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I came across one circuit for making fuse using transistors. In the circuit my doubt is how the voltage is dropping across Q2 when I changes the load resistance value (hence current). Usually the pnp transistor is suppose to have constant 0.2V drop across EC.

Please help me to understand how the below circuit is working, (This circuit is working fine as per simulation results)

Fuse Circuit

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  • \$\begingroup\$ It's a type of 2 transistor schmitt trigger. As more current flows through R3, less current flows through R2/R4. Voltage at V3 falls & Q1 starts to turn off. This lowers the voltage at Vout, which lowers V3 and this cycle of +ve feedback snaps Q1 off. See this excellent reference (johnhearfield.com/Eng/Schmitt.htm). \$\endgroup\$
    – Buck8pe
    Jun 8, 2020 at 9:04

2 Answers 2

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The current limit as shown is determined by the (reverse) hFE of the PNP transistor - and the supply voltage and R2+R4, so it is pretty crude. When Q2 goes out of saturation the there is enough voltage to turn Q1 on and latch the output off. Typical reverse hFE of a normal transistor is of the order of 5-10.

Edit: This is an interesting circuit with some subtlety, thanks to @jonk for taking a closer look at it. Below is some hand-waving, I've not worked out the exact design equations. Note that Q2 is operated in reverse, which has two main consequences- low current gain (hFE) and low breakdown voltage (Vbe, which becomes Vce in reverse). Most jellybean transistors are rated at 5V and can withstand somewhat more than that.

For start-up to properly take place R2 + R4 is low enough that the current flowing through Q2 raises the output voltage fast enough that Q1 cannot steal enough base current to keep Q2 from turning on. The critical input voltage is about 1 Vbe at the input.

For the output latching to take place properly Q1 must be able to divert enough base current from Q2 to keep it off, so the hFE of Q1 must be much higher than that of Q2.

enter image description here

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  • \$\begingroup\$ Q2 is operating in reverse mode. I'm sure you noticed, but didn't mention it. The circuit as shown will not function if Q2 is operating in forward mode. I think the OP may be asking for more details than just "hFE", about why it works one way and not the other. (The reverse mode has far, far smaller hFE, of course.) There is also the question about the magnitude of VCE in reverse mode, I suppose. \$\endgroup\$
    – jonk
    Jun 8, 2020 at 18:16
  • \$\begingroup\$ Oh, and there is an issue with respect to VBE breakdown. Just crossed my mind, as well. \$\endgroup\$
    – jonk
    Jun 8, 2020 at 18:26
  • \$\begingroup\$ You're right, it's an overly specific answer to "how the voltage is dropping across Q2 when I changes the load resistance value (hence current)". Maybe I can improve it later. \$\endgroup\$ Jun 8, 2020 at 19:30
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    \$\begingroup\$ V-src:$$R_{_\text{LOAD}}\ge\frac{V_{_\text{BAT}}-V_{_{\text{EC}_\text{SAT}}}}{V_{_\text{BAT}}-V_{_\text{CB}}}\cdot\frac{R_2+R_4}{\beta_{_{R_\text{MAX}}}}$$I-src:$$\frac{V_{_\text{BAT}}-V_{_{\text{BE}_\text{Q2}}}}{V_{_\text{BAT}}-V_{_\text{CB}}}\cdot\frac{R_2+R_4}{\beta_{_{R_\text{MAX}}}}\le R_{_\text{LOAD}}\lt\frac{V_{_\text{BAT}}-V_{_{\text{EC}_\text{SAT}}}}{V_{_\text{BAT}}-V_{_\text{CB}}}\cdot\frac{R_2+R_4}{\beta_{_{R_\text{MAX}}}}$$...and:$$R_{_\text{LOAD}}\lt \frac{V_{_\text{BAT}}-V_{_{\text{BE}_\text{Q2}}}}{V_{_\text{BAT}}-V_{_\text{CB}}}\cdot\frac{R_2+R_4}{\beta_{_{R_\text{MAX}}}}$$ \$\endgroup\$
    – jonk
    Jun 9, 2020 at 19:18
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    \$\begingroup\$ That's what I think, anyway. The first is voltage source mode, the second is current source mode, and the last one is the variable resistance (positive or negative) mode. [The reversed lettering refers to the reverse-mode BJT (which I think of as \$Q_1\$) and the other BJT is \$Q_2\$.] \$\endgroup\$
    – jonk
    Jun 9, 2020 at 19:18
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Overview

Interesting circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Note that \$Q_1\$ appears to be arranged to operate in reverse mode. This seriously degrades the effective \$\beta_1\$ (which now I'll call \$\beta_{_R}=\frac{I_\text{E}}{I_\text{B}}\le 5\$.) Meanwhile, \$Q_2\$ when active and not off is operating in forward mode will have \$\beta_2=\frac{I_\text{C}}{I_\text{B}}\ge 100\$. So, below, \$\beta_{_R}\$ and \$\beta_{_{R_\text{MAX}}}\$ always refer to \$Q_1\$. Since \$V_{_{\text{EC}_\text{SAT}}}\$ (note the reversed lettering here) only matters for \$Q_1\$, that term also always refers to \$Q_1\$. Finally, since \$Q_1\$ is the only BJT operating in reverse mode, In normal operation, \$V_{_\text{BE}}\$ refers to the forward-biased BE junction. But here I'll use \$V_{_\text{CB}}\$ to refer to \$Q_1\$'s forward-biased "reverse-mode" "CB" junction, which performs a similar function now. In fact, if you see reversed lettering, it's almost certain I'm referring to \$Q_1\$. I hope that will remove some possible confusion below as I lay out some thoughts.

The above circuit has three primary regions of operation:

  1. Voltage Source Mode: \$R_{_\text{LOAD}}\ge\left[\frac{V_{_\text{BAT}}-V_{_{\text{EC}_\text{SAT}}}}{V_{_\text{BAT}}-V_{_\text{CB}}}\cdot\frac{R_2+R_4}{\beta_{_{R_\text{MAX}}}}\right]\$

    In this mode, \$Q_1\$ is heavily saturated, \$Q_2\$ is off, \$0 \le \beta_{_R}\lt \beta_{_{R_\text{MAX}}}\$, \$I_{_B}=\frac{V_{_\text{BAT}}-V_{_\text{CB}}}{R_2+R_4}\$, \$R_3\$ carries no current, and \$V_{_\text{BAT}}-V_{_{\text{EC}_\text{SAT}}}\le V_{_\text{LOAD}}\le V_{_\text{BAT}}\$.

    \$\$

  2. Current Source Mode: \$\left[\frac{V_{_\text{BAT}}-V_{_{\text{BE}_\text{Q2}}}}{V_{_\text{BAT}}-V_{_\text{CB}}}\cdot\frac{R_2+R_4}{\beta_{_{R_\text{MAX}}}}\right]\le R_{_\text{LOAD}}\lt\left[\frac{V_{_\text{BAT}}-V_{_{\text{EC}_\text{SAT}}}}{V_{_\text{BAT}}-V_{_\text{CB}}}\cdot\frac{R_2+R_4}{\beta_{_{R_\text{MAX}}}}\right]\$

    In this mode, \$Q_1\$ is only lightly saturated or almost active, \$Q_2\$ is still off or so lightly active that it doesn't matter (\$Q_2\$'s collector doesn't supply enough current towards \$R_4\$ to matter, yet), \$\beta_{_R}\to \beta_{_{R_\text{MAX}}}\$, \$I_{_B}=\frac{V_{_\text{BAT}}-V_{_\text{CB}}}{R_2+R_4}\$, \$R_3\$ has negligible current, and \$V_{_\text{BAT}}-V_{_{\text{BE}_\text{Q2}}}\le V_{_\text{LOAD}}\lt V_{_\text{BAT}}-V_{_{\text{CE}_\text{SAT}}}\$. (The onset of \$Q_2\$ turning on occurs at \$V_{_{\text{BE}_\text{Q2}}}\$.)

    \$\$

  3. Negative or Positive Resistance Mode: \$R_{_\text{LOAD}}\lt\left[\frac{V_{_\text{BAT}}-V_{_{\text{EC}_\text{SAT}}}}{V_{_\text{BAT}}-V_{_\text{CB}}}\cdot\frac{R_2+R_4}{\beta_{_{R_\text{MAX}}}}\right]\$

    In this mode, \$Q_1\$ is going active or is active, \$Q_2\$ is turning on or is on, \$\beta_{_R}= \beta_{_{R_\text{MAX}}}\$, and \$V_{_\text{LOAD}}\lt V_{_\text{BAT}}-V_{_{\text{BE}_\text{Q2}}}\$. Additional load current must come via \$R_3\$. But pulling that additional load current via that direction has further implications because it reduces the voltage across \$R_2\$ and thereby lowers the base current in \$Q_1\$. That reduces the available current arriving to the load via \$Q_1\$, increasing still further the load's draw through \$R_3\$.

    What's really interesting about this mode of this circuit is that the behavior can vary in direction and can have either a negative resistance or a positive resistance behavior.

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