# need correction about linear regulator circuit

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#### zhi_yi

##### Full Member level 4

thank you very much

#### VVV

##### Advanced Member level 5
First of all, let me say that you need a resistor between the output of the error amplifier and the base of the transistor.

When the output voltage drops, you are right, the opamp's output gets higher and forces the transistor to deliver more current.
But that happens by LOWERING the voltage between the collector and emitter, in order to compensate.

Nwo when you short the output, the error amp will drive the base high, attempting to restore the output voltage. But with a dead-short that is impossible, so the entire input voltage will be across the transistor and the short-circuit current will only be limited by whatever resistances there are in the circuit: wires, traces, etc. That means a high current through the transistor, with a high voltage across it= high power dissipation in the transistor.

Without a fuse or some other means to limit the current or open the circuit, the transistor may fail, if the heatsink is not adequate.

#### zhi_yi

##### Full Member level 4
okay, thank you very much

is it the resistor between the output of the error amplifier and the base of the transistor is for limiting the exceeding current flow to the base of the transistor?

#### IanP

##### Advanced Member level 5
I don't think you need a resistor between error amplifier and transistor (see examples below).
You can connect one but as far as current limit is concerned there are other more efficient methods of limiting currents in power supplies ..
In fact if you use an opamp as an error amplifier, it has current limit implemented in its output stage ..
Regards,
IanP

### zhi_yi

Points: 2

#### zhi_yi

##### Full Member level 4
thank you very much please tell me how do the current limitter circuit work in the first schematic? what is the function of the diode there? please tell me what will happen to that circuit if the load decreases and makes the exceeding current to flow, it will makes the output voltage drop, right?and then, the noninverting input would be more positive than the inverting input, so, the output from the error amplifier would be positive, and then, what will happen to the diode?

in normal operation, there are no current flow through the 2n3055, 2n3904 and 2n3053, and to the 3K resistor, because the current would flow through the shorted wire from collector in 2n3055 to the emitter of 3904, is it rigth?

thank you

#### Learner

##### Full Member level 2
zhi_yi said:
thank you very much please tell me how do the current limitter circuit work in the first schematic?

The 0.3 ohm 5W resistor is a current sensing resistor which develops a voltage between the emitter and the base for the 2N3904 transistor. When the load has droped in resistance and draws more current, Vbe of the 2N3904 transistor increase which decrease its collector emitter resistance Rce and sinks more current.

what is the function of the diode there? please tell me what will happen to that circuit if the load decreases and makes the exceeding current to flow, it will makes the output voltage drop, right?and then, the noninverting input would be more positive than the inverting input, so, the output from the error amplifier would be positive, and then, what will happen to the diode?

When the output of the error amplifier becomes more positive it will sink the current from the 2N3904 transistor.

The diode is there to ensure that the current flows 1 way, in case of the output of the error amplifier becomes negative which sources current.

in normal operation, there are no current flow through the 2n3055, 2n3904 and 2n3053, and to the 3K resistor, because the current would flow through the shorted wire from collector in 2n3055 to the emitter of 3904, is it rigth?

thank you

The" shorted wire" you are refering to is connected to the Vcc+ to power the error amp, it is not connected to the output.

### zhi_yi

Points: 2

#### zhi_yi

##### Full Member level 4
quote : The diode is there to ensure that the current flows 1 way, in case of the output of the error amplifier becomes negative which sources current.

which way is it? the diode will conduct only if the output from error amplifier is negative? so, the current flow through the diode would be negative?

thank you

#### IanP

##### Advanced Member level 5
Power transistors are positively biased by 3kΩ resistor, and the error amplifier (through the diode) pulls this voltage down to required level (≈3V lower than the output volage).
The function of the diode is to provide OR function for voltage control from the opamp and current limit from 2N3904 transistor/0.3Ω resistor ..
Regards,
IanP

#### Borber

##### Advanced Member level 5
Purpose of diode in 3_24V_3A_PS.gif schematics is to protect error amplifier when current limiting transistor (rightmost 2N3904) is on. In that case error amplifier output voltage will try to rise regulator output voltage and excessive current may flow out of it. Current limit when activated normally lowers output voltage and it makes no sense that error amplifier tends to correct it.

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