# confused about transistor in circuit

Discussion in 'General Electronics Chat' started by count_volta, Jul 16, 2013.

1. ### count_volta Thread Starter Active Member

Feb 4, 2009
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Hey guys, I have been learning power electronics recently. Specifically switching power supplies, inverters, and etc. One thing which I seen in all books and everywhere online has been bothering me. I don't understand how this works. Please see the picture below.

This is a linear power supply diagram. My confusion is this. You see that the transistor is in the "emitter follower" configuration. I think the transistor in the diagram is an IGBT but it should be similar to an NPN BJT. So I will talk about it as if though its a BJT.

In the emitter follower, the voltage at the emitter is equal to the voltage at the base minus 0.7V Vbe voltage drop. If this is true, in the diagram above, the voltage at the emitter and hence at the load is equal to the controller voltage voltage minus 0.7V.

But the video where I got this diagram, and all books I seen on this subject treat this transistor like a variable resistor. Meaning that the voltage at the load can vary from Vdc to 0. This would be true if the voltage the controller is outputting is equal to Vdc. But that is not usually the case.

Usually the voltage from the controller is 5V from a uC, and we are trying to drive 15V or higher for example at the collector for example to power a motor. So Vdc =15V. Vbase = 5V. And somehow they are telling me I can end up with 15V at the emitter. This would be true if the transistor is a literal switch. But its not. Do you get what I am saying?

Can someone please clear up the confusion for me. Thanks.
-Gene

Edit: In fact the transistor WOULD be a literal switch if the load was connected between Vdc and the collector. But the load is connected at the emitter.The voltage at the load HAS to be Vbase -0.7V. Or V_controller - 0.7V.

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Last edited: Jul 16, 2013
2. ### bug13 Well-Known Member

Feb 13, 2012
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Maybe the controller means uC + DAC + Op Amps, or simply means a trim pot? I could be totally wrong tho, at least that's what I would assume, hope it helps

3. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
Maybe. But that would be kind of stupid. The signal applied at the base is the control signal. It should be small. You are using the transistor as a switch right. The signal to open and close the switch should be tiny and the signal you are trying to control is big. Its like controlling a giant water dam with a valve which requires you to lightly spin a knob. The giant water dam is at the collector-emitter, and the valve/knob is at the base.

By using the emitter follower configuration as a switch, its like having to use a dam of water to control another dam of water.

Last edited: Jul 16, 2013
4. ### studiot AAC Fanatic!

Nov 9, 2007
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I'll go along with this half of your reasoning.

So the emitter ouputs the positive terminal of the power supply (for the polarity of device you have shown).

And the voltage at the base is always 0.7 volts (actually for a power device it may be less than this) higher (more positive) than the base.

The problem lies in your connections of the control circuitry.

The control circuitry output to the base needs another connection usually through a resistor ( it may be a constant current circuit) to the collector of the series pass element.

The controller is a differential amplifier of some sort that receives two inputs and adjusts the voltage of its output to the base of the pass device in accordance with the differnce between these two inputs.

One input that you have not shown is taken from a stable reference.

The other input is taken from a sample of the output voltage that is set by a resistor/potentiometer chain.

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5. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
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Thanks. I got this from a power electronics lecture on youtube. I think the professor is trying to simplify the linear regulator into a semi black box in order to explain it easier with the short amount of time he has.

But okay. Even if the control system that creates the base voltage is missing from the diagram.

Answer this. Whatever appears across the load is the output of the controller minus Vbe. Is this correct?

If this is correct, then the controller must output a voltage on the same order of magnitude to the voltage connected to the collector. Like if 15V is connected to the collector, approximately 15V must be output from the controller to the base. If the object of this circuit is to maintain a constant 15V DC for example.

6. ### shortbus AAC Fanatic!

Sep 30, 2009
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An IGBT is like a BJT only in the Collector and Emitter. The gate is like a Mosfet, basically a capacitor. It is either on or off, no control like a Base in a BJT.

7. ### t06afre AAC Fanatic!

May 11, 2009
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The "Controller" unit is in basic what is named a proportional regulator. This shows a more detailed conceptual power supply drawing

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8. ### studiot AAC Fanatic!

Nov 9, 2007
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Yes, that's correct in your configuration.

That represents one of the issues you have to resolve in a power supply design.

How to arrange things so that your controller can output high voltage or low voltage. You have to design the controller so that the correct bias voltages appear on its components. You will find this limits the available output range.

I am sorry I am short of time for sketches at the moment. These make the issue clearer.

I also agree with T06afre that we should distinguish between different series pass elements. Are you happy with this terminology?

9. ### LDC3 Active Member

Apr 27, 2013
920
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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.

10. ### bug13 Well-Known Member

Feb 13, 2012
1,208
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if you are looking for control high power high voltage DC, maybe try SCR control rectifier? you only need 5V from your uC*, through a pulsing transformer to control >450VDC and in the magnitude of KW.

edit: *by controlling the firing angle of the SCR

11. ### Ron H AAC Fanatic!

Apr 14, 2005
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You might want to clarify that. Vce is never infinite.
Also, since this is a linear regulator, PWM would not be used unless it were heavily filtered before being applied.

12. ### LDC3 Active Member

Apr 27, 2013
920
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OK, nit-pik if you want.

Vce is over 10M when the transistor is not conducting.

I just assume that the controller used PWM since the transistor is saturated. If the transistor is in the linear region, then the voltage wouldn't be very stable.

13. ### Ron H AAC Fanatic!

Apr 14, 2005
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Vce is a VOLTAGE !!! It will not be higher than the supply voltage.

And all linear regulators have non-saturating pass elements. Commercially available ones are very stable, so long as they are applied using practices recommended in their datasheets.

Last edited: Jul 16, 2013
14. ### LDC3 Active Member

Apr 27, 2013
920
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OK, I wasn't paying attention. Am I going to detention now?

15. ### Ron H AAC Fanatic!

Apr 14, 2005
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657
Nah. Just go stand in the corner for an hour.

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16. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Every practical linear regulator is probably going to be designed with some voltage headroom to allow for the issue the OP has raised.

The 3-terminal regulators such as the 78xx series all have a minimum dropout voltage. The designer takes this into account when using the device.

17. ### LDC3 Active Member

Apr 27, 2013
920
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zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

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18. ### t06afre AAC Fanatic!

May 11, 2009
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It look more to me that you miss some basic understanding, and you try to compensate for that by being cocky. Here is a good link http://sva.ti.com/assets/en/appnotes/f4.pdf. That will bring some understanding to you about the topic

Last edited: Jul 17, 2013
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19. ### WBahn Moderator

Mar 31, 2012
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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.

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20. ### WBahn Moderator

Mar 31, 2012
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No, the signal applied to the base is not necessarily small. As you pointed out, it is Vbe greater than the output voltage. So if you are trying to output 12V, the signal applied to the base will be about 12.7V.

Also, you most definitely are NOT using the transistor as a switch. If you are, then this would NOT be a LINEAR power supply!

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