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So, i got this Vccs opamp that I want to use. I know that the Vcontrol is the voltage that will appear in my Rsense and knowing my Rsense I can find the Ie. What I dont get is the role of the Vc and whether it determines the Ib and Ic. I got the datasheet for both the opamp and the transistor which do have the different Hfe values. I mean I will have to supply the Vc.

What I want is to know exactly the values of all the currents and voltages in paper before trying it out on the actual circuit.

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  • \$\begingroup\$ "Role of the Vc" - do you mean Vcc+ and Vcc- that power the op-amp? \$\endgroup\$ – glen_geek Oct 26 '17 at 16:14
  • \$\begingroup\$ Well, the Ic is Ic = Ie*(hfe/(hfe+1)) and Ib = Ie/(hfe +1) and Vc should be larger than Vsens +0.2V + Iload *Rload. \$\endgroup\$ – G36 Oct 26 '17 at 16:15
  • \$\begingroup\$ As G32 points out there is a base current error, you can use a fet rather than a bjt if the ~1% error is a problem. \$\endgroup\$ – George Herold Oct 26 '17 at 16:20
  • \$\begingroup\$ @glen_geek No i mean the "V" as shown in the figure.Vc with c refering to the collector \$\endgroup\$ – Dimitris Dirokaltsis Oct 26 '17 at 17:25
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Vcc need to be chosen based on a few things.

You need to examine the spec-sheet of the particular op-amp.

First: The device itself will have a minimum voltage at which it will operate so your VCC needs to be at least that large.

Second: the op-amp needs to output a voltage at least as large as the maximum control voltage plus the base emitter diode drop in that transistor. Op-Amps have a voltage swing that does not go all the way up to Vcc. Look at the spec-sheet and see how much greater Vcc needs to be over the voltage you just calculated.

Whichever of those voltages is greater is your minimum Vcc. Chose a suitable supply greater than that value.

Vcc- could be ground in this circuit since you do not need negative voltages. However, not all op-amps will work rail to rail and not all op-amps will drive all the way to the bottom rail. Again, you need to check the data sheet to see what you need on this supply so your output can reach zero.

In this circuit, since the current through the sense resistor is \$Ic + Ib\$ then collector current is actually governed by the equation

\$Ic = (V_{Control}/R_{Sense}) * H_{fe}/(H_{fe}+1) \$

As such, since \$H_{fe}\$ is quite variable, it is not that great a circuit.

ADDITION:

To calculate the \$V\$ value, you need to know the load resistance. This is where your example schematic is incomplete. At this point you have to assume the implied components shown below exist.

schematic

simulate this circuit – Schematic created using CircuitLab

\$V_{min} = V_{ControlMax} + I_{cMax} * R_{Load} + V_{CESat}\$

However it can not be grater than the max \$V_{CE}\$ of the transistor nor high enough to cause the transistor to overheat from \$P = V_{CE} * I_{cMax}\$

\$V_C\$ will of course be \$V - (R_{LOAD} * I_C)\$

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  • \$\begingroup\$ Sorry for not being clear.I meant the "V" as shown in the figure.Is it correlated with Hfe, Ic and Ib and how? \$\endgroup\$ – Dimitris Dirokaltsis Oct 26 '17 at 17:38
  • \$\begingroup\$ @DimitrisDirokaltsis see addition \$\endgroup\$ – Trevor_G Oct 26 '17 at 17:49
  • \$\begingroup\$ As you mentioned Hfe is variable and according to the spec sheet of this PN2222A npn it depends on the values of Ic and Vce.So how can i figure out what the Hfe will be in order to calculate Ic? \$\endgroup\$ – Dimitris Dirokaltsis Oct 26 '17 at 18:11
  • \$\begingroup\$ @DimitrisDirokaltsis you basically can't. You can estimate a range and accuracy based on the values given though by iterating though the values. Spreadsheet time :) As others have mentioned a fet or mosfet design would be better in this configuration. \$\endgroup\$ – Trevor_G Oct 26 '17 at 18:14
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    \$\begingroup\$ But you do not need to worry about Hfe much. Even if hfe is low as 50 the Ic =50/51*Ie = 0.98*Ie (98 percent of a Ie current) \$\endgroup\$ – G36 Oct 26 '17 at 18:16

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