Voltage Peak detector
1. Vcc = 12v.
2. Choose VCQ1 output, = 9v.
3. Choose VBQ1 = 8v.
4. VEQ1 = 7.3v.
5. choose RCQ1 = 470 ohms. (to drive a LED output)
6. IRCQ1 = (Vcc - VCQ1) / RCQ1 = ~ 6.38 mA.
7. RE is a split resistance, between the emitter of Q1 and Q2.
RE = ( REQ1 + REQ2) = (VEQ1 / IRCQ1) = ~ 1143 ohms.
So make REQ1 = REQ2 = 560 ohms each.
8. Now VEQ2 = ~ 3.65v.
9. Now with around 1100 ohms for RE, then make RB1Q1 around 2 times RE =~ 2K ohms.
10 IRB1Q1 = (VBQ1 / RB1Q1) = 4mA.
11. RB2Q1 = (Vcc - VBQ1) / IRB1Q1 = 1K ohms.
By varying the resistor value of REQ2, to a higher value, will increase VEQ2, which in turn will then detect a higher PK input voltage.
This will also raise the idle (standing voltage) at the output too.
Idle output voltage.
Input @ 4v.
No trigger of circuit.
Input voltagte @ 4.3V.
The circuit is triggered.
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