I made a circuit about motor short circuit protection today, referring to the classic circuit:
This is a self-locking protection circuit. When short-circuiting: Q3 pole is pulled low, Q2 is turned on, forming self-locking, forcing Q3 to cut off, Q3 is not loaded with voltage after the end, then there is no load at all, so even take There is no output from the load. To recover the output after removing the load, you can connect a resistor to the C E junction of Q3 and take about 1K.
C2 and c3 are very important. After self-locking, the circuit is restarted by these two capacitors, otherwise the startup fails. The principle is that the voltage across the capacitor cannot be abrupt when power-on, and C2 keeps the base of Q2 high at the moment of power-on, making Q2 non-conducting. C3 makes the base of Q3 remain low at the moment of power-on, so that Q3 turns on Vout with voltage. Thus, the R5 bit is high and the conduction is blocked.
But when I quoted it, there was a problem: I took the R4 with a 1K. The problem is: VOUT load capacity is getting worse.
The reason is: R4 becomes smaller, and Ib of Q2 becomes larger, so that Q2 becomes easier to conduct. It also pulled Q3 down.
So take a deeper reason: How do we calculate the value of each component?
For good calculation, R3 takes 1k,
Assuming VCC=5v, considering that the voltage will be pulled low, VCC takes 4.5V in the short circuit. To make Q3 cut off, the base of Q3 takes 3.9V or more. Then Ic=3.9/1K=3.9ma of Q3.
Q3Q2 uses 8550 to find the specification (beware that there is something wrong on the Internet).
Take any one of the calculated B values ​​(not much) B=200,
Then, IB=3.9/200=0.0195MA of Q2 is obtained.
To turn Q2 on, the base voltage of Q2 is 4.5-0.6=3.9.
Q2 has the base voltage and current is available, then R4=3.9/0.0195=200K
Of course, it is ok to take 10k for R4, except that the Ib bias current of Q2 is large. The current of the IC is also large, so that the R3 voltage rises about a block, and the base voltage of Q3 is more easily pulled up, so R4 is the adjustment sensitivity.
This is its 1, the most critical is R3, if you want Q3 to enter the cutoff, then the state of Q2 is determined. The larger R3 is, the smaller the IC of Q2 is, the easier it is for Q2 to enter saturation state.
There is some abstraction here, so I will understand it by drawing a picture. When the triode is turned off, the voltage across CE is the largest. We set it to 4.5V. During the process of increasing IB, VCE is getting smaller, and Ic in saturation is determined by RC. Please see the figure:
The smaller the load (RC) is, the larger the saturation current is, the larger the VCE is, the smaller the saturation voltage difference from 4.5V to VCE is, the smaller the conduction time is, and the most important is that the corresponding VR3=4.5-VCE (saturation), The lower the voltage at both ends of VR3, the smaller the Q3 can't reach 3.9V, the Q3 can't be cut off. This is the reason for the impact of R3 on sensitivity.
Look at the impact of R4:
At the end of Q2, the base of Q2 is current, and this current flows from E to R4 and R5 to ground. The base current IB of Q2 gradually increases on the basis of this initial current. As seen from the green line above, it is known that the red line perpendicular to the red line reaches the protective base current IB(sta) of Q2.
Therefore, the larger the initial base current IB, the faster the rise rises to the saturated base current IB(sta).
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