I'm trying to get a 27V 120Hz square-wave, so I tried this circuit :

enter image description here

I get a decent 128Hz signal with 9V, which is close enough. However, if I increase the input voltage past a certain point, the frequency increases too. For 27V, it is about double, 250Hz. I get no such increase in LTspice simulation so I guess it is due to some real-world physical limitation of the components.

What is happening and why? Can I/should I use such a circuit with 27V or is it too high?

  • 3
    \$\begingroup\$ To remediate the mentioned reverse-biased base-emitter junctions, apply a reversed diode at the BE junction. This also changes the frequency. (But makes it less dependent on the BJTs, too.) One should also be able to provide you with a reasonably well-developed prediction from theory about the expected rate when you change voltages (hints here) and how to re-design to achieve about the same rate when changing voltage. Out of that study, there may also be some interesting ideas about how to make the rate relatively immune to Vcc. \$\endgroup\$
    – jonk
    Dec 5, 2021 at 23:10
  • \$\begingroup\$ Try adding zeners from B-E on the transistors to get the simulation to more closely match reality. Eg. 8.2V KDZ8_2B in LTspice, in series with 1N4148 diodes to avoid forward biasing them. \$\endgroup\$ Dec 6, 2021 at 3:04
  • \$\begingroup\$ Bastien... One way that comes to mind right now is that jacking up both BJTs on a shared emitter resistor will definitely reduce the dependence upon Vcc. The problem is that the output voltage will no longer reach ground level. It will ride on a large DC base, instead. But if you are looking to turn this very circuit into something that is Vcc independent, that would be the direction I'd head. It's just one more cheap resistor and it moves in the right direction of Vcc-independence of the oscillating frequency. The idea here is that you want a constant current source at the shared emitters. \$\endgroup\$
    – jonk
    Dec 6, 2021 at 3:39
  • \$\begingroup\$ Bastien... If interested, I'll elaborate. It also completely solves the base-emitter avalanche problem without needing added diodes or zeners. However, keep in mind that this does mean you may need an added stage to get the rail-to-rail output you may want out of it. \$\endgroup\$
    – jonk
    Dec 6, 2021 at 3:40
  • 1
    \$\begingroup\$ Did you try my suggested solution with same transistors and LEDs? \$\endgroup\$ Dec 7, 2021 at 17:53

4 Answers 4


The reason is that during switching cycles, the base of each NPN is pumped below ground. Most NPN base-emitter junctions will break down and degrade with 6-9 V reverse bias.

LTSpice and other simulators don't simulate this breakdown. This is why simulation results are different from measurement.

For 27 V output, it's best to run this from 6 V, and then amplify to get 27 V. You can do that with another NPN driven from one of the collectors with about 10 k. This will also isolate the oscillator from variable loading effects of the 27 V signal.

Your NPNs may have been slightly damaged by running at 27 V.

You could instead replace the 47 nF with 10 nF and connect 47 nF from each NPN base to ground. This will attenuate the 27 V signal by about 5x and 'save' the NPNs. The frequency will now depend on (47nF + 10 nF).

  • \$\begingroup\$ Thanks, i'll try that. Shouldn't a simple diode on the base work too? Which would be better? \$\endgroup\$
    – Bastien
    Dec 6, 2021 at 22:33
  • \$\begingroup\$ a zener or LED < 5V would be better to reduce C \$\endgroup\$ Dec 7, 2021 at 0:50
  • \$\begingroup\$ A simple diode works but still gets huge currents spikes from e.g. left side Rc=1k supplies 26 mA to right side base which is amplified by hFE and cross coupled to the base on the left side reverse diode causing > 1A thus splitting Rc into two R's limits the spike current for a smooth sawtooth as I showed in my answer. But yes an AC couple "half- bridge amplifier" would work better from a lower signal, but all depends on bridge logic input levels and output currents needed from load. \$\endgroup\$ Dec 7, 2021 at 19:42

You are likely running afoul of too much reverse voltage from BASE-to-emitter: (from ON semiconductor data sheet

PN2222A data sheet

The base-emitter junction looks like a zener diode when you reverse-bias it. You shouldn't do this - it can damage the transistor in subtle ways, given time.
The circuit should be safe for low supply voltages, and should give a stable frequency that is not supply-voltage dependent.

  • \$\begingroup\$ It is true that when VCEO and VEBO are driven by voltage sources the MAX RATINGS will cause early failure, yet when current limited as in this example by V+/Rb=Ib and Ib * ~10% of hFE when saturated , the collector is current limited so the power in VEBO * IEBO = Pbe is also limited and acts like a zener with the avalanche effect. The same is true for VCEO*Ic=Pd so it can act as a Zener something > 40V with low power <50% of rating and still work in most cases. \$\endgroup\$ Dec 7, 2021 at 17:51
  • \$\begingroup\$ @TonyStewartEE75 Those base-emitters get quite a large pulse of current when "zener-ing". Sure, its a short pulse, but current density is high (LTspice says max pulse amplitude is 1.5A in OP's circuit). I'd still be wary of long-term damage. \$\endgroup\$
    – glen_geek
    Dec 7, 2021 at 18:18
  • \$\begingroup\$ yes I forgot, the other Rc=1k is what drives Ib to controls its own a large -IEBO in the other device and not the Rb offstate 120k ramp.. i.e R1 boosts Ib1 in Q1 that drives Ic1 into -Ib2. What did LTSpice say was the average Pd in B-E during spike? \$\endgroup\$ Dec 7, 2021 at 19:19

Define your real objectives in specs 1st; Frequency tolerance, Rise/fall time limits, output impedance or load impedance, capacitance. THIS PROCESS OF LISTING DESIGN SPECS IS MANDATORY FOR ALL GREAT DESIGNS. (Even if you change them after testing to correct poor assumptions) Then you can choose the best topology.

The flaws in this design are obvious to any senior designer and are well listed in comments and other answers. But the real goal is how to make it better by good specs, then better choices. However, I will show how to stabilize this primitive design and protect Veb and Pd max of each part.

It can be made safe using a reverse diodes or reversed current limited LEDs across Vbe , but then it ramps up faster with a small ramp voltage from -0.7 to 0.6V so Rb may need to be reduced < 50% and C must be increased 20x to get full voltage.

Without the added Rcl, current limit the Transistor acts as a 1 ohm switch with 26 V charged on large caps , displacing large current negative spikes quickly but safely > -6V if rated large enough.

Rb/Rc must be greater than hfE to get full swing to a non saturated Vce around 1 to 2 V and C must be increased to lower f.

Then the 1k resistors must be made from 4 x 1k 1/4W Rs in 2S2P to dissipate and share ~ 670 mW.

For even a better solution using BJT's only, use White , Blue or Green LEDs and current limit with Rcl= 680 ohms on the collectors to just above 30 mA with same 1k pullup Rpu, for the pulse current to be safe and also more stable. and reduce the 1k power dissipation to 250 mW 50% d.c. at 85'C . (hot)

Proof of concept

enter image description here

The power diode version runs faster as explained.


The multi-vibrator works well with Vcc=5V providing a -5V base Sawtooth to ramp up to Vbe= 0.6V then switch again. But higher Vcc's demand a clamped negative Vbe voltage such as a 3V LED or 4.7V zener. (reversed to clamp negative voltages). or even a two 2V LED's in series. They won't illuminate as the current limit Rc and low duty cycle only clamp for a very low duty cycle then the RC ramp to Vbe controls the cycle time. People get away with higher 9V Vcc's because the average power in Veb reverse current dump is < 150 mW, so it appears to survive but can wound the junction to fail sooner than expected.

  • \$\begingroup\$ That would lead to a circuit with a very strong variation of frequency with temperature -- increase by about 2m/700m = 0.3 % per °C. Agree with the 1k's -- better to replace with 10k. \$\endgroup\$
    – jp314
    Dec 6, 2021 at 0:30
  • \$\begingroup\$ @jp314 I don't think so, the BJT's just switch low current across Vce for < 50 mW at 200'C/W is < 10'C rise and Vbe won't change much. It's certainly not how I would generate 27V 120Hz, but what the heck . this outta be close enuf for gov't work. The diode power is also VERY low duty cycle, so these LED's won't even be bright. \$\endgroup\$ Dec 6, 2021 at 1:13
  • \$\begingroup\$ THe power diode version won't even get warm but, you are right about the high current but very fast fall and slow rise. tinyurl.com/y32mr7aj FWIW @jp314, not perfect but then no real specs, so it's perfect with what all was requested. \$\endgroup\$ Dec 6, 2021 at 1:28
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    \$\begingroup\$ This is way above my head. I'm just trying to light up a christmas leds string so my requirements are pretty basic and tolerance is high to say the least. But thank you nonetheless. \$\endgroup\$
    – Bastien
    Dec 6, 2021 at 22:16
  • 1
    \$\begingroup\$ This is so heavy-handed about all the 'spec it completely/TBD first' stuff, heading towards evangelism of it. No, full specifications are not mandatory and are a rubbish sole way for a project to progress and for learning. It's a mix. Doing a personal project, experimenting, learning from mistakes like @Bastien is such a valuable way to learn and strongly encouraged. Like anything, engineering has to appeal to a learner and boring oneself in lengthy study and planning before doing is not the way. I really clearly get the point you're making, all the pros and cons of it all, but it's excessive. \$\endgroup\$
    – TonyM
    Dec 7, 2021 at 13:52

Increasing the supply voltage increases the rate at which C1 and C2 charge through the resistors connected to them, which causes the trasnstors to turn on sooner, increasing frequency. You should see this in LTSpice.

For example, when Q2 is on, the left side of C1 is near ground, the right side is charging through R2. When C1 has charged long enough that the right side is about 0.6V, this will turn on Q1, initiating the next portion of the clock cycle.

  • 1
    \$\begingroup\$ Not the correct reason - yes, the rate of charging n erases, but so does the voltage change needed. These cancel resulting in substantially constant frequency. \$\endgroup\$
    – jp314
    Dec 5, 2021 at 22:55
  • \$\begingroup\$ I overlooked that, thanks for the correction! \$\endgroup\$
    – 65Roadster
    Dec 12, 2021 at 15:12

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