vce is infinite? Huh?

Why would you say that the controller of a LINEAR supply uses PWM?

Perhaps everyone else aren't the only ones guessing.

count_volta

Joined Feb 4, 2009
435
Thank you WBahn, you cleared it up for me. That does actually make sense.

Why do you say that? You just got done saying that the base voltage would always be 0.7V higher than the emitter voltage. That is indepedent of what the collector voltage is. Having said that, "on the same order of magnitude" covers a lot of gound. If Vcc is 15V, then the base voltage could be anywhere between 1.5V and 150V and be within an order of magnitude.

If the collector is 15V, then the output of the circuit is not going to be able to maintain 15V at the load. You need some overhead.

The purpose of the controller, which can be very simple or very complex, is to take a reference signal and a signal related to the actual output being controlled and, based on the difference between the two, produce a change in the controller output signal that moves the output closer to the desired value. In the case of a circuit like this, the controller output value will always be Vout+Vbe. However, if the load increases (resistance goes down), more current must flow through the pass transistor to keep Vout from dropping. But more current means a greater Vbe. Not by much -- a change of only about 20mV will double the collector current. But still, to maintain the same Vout with a higher Vbe requires that the controller output a higher voltage applied to the base.

shortbus

Joined Sep 30, 2009
8,952
Maybe I missed something in this discussion. How is an IGBT, which is clearly shown in the original post, considered a BJT?

Wouldn't the gate of the IGBT need to be PWM for it to work? Applying less than the full turn on gate voltage would cause a high level of heat in the IGBT, wouldn't it?

count_volta

Joined Feb 4, 2009
435
Maybe I missed something in this discussion. How is an IGBT, which is clearly shown in the original post, considered a BJT?

Wouldn't the gate of the IGBT need to be PWM for it to work? Applying less than the full turn on gate voltage would cause a high level of heat in the IGBT, wouldn't it?
Obviously there are differences between a BJT and IGBT, but I meant to understand the general behavior of the circuit its conceptually the same. There is no gate current in an IGBT at steady state (unlike base current in a BJT), but when connected as an emitter follower, whatever appears at the gate will appear at the emitter minus the turn on voltage. Its the same for a Mosfet. At least I think I'm correct.

shortbus

Joined Sep 30, 2009
8,952
Obviously there are differences between a BJT and IGBT, but I meant to understand the general behavior of the circuit its conceptually the same. There is no gate current in an IGBT at steady state (unlike base current in a BJT), but when connected as an emitter follower, whatever appears at the gate will appear at the emitter minus the turn on voltage. Its the same for a Mosfet. At least I think I'm correct.
When there is no current in an IGBT gate, it is turned off. Just like a N mosfet.

bug13

Joined Feb 13, 2012
1,996
When there is no current in an IGBT gate, it is turned off. Just like a N mosfet.
I haven't use an IGBT, but isn't it voltage driven?? (let's ignore the leakage current here)

shortbus

Joined Sep 30, 2009
8,952
IGBT can also work in the linear (active) region.
You can treat IGBT as a "Sziklai pair". When the first BJT is replaced with a MOSFET.
The image shows the "equivalent" circuit for a IGBT. Wikipedia says, "The Isolated-gate bipolar transistor or IGBT is a three-terminal power semiconductor device primarily used as an electronic switch"
https://en.wikipedia.org/wiki/Insulated-gate_bipolar_transistor

WBahn

Joined Mar 31, 2012
26,398
When there is no current in an IGBT gate, it is turned off. Just like a N mosfet.
If an IGBT is only on when there is a gate current, then it most definitely is NOT just like an NFET.

Ron H

Joined Apr 14, 2005
7,014
If an IGBT is only on when there is a gate current, then it most definitely is NOT just like an NFET.
If an IGBT has significant gate current, then it is just a lump of silicon.

shortbus

Joined Sep 30, 2009
8,952
@ WBhan and Ron H, I know that the gates are voltage driven, was just adopting the language of the members I was replying to.

What I am getting at is that according to everything I've read, IGBTs and Mosfets don't live long in the linear/resistive mode. They are made for switching and don't work well as a resistor. Even in an amplifier they are switched on and off in a PWM like method, unlike a BJT.

shortbus

Joined Sep 30, 2009
8,952
Sorry duplicate post. Mouse is acting up.

Ron H

Joined Apr 14, 2005
7,014
@ WBhan and Ron H, I know that the gates are voltage driven, was just adopting the language of the members I was replying to.

What I am getting at is that according to everything I've read, IGBTs and Mosfets don't live long in the linear/resistive mode. They are made for switching and don't work well as a resistor. Even in an amplifier they are switched on and off in a PWM like method, unlike a BJT.
I don't know much about IGBTs, but there are CMOS linear voltage regulators available.

WBahn

Joined Mar 31, 2012
26,398
@ WBhan and Ron H, I know that the gates are voltage driven, was just adopting the language of the members I was replying to.

What I am getting at is that according to everything I've read, IGBTs and Mosfets don't live long in the linear/resistive mode. They are made for switching and don't work well as a resistor. Even in an amplifier they are switched on and off in a PWM like method, unlike a BJT.
Not true. MOSFETs work just fine in analog circuits. I spent over a decade designing mixed-signal CMOS ASICs. I've never worked with IGBTs, but I suspect they will work in the active region as well.

I think what you are thinking of is that, in a circuit intended for switching operations, the transistors will not survive long if placed in the active region. But this has nothing to do with the fact that it is a MOSFET. It's simply that the circuit isn't designed to limit the current and so you get excessive power dissipation in the transistor. The same would be true of BJTs in similar circuits.

shortbus

Joined Sep 30, 2009
8,952
@ WBahn and Ron H, not arguing with you, just trying to learn/understand better.

Isn't CMOS a totally different animal here? The power levels involved alone would make a big difference.

Ron H

Joined Apr 14, 2005
7,014
@ WBahn and Ron H, not arguing with you, just trying to learn/understand better.

Isn't CMOS a totally different animal here? The power levels involved alone would make a big difference.
CMOS is just NMOS and PMOS on the same chip. There is no reason that an analog CMOS cannot handle as much power as a bipolar one.

count_volta

Joined Feb 4, 2009
435
I have a related question. This has also been bothering me, and unlike a linear regulator, this circuit is using the transistors as a switch. See picture below.

I have used this circuit before to drive a motor for my robot. I know I am correct when I say this because I measured these voltages before in real life. Its not just a theoretical circuit to me.

The top transistor in Leg A (s1) is in the emitter follower configuration. The bottom transistor in Leg B (s4) is in the common emitter configuration.

If S1 is on, the voltage at node A with respect to the negative terminal of the capacitors is Voltage at gate of S1 minus the turn on voltage Vbe or VGE, whatever. I'm treating it as a BJT again because it doesn't matter very much. Lets just say its Vbe.

If S4 is on, the voltage at node B is very low and close to 0. I am assuming S4 is in saturation because we are using it as a switch.

So the voltage across the load is (V_base - Vbe) - voltage of node B. Assuming node B is 0 (ideally), the voltage across the load is (V_base - Vbe).

Now here comes my confusion. The professor in this lecture is saying that Vab(t) or the (voltage across the load) is VDC when S1 is on and S4 is on. The only way this could happen is if the voltage at the base or gate or whatever is equal to VDC and we ignore the turn on voltage Vbe which the professor is doing. He is assuming its a perfect switch.

So again, the same question as before. Is the voltage at the gate/base the same as the voltage at the collectors of S1 and S3? I am assuming we ignore the Vbe voltage drop. Because if the base/gate voltage is something much lower than VDC, what the professor is saying is a bunch of nonsense.

I hope you guys understand what my confusion is. Usually the base/gate signal is the output of a uC which is 5V TTL logic signal. This is especially true if the circuit is being driven with PWM. This means VDC must also be 5V maximum (ignoring the Vbe turn on voltage). But if its not, the professor, and the three other power electronics books I read which show the same thing are lying to me.

The professor is using the transistor as a literal switch. I mean he is assuming if the switch is on, the collector voltage directly appears at the emitter. This is not true however. Especially not in emitter follower configuration.

I hope you guys get what I am saying. WBahn I really liked your reply. You seem to get what my confusion is and what I am asking. Maybe you can help me understand this also? Anyone else too. Thanks guys.

I am using these inverters at work and I am just blindly accepting the results in that picture, but its really bothering me because it doesn't make sense to me.

At work, our inverter is being PWM driven with a uC + opto coupler type circuit, but the collector voltage (VDC) is at hundreds of volts!!!! But according to what I am saying, the voltage across the load wouldnt be hundreds of volts. It would be the tiny signal coming from the uC + opto coupler - the turn on voltage.

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Jony130

Joined Feb 17, 2009
5,250
For this type of a circuit we use "high side drivers" to drivers the top and "low side driver" for the bottom transistors.
Here you have some example.
The simplest bootstrap type high side drivers.

Try analysis this simple circuit very carefully.
http://www.fairchildsemi.com/an/AN/AN-6076.pdf

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count_volta

Joined Feb 4, 2009
435
For this type of a circuit we use "high side drivers" to drivers the top and "low side driver" for the bottom transistors.
Here you have some example.
The simplest bootstrap type high side drivers.

Try analysis this simple circuit very carefully.
http://www.fairchildsemi.com/an/AN/AN-6076.pdf
So you are saying the signal from the uC PWM is converted to a much higher voltage close to the collector voltage and this drives the gate?