The astable multivibrator circuit has two higher-resistance resistors (47k on the pic). What are they there for? Wouldn't it work without? And what is exactly their function? I spent a lot of time trying to figure it out using this interactive visualization but i still don't understand it fully...

enter image description here

  • \$\begingroup\$ Try deleting both of those higher-value resistors from the simulation and click reset, see what happens. \$\endgroup\$
    – Hearth
    Jan 15, 2019 at 2:42

1 Answer 1


The purpose of the Base resistors is to integrate the supply voltage ( when Collector is near 0V and pullup the cap voltage from -Vx to +Vbe to start conduction on the other side then switch ON Q (toggle outputs) and repeat for the other side.

This design works best on low voltage.Here with a variable Battery and reduced collector R to raise LED current at lower voltage. 3 to 5V enter image description here

When Q1 conducts hard from C2 it dumps a negative pulse to Vbe2 thru C1 to reverse bias Q2 for a period reduced by T=R1C1 but then ramps up slower by T=R2C1 until Vbe reaches the conduction threshold.

Due to the current gain each device switches hard then the base current is slowly reduced. But if the Rb/Rc ratio is too high the LED current will decay quickly and the clock slows down.

The base current is integrated so RbC affects the slew rate and cycle frequency so V+/RbC is the major factor for determining the frequency . However Vbe has a max reverse voltage often =5V and if V+ is > 6V then there is a damage risk to the above design to Vbe reverse limits. So a reverse diode was often added to prevent this failure. But that also reduces the ramp swing and thus raises the frequency.

A better simpler design now uses CMOS Schmitt triggers. This can work with prudent choices of R=33M and C values with low leakage caps to make cycle times ~ 1 Minute possible or > 50 MHz with 10pF, 1K

Both use the negative feedback of input current to reach a conduction threshold then toggle the integration of voltage to reach threshold to toggle switching again.

note the polarized caps can usually handle up to 10% of its voltage rating in reverse voltage but is never published as such. Here they see the Vbe voltage in reverse across the cap for a half cycle.

Vbe however must never be reverse biased below -5V as indicated as Vr MAX in all datasheets.

  • \$\begingroup\$ Maybe this shows it better tinyurl.com/y9l69hgj SHow Base voltage then above Collector Voltage then Capacitor spike current.above that. \$\endgroup\$ Jan 15, 2019 at 3:18
  • \$\begingroup\$ It might actually kinda-sorta work with those values because of the LEDs. They would have an effective impedance of essentially infinity with no current, dropping down to whatever low value as they lit up. It's not exactly a circuit that you can linearize, though. \$\endgroup\$
    – TimWescott
    Jan 15, 2019 at 4:05
  • \$\begingroup\$ The Collector becomes open circuit when Vbe is negative to 0.5V as well as the LED , but when conducting, the LED equiv cct= Vf+15 Ohms for a 5mm part. so R is irrelevant and Vf is irrelevant except for reducing the collector current by 9V-Vf. So the LED's have the same effect as say 2V/20ma= 100 Ohms or 3V/20mA= an additional 150 ohms roughly but does not affect the frequency., It is the Rb C and supply voltage that affects f . The LEDs only affect the f a few % \$\endgroup\$ Jan 15, 2019 at 4:12
  • \$\begingroup\$ Also low hFE raises the f greatly until it stops \$\endgroup\$ Jan 15, 2019 at 4:18
  • \$\begingroup\$ see here tinyurl.com/ycvy74y9 it oscillates but Vbe<< -5V the absolute max so the transistors burn out the base emitter junction from exceeding the reverse voltage \$\endgroup\$ Jan 15, 2019 at 4:22

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