need correction about linear regulator circuit

hi there, please help me, i got a little confused with this circuit, please give a correction, is it from the circuit would be like this? if the load increases, then the current flow through the load would be decreased because of the capacitor would be discharging its charge slowly (Vc = Vload = Vin.e^-t/RC), and the voltage through the load would just fall a little, so the difference between Vref and the output divider would just fall a little too, so the output from the error amplifier would not be so high, so, the headroom voltage would be lower, its mean that the resistance in collector emitter is low, and will providing more current through the load, and if the load is decreased, for example we short the two terminal of the capacitor, then, the load current would be exceeded, and the output voltage will fall, so, the voltage in the divider output that connect to the inverting will drop too, so, the difference between the Vref and the Voltage divider would be high, and the output from the error amplifier will be high too because Vout = A(V+ - V-), and it would makes the Collector emitter voltage rise, and its mean that the resistance in the collector emitter would be rise too, and limit the exceeded current to flow to the load, is it true? please correct me...

thank you very much

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.

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?

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

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

zhi_yi said:

thank you very much please tell me how do the current limitter circuit work in the first schematic?

Click to expand...

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?

Click to expand...

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

Click to expand...

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

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

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

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.