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I was searching around for circuits that could generate a square wave and I noticed that they all have only one output and can't generate a current. For example: enter image description here

Now, measuring across the positive side of C2 and ground you get a voltage waveform that looks about right: enter image description here As you can see, the waveform is ok for such a simple circuit. But, the moment you try to use this voltage as a current, for example to power a transistor, the circuit shorts and you lose all chance of a square wave. This is so bad, you can place a 100M ohm resistor across C2 and ground and you lose the square wave. Now, I don't think this is how this circuit is used (it would be used in a MOSFET), but pretty much all square wave generators create voltage differences that short when you try to use them to power a current. So my question is, what are some square wave generators that can generate a current, or do they all generate voltages and you have to use a MOSFET?

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    \$\begingroup\$ What do you need to drive that requires current? How much current compliance do you need? There are plenty of ways to use 2 BJTs in an oscillator that will drive LED levels of current, for example. But perhaps you need a lot more than that. Also, please disclose your voltage source, as well. \$\endgroup\$
    – jonk
    Commented Jul 8, 2021 at 16:56
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    \$\begingroup\$ @jonk I am trying to create a boost convertor and I only really need a couple of mAmps of current. I know that boost convertors are usually run with MOSFETS and in that scenario, this circuit would work but I don't have any and I don't want to wait to start building. \$\endgroup\$
    – Fateh A.
    Commented Jul 8, 2021 at 17:00
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    \$\begingroup\$ I doubt your statement "you can place a 100M ohm resistor across C2 and ground" and it stops. Make sure your simulator does not interpret 100M as 100m\$\Omega\$. \$\endgroup\$ Commented Jul 8, 2021 at 17:06
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    \$\begingroup\$ @jonk I am hoping to use the design in the Wikipedia page for a boost convertor. link \$\endgroup\$
    – Fateh A.
    Commented Jul 8, 2021 at 17:08
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    \$\begingroup\$ OP's circuit is entirely symmetrical, which should give equal switching on/off time. Its not - something is wrong or not shown. \$\endgroup\$
    – glen_geek
    Commented Jul 8, 2021 at 17:22

2 Answers 2

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Rather than making a switched current source, it appears that your actual goal is to drive a load with your multivibrator circuit. In that case you just need to buffer the output using an extra transistor.

Without the transistor the output impedance is 1K ohms, so you could only drive very small loads (like 10K resistors or similar). By adding a transistor the output impedance is divided by B (the transistor gain). So for the case of the 3904 (with B=100) your output impedance now becomes 10 ohms and you can drive much larger loads.

Note that if you want to drive an LED as the load you will still need a series resistor (say 100 ohms or so).

You can get even more output power by using a transistor with a higher gain, a Darlington transistor, two transistors, or even a MOSFET transistor.

enter image description here

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Generating a current requires a load with a source that is much lower resistance so that it is not affected by the load.

For an Astable Oscillator to be isolated from the load, you need a current buffer or impedance divider that can reduce the impedance and therefore the effects of loading back to the source. In all cases with feedback, where the loop gain = 1 the phase must be positive for feedback for voltage but must also be buffered with current gain to generate current.

Active circuits with high negative feedback reduce both the input and output impedance by the feedback gain. Such as the case with virtual grounds between Op Amps where the small signal output impedance is nearly 0 yet active current limiting circuits will raise the impedance when this occurs and the gain reduces to 0 when it saturates to either rail.

Now the classic Astable Multivibrator in your question has very high resistor values and as we know the current gain drops rapidly when saturated, we rate the Vce(sat) typically at Ic/Ib=10 could be done for a ratio of 20 or 50, but that requires a linear hFE that is 10x this ratio. The output impedance can be seen as Rb/ hFE but keep in mind hFE drops to near 10% of the linear hFE, so an external buffer like a Darlington Emitter Follower or the complementary NPN+PNP of the same and then with negative feedback at unity gain to increase the BW and current out with lower R out.

This is easy to see in logic with the high input impedance of FET's and the low RdsOn which can be used in logic. For example 74HCxxx logic at 5V is typically 50 Ohms nominal. A power FET Half bridge has a similar P+N complementary driver but at much lower RdsOn or may be dual Nch totem pole drivers with the lower side creating a PWM signal to rectify with a cap and diode to create a boost voltage for the high side Nch to be greater than the Vdd supply.

Considering Switched Mode Power Supply (SMPS) design, FET's may be chosen with RdsOn in the sub-milliohm range to gain current, but the tradeoff is when "Ron" is reduced the junction switching capacitance is also increased and this can pose conflicts in the scaleability of RLC parts in SMPS designs as the FET Ciss and Coss affects the transition slew rate that occurs must be considered.

After you learn some theory of all the various configurations of Buck and Boost, it is easier to learn how designs are achieved from websites with custom solutions and a few good books by SMPS experts.

https://webench.ti.com/power-designer/switching-regulator/select

https://www.onsemi.com/support/webdesigner+

https://www.analog.com/en/design-center/design-tools-and-calculators/power-management-tools.html

https://www.microchip.com/en-us/solutions/power-management-and-conversion/intelligent-power

https://www.yumpu.com/es/document/view/19291874/digitally-controlled-power-supply-design-wizard-power-electronics

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