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schematic

simulate this circuit – Schematic created using CircuitLab

During my bachelor thesis, I am learning IGBT gate driver circuits. I am struggling with understanding the following circuit. It is a current-sourcing part of a IGBT gate push/pull driver and it's sinking part is a replica of it with analogous P-type devices. I am simulating only the current-sourcing part to learn the current-control mechanism of this gate driver.

Functionality:

The voltage at Vin is adjusted in such a way that V- = V+ = 5 V, so that the op-amp always operates in linear mode. For example, 5 A at output:

if Rf = R1, (15 - Iout) - V+ = V+ - Vin

Vin = 0

My problem is:

Why is a negative voltage supply is being used to maintain a specific negative voltage on the emitter of Q1? And how can the values of Re1 and Re2 be calculated?

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  • \$\begingroup\$ Just to be sure: Is this supposed to be a voltage controlled current source (VCCS) that provides \$0-12\:\textrm{A}\$, given a control voltage? And where you will have to dissipate close to \$3\:\textrm{V}\$ at full current with your \$Q_2\$ and \$Q_3\$ (tied to ground, not \$-15\:\textrm{V}\$)? And where you are wanting to possibly jack up the opamp inputs to a common mode voltage of \$+5\:\textrm{V}\$? \$\endgroup\$
    – jonk
    Apr 27, 2017 at 17:48
  • \$\begingroup\$ Yes this is a VCCS. Control voltage you can find the way I computed it for one operating point 12A. Dissipation is done in capacitance of a Transistor that needs to be connected to Emitter of Iout later. No idea 100%. that one is out of scope of my study. Input voltage Vin is adjusted in such a way that V- = V+ to obtain desired output current. Check the basic form of this circuit electronics.stackexchange.com/questions/299110/… \$\endgroup\$ Apr 28, 2017 at 6:17

4 Answers 4

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The voltage at Q1's collector changes in the opposite direction to the op-amp's output, due to the inverting action of Q1. Since feedback is taken from the collector, to keep feedback negative, the inverting and non-inverting op-amp inputs have been swapped.

With feedback taken from the collector, that node may considered an intermediate "output", and the closed-loop part of the system can be reduced to this (where node C is Q1's collector):

schematic

simulate this circuit – Schematic created using CircuitLab

Intuitively, gain here (with respect to \$V_{IN}\$) is clearly \$-\frac{R_F}{R_1}\$, but there's an offset of +5V. This system produces the input/output relationship:

$$ V_C = 5V - \frac{R_F}{R_1}(V_{IN}-5V) $$

Collector \$I_C\$ current flowing via resistor \$R_C=1\Omega\$ will be:

$$ \begin{aligned} I_C &= \frac{+15V - V_C}{R_C} \\ \\ &= \frac{15V - \left(5V - \frac{R_F}{R_1}(V_{IN}-5V)\right)}{R_C} \\ \\ &= \frac{10V + \frac{R_F}{R_1}(V_{IN}-5V)}{1\Omega} \\ \\ &= 5\left(2-\frac{R_F}{R_1}\right) + V_{IN}\frac{R_F}{R_1} \end{aligned} $$

Assuming base current \$I_B\$ to be negligibly small, we have the same expression for emitter current \$I_E\$. That current is split into two paths \$I_{E1}\$ and \$I_{E2}\$, so that by KCL:

$$ I_{E1} + I_{E2} = 5\left(2-\frac{R_F}{R_1}\right) + V_{IN}\frac{R_F}{R_1} $$

Both paths share the same emitter potential \$V_E\$, so by KVL we also have:

$$ I_{E1}R_{E1} - 15V = I_{E2}R_{E2} $$

I'll leave you to do the algebra to find \$I_{E1}\$ and \$I_{E2}\$ in terms of \$V_{IN}\$, because the resulting expressions won't make any sense to us. We don't have the rest of the circuit for context.

In other words, we can't answer the question "why a negative voltage supply is being used" (the purpose of that last \$-15V\$ term) without more context (such as what are \$R_{E1}\$ and \$R_{E2}\$), but at least now you have a clear relationship with \$V_{IN}\$ that you may be able to get some meaning from.

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Hint: start from the end: determine the range of Ib2 in relation to Iout. Then figure out the range of Vb at Q2. Only then you could figure out the resistors around Q1.

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  • \$\begingroup\$ I already mention my Vb which is (15 - Iout) - 1.4.. so for 1 - 12 A Iout for example, you have 12.6 - 1.6 V. And then the RE ? \$\endgroup\$ Apr 28, 2017 at 6:09
  • \$\begingroup\$ Never mix units! Just because your output resistor is 1R does not mean you should think that a voltage is 15 - Iout! Values that reduce are only to be reduced in the final equation. Hint 2: you have a voltage controlled current source. What is the range within which you want to control the current and what input voltage should control this current? \$\endgroup\$
    – electrobob
    Apr 28, 2017 at 23:35
  • \$\begingroup\$ Given that R1 = Rf, input voltage is 12V range for controlling 1 - 12 A output current. \$\endgroup\$ May 2, 2017 at 6:45
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Without negative feedback around the opamp, it will always be in saturation and its output will be either + or - 15 volts depending on whether V+ is slightly above or below V- at the opamp input. Can't see how this can work.

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    \$\begingroup\$ this is already a negative feedback circuit ! \$\endgroup\$ Apr 28, 2017 at 6:07
  • \$\begingroup\$ @HerrderElektronik are you sure? The feedback resistor is going to the positive terminal. \$\endgroup\$
    – Reinderien
    May 29, 2018 at 18:31
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    \$\begingroup\$ The transistor stage is inverting, so the overall feedback is negative. I'm not sure that this circuit would have great stability margins, though -- generally when you add a gain stage outside of the op-amp you need to add compensation. \$\endgroup\$
    – TimWescott
    Dec 28, 2019 at 1:33
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if I assume that 12 A is my maximum current so Q1 might be near saturation, assuming Vce = 0.5, Re I calculate to be

Re = 1.6 - Vce (assumed 0.5) / 50mA = 22 Ohm

With 12A through the 1Ω sense resistor the voltage at Q2 Emitter is 15V-12V = 3V. This is 2V below 5V, which balances the op amp inputs with the control input at 7V (2V above 5V) when R1 = Rf.

Assuming 1.4V Q2 B-E bias, Q1 Collector must be 3V-1.4V = 1.6V. Subtracting 0.5V gets 1.1V across Re. Q1 Collector and Emitter currents are almost the same, so Re = 1.1V / 50mA = 22Ω.

This value of Re doesn't work for 12A

Your calculations are correct, so the circuit should 'work'. However Iout goes directly to Ground so it doesn't do anything useful. Circuit operation could be quite different when a load is connected.

Iout goes into IGBT gate.

There is no IGBT in your circuit, nor do you show where it would be connected. IGBT drivers usually have push-pull outputs, whereas your circuit appears to be class A. How is it meant to work?

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  • \$\begingroup\$ Thanks alot for taking your time. Yes, actually this circuit is also a Sourcing Current part of Push-Pull gate driver circuit. the another sinking part is same replica of shown circuit with analogous P- Channel devices. Signal Vin is fed from a uController logic and Iout feeds gate of IGBT. I simulate only one part of it to understand the basics. What I missed before is that I did not connect -6V to emitter at middle stage BJT. Do you know which effect it can introduce or if it helps to calculate both RE1 and RE2 ? \$\endgroup\$ May 9, 2017 at 11:34
  • \$\begingroup\$ I understand, after this gate driver is connected to IGBT which will have its own resistance, the output current limit should Fall below 12 A. but i wanted to analyze circuit first without IGBT effect to have feeling about how it is working. I also consider that IGBT should have internal equivalent gate Resistance greater than 1 Ohm Rc with Q3 to avoid oscillation of output current have not to have loopgain>1 as explained by you in my last question ! \$\endgroup\$ May 9, 2017 at 11:38

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