Background: I have a μBITx v6 HF transceiver. It has an audio input and output, and a mechanism where the transmitter is keyed up when the ring connector (called PTT, push-to-talk) from the audio input is shorted to ground.

In the radio hobby, there are methods of communicating called digital modes. In essence, you connect the radio to a computer, which acts as a modem that decodes and encodes digital signals into audio, that are fed into the radio. However, it's not as simple as just plugging two aux cords into the radio, I need some control circuit to tell the transmitter when to key up.

The simplest and cheapest method of doing this is to create a VOX circuit. When the circuit detects audio at the input, the PTT is shorted to ground and the transmitter is turned on. When the audio is not there anymore, after a short delay of a few milliseconds, PTT is no longer shorted to ground, and the transmitter stops, allowing the transceiver to receive audio from the airwaves again.

Steps I tried:

I had a breadboard handy, and prototyped a VOX circuit according to this design.

Using a bench DC supply at 9V (switched mode), I powered up the VOX circuit. I also had a 12V lead acid battery, which I used to power the radio. I used an alligator clip to connect both grounds together.

When I connected the audio input of the radio to the VOX circuit, it immediately keyed up regardless of what setting I had at the potentiometer (R6). This happened with no audio source connected. When I disconnected the base of Q1 from the rest of the circuit, the transmitter stopped keying up, but immediately started when I reconnected the base to the circuit. This tells me that something is causing Q1 to saturate when it shouldn't be saturated.

Can anyone provide other insights into what the problem could be? Again, the issue is that Q1 is allowing PTT to short to ground when I did not expect that to happen. No audio was applied to the input.

Just to make sure that my transistor wasn't faulty, I used a multimeter in diode mode to test the transistor. Everything looked normal. I don't think that's enough to rule out a problem with the transistor, but I tried substituting that transistor with other ones of the same part number. The same issue occurred.

I also tried to use the battery for both the radio and the VOX circuit, skipping the power supply altogether. This did not fix the issue.

I noticed that my diode, which has part number 1N4007, does not match what the schematic specified (1N914). Could that be an issue?

EDIT 1: I tried jumping R5 according to a comment, so Vcc is directly connected to the potentiometer. This did something, but not exactly what I was looking for. Now, if I connect the radio to the circuit, PTT does not trigger, but the moment I connect the computer to the circuit, it triggers. It also caused some RF feedback, which I solved with some ferrite chokes on the line. I'm going to try powering this with a 9V battery instead to see if the RF feedback problem disappears entirely.

EDIT 2: Using the 9V battery worked. However, the circuit still keys up. I tried connecting this ground loop isolator I purchased online a while back, and that caused the PTT to trigger as well. This means that any audio device will cause a short to ground when it is undesired. To counteract this, I will have to access the PTT line on the radio itself and add another coupling capacitor similar to C1 for the connection to the radio. I will post another edit on the results.

  • \$\begingroup\$ Which op-amp you used? I don't think the diode type matters much here, even though 1N914 is a small-signal diode and the 1N4007 is a mains frequency rectifier. \$\endgroup\$
    – Justme
    Commented Jan 31, 2021 at 20:16
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 I'm going to try to jump R5, will post an edit later. \$\endgroup\$
    – Expectator
    Commented Jan 31, 2021 at 20:21
  • \$\begingroup\$ @Justme The text on it says: M28AT LM1458N \$\endgroup\$
    – Expectator
    Commented Jan 31, 2021 at 20:22
  • \$\begingroup\$ The circuit says it is for microphone use. Microphone output voltage levels are extremely small compared to line level output voltages from computers, and the circuit had a huge 100x gain. Try adjusting the gain lower, replace the 100k feedback resistor to like 1k or maybe a 10k potentiometer to adjust gain. \$\endgroup\$
    – Justme
    Commented Jan 31, 2021 at 20:26
  • \$\begingroup\$ @Expectator The opamp at left is AC-coupled. Without input present, that opamp should present about half the supply voltage at its output. Leave the input unconnected to a sound source. Measure its output with a VOM to verify that the output is about half the supply value. Add an LED and series current-limit resistor between PTT and the (+) supply. Now adjust the potentiometer. You should be able to make the LED go on and also off. If that behavior doesn't happen, then something is wrong with the circuit itself. If it does, then Justme is probably right. \$\endgroup\$
    – jonk
    Commented Jan 31, 2021 at 22:01

3 Answers 3


The answer is considerably simpler than other answers have indicated.

In this circuit, the second 1458, which is used as a comparator, really needs to be powered by a negative voltage on pin 4. Once you do that, you should ground pin 3.

The problem is that the 1458 (which is basically a dual-unit version of the now-ancient 741) was intended for split-supply operation. As a result, its output can drive around the midpoint of the supply voltages, but was never intended to drive close to the supply voltages themselves. Not only that, but the behavior of the IC is not really specified for voltages other than +/- 15 volts, even though it will typically operate down to +/- 5 volts. But look at the data sheet, especially VOM, the maximum voltage swing. Notice that for large output resistances (>10k) it is only guaranteed to get to +/- 12 volts, which is 3 volts away from the power supplies. For smaller resistances (>2k) the swing is only guaranteed to be within 5 volts of the supply.

This means that, even when the amp is driving as low as it can, it will still be several volts above the negative supply (ground in this case), so it still will provide voltage through D1 to turn on the transistor.

To make life more complicated, there's no guarantee that the 1458s you have will behave exactly this way - you're using them way outside the nominal range, and various manufacturers can design their circuits differently.

In this case, there's an easy test to see if this is happening. Assume that your power supply is at 12 volts. Ground the input, crank R6 all the way up. you should see about 6 volts on pin 5, and about 7 volts on pin 6. Now look at pin 7. You'll probably see a couple of volts or so. The transistor will be on. Now crank down R6 so that pin 6 drops to about 4 volts, or at any rate less than pin 5. Now you should see 9 or 10 volts on pin 7. The thing is, either voltage is enough to turn on your transistor.

There are two approaches you can use.

First, provide split supplies to the op amp. Two 9-volt batteries will work, if connected like


simulate this circuit – Schematic created using CircuitLab

Also, as previously mentioned, tie pin 3 to ground and get rid of the two resistors. In this case, for +/- 9 volts, the output of the second op amp will swing between about + and - 6 volts. The +6 will turn on the transistor, while the -6 will be blocked by the diode.l

A second, rather easier approach is to give the op amp a bit of help. Don't bother with dual batteries - keep what you've got. Get a 5-volt zener diode and connect it to the second op amp like this


simulate this circuit

What this will do is prevent any current from flowing into the transistor unless the op amp output is greater than about 7 volts. Note that this may not occur if you're using a 9 volt supply/battery, and it may be necessary to get a lower-voltage zener, probably about 3.3 volts.

Finally, you need to add some capacitors. Add something in the range of 10 to 100 uF between pins 4 and 8 of the op amp, and 1 to 10 uF across R4. Currently I expect you're seeing a lot of noise on the outputs, and this should help considerably.

  • \$\begingroup\$ I didn't have a zener diode on hand, and got rid of both resistors instead (it's a voltage divider, right?) Since I got rid of R4, where am I supposed to put the 10 uF capacitor? In parallel with the short to ground? Also, your solution is working to an extent! When I attach an audio source, it still accidentally triggers, but reconnecting the base turns it off. When I apply a louder audio signal, the transceiver starts TXing, so clearly the comparator stage is starting to work. \$\endgroup\$
    – Expectator
    Commented Feb 2, 2021 at 20:27
  • \$\begingroup\$ @Expectator. With R2/R4 gone, there is no need for a 10 uF cap. That was just to smooth out any noise which the Vcc might produce, and give a cleaner signal. Although, in looking again, connecting a 10 uF from pin 6 to ground is a good idea. This will clean up any noise on the threshold level at the comparator. \$\endgroup\$ Commented Feb 3, 2021 at 13:26
  • \$\begingroup\$ @Expectator - Also, I overlooked a point. If you've gone to a split supply, you should short out R7, and connect the bottom of R6 to ground. This will allow the comparator to trigger on lower-level signals. Depending on your signal source, you may have to play around with the value of R5 to get a useful sensitivity range when you adjust R6. My apologies - I should have picked that up the first time around. \$\endgroup\$ Commented Feb 3, 2021 at 13:32

It looks to me like the mic preamp (3904 circuit) has a gain of approx. 20, or around 26 dB. Seems a bit low to me, but I didn't delve into the modulator.

Also, an LM358 is (((not))) a dual 741. The circuit page is incorrect. The 358 has more noise and more output stage distortion, but also has what were in 1971 amazing input and output stage operating ranges. It is optimized for single-supply "industrial" circuits, and is a good call for this one: fidelity is irrelevant, but swinging close enough to its negative rail to turn off Q1 is critical.

My guess is that the VOX circuit is designed to take the mic signal directly, not from a preamp, and that the extra 26 dB of gain is overdriving the comparator stage. Decrease R3 to around 10 K and see of things improve.

At Vcc = 12 V, the output of the first opamp is resting at 6 V. The comparator trip point adjustment range of R6 is from 5.2 V to 6.8 V. For the circuit to rest in the "off" state, the trip point must be slightly greater than 6 V plus any error caused by the tolerances of R2 and R4.

With the pot at the high end of its range and the input shorted to GND, Q2 should be off. Verify that pin 5 is around 6 V and pin 6 is around 6.8 V. If not, check your wiring (again), replace the opamp, and try again.

I think the VOX circuit is designed to take the mic signal directly, not after a preamp. That extra 26 dB of gain might be overdriving the comparator stage. To test this, reduce R3 to 10 K to see if the adjustment range now has an effect.


Microphone Signal Level

The schematics are there to read. The microphone input looks like this:

enter image description here

  1. \$R_{66}\$ and \$C_{64}\$ help isolate the circuit from the main \$+12\:\text{V}\$ power rail. Think of \$R_{66}\$ kind of like a set of bed-springs for the circuit.
  2. \$R_{60}\$ tells me this input supports electret microphones. They pull up to about \$1\:\text{mA}\$.
  3. The \$R_{61}\$&\$R_{62}\$ divider pair look to be dragging almost \$1\:\text{mA}\$ and also setting the base to \$\lt 2\:\text{V}\$.
  4. That now suggests that the emitter current is also going to be \$\approx 1\:\text{mA}\$. (This is looking like \$1\:\text{mA}\$ everywhere!)
  5. That means there's a quiescent \$1\:\text{V}\$ drop across \$R_{65}\$ and that probably means that the output of this circuit should be limited to about \$\pm 500\:\text{mV}\$.
  6. Input impedance is \$\le 1.5\:\text{k}\Omega\$. This site shows values about in that area, too. The JIS C-5502 standard has as one of its values \$600\:\Omega\$ which also just happens to be a commonly found audio transformer value, too. So about half the microphone output is expected to be tossed and I'm concluding a total gain: \$A_v\approx 6\$.
  7. Since I know the circuit output shouldn't exceed \$\approx \pm 500\:\text{mV}\$ I can now say that the input shouldn't exceed about \$\pm 80\:\text{mV}\$. Or \$\approx -25\:\text{dBV}\$, thereabouts.

That said, the above tells me that the expected signal level isn't tiny.

(I had no choice but to look at the unit you have to see what it expects on the assumption that you have a working microphone. All that because you didn't specify your microphone details.)

I've learned something else, of course. I now know for certain that you are using a transceiver that accepts and powers electret microphones as well as accepting a relatively wide variety of other sources. I also know that the circuit you are trying to build and test does not supply power for electret microphones.

So if you are using an electret microphone that requires an external supply, then it will work just fine with your transceiver but will not work fine with the circuit you are working on.

My ignorance about your microphone has now become a problem in my trying to be of help.

VOX Module

The 741 opamp is a pretty poor choice, these days. It was designed back when only a few transistors could be affordably placed on an IC. I still have some 1458's from 40 years ago (and more) sitting in one of my boxes. That's how far back they go. But I understand why you'd resist replacing them without knowing circuit details.

This circuit works on a pretty simple principle. It includes a first stage giving \$A_v=100\$ (\$+40\:\text{dB}\$) and having \$1\:\text{k}\Omega\$ input impedance. That impedance isn't too far from the one I'm seeing for your transceiver circuit. In that sense, there's a match-up. (Of course, there's still my question about whether or not your microphone requires a small current to operate it.)

Given what I believe I understand about the transceiver's microphone input circuit, the voltage gain seems quite high. But I don't think that's necessarily a problem. It just means that the output may saturate near its rails, chattering between them, when you speak into a suitable microphone. Lower sound pressure levels may very well avoid it railing like that and instead deliver an amplified but largely undistorted result, too.

So I can't say this is a design flaw. I'd need to know more about the reasonable range of appropriate microphone SPLs and the kinds of levels of voicing that need to be included in activating the circuit. With sufficient research into appropriate microphones likely to be used with your transceiver and enough time to learn about different human voicing levels, I might be able to make a credible argument one way or another. But I'm not in the mood to even attempt that effort. So I must say, "I don't know. Perhaps this voltage gain is just fine."

What I can say is that there is a potentiometer in your circuit. It has a certain range of motion providing a certain range of input voltage to one side of the second stage opamp. If this range, together with the voltage gain and the vagaries of appropriate microphones and voicing levels, allows successful use then the circuit is fine. On the other hand, if just ambient background noise and electronic noise in typical microphones happens to cause the first stage to swing around too widely, then it's just never going to work because it will always be triggering the second stage. So one possible suggestion is to consider reducing the first stage's voltage gain by changing the \$100\:\text{k}\Omega\$ resistor to a smaller value.

But that's a maybe. Again, I'm too ignorant to know for sure.

The idea of the second stage is to just operate it completely open-loop. This just means that it's output will have a very, very high gain that multiplies the difference of its two inputs. Only the smallest of differences might yield an analog output that is between its two rails. Too small of a difference to be rationally expected to happen. So without any applied input, this second stage will either be always at one rail or the other, saturated out.

The idea here is that this is okay. The output of the first opamp should be centered midway between the voltage rails. If the potentiometer sets its voltage sufficiently above or below that midpoint level, then very low level audio signals, output from the first stage, will always stay either above it or below it, meaning that the PTT is either always active or always inactive. You will want to set it so that at these very low levels the PTT is always inactive, of course. (But the circuit allows you to set it the wrong way, too.)

If the first stage gain is acceptable, then you should be able to move the potentiometer to one side or the other and get these two behaviors with very low levels of input. But if the ambient (and other) background, and/or low levels of audio that you don't want to trigger the circuit, magnified by the first stage leaves you with no possible position of the potentiometer where you can inactivate the PTT, then the voltage gain of the first stage is simply too high and you'll need to remedy that fact.


I've already suggested that you try leaving your microphone disconnected from the input and using a VOM to measure the first stage's output voltage. Although you might prefer to use an oscilloscope to see how much the noise swings the output around, I'm gathering you don't have access to one. A VOM should at least show you the average value and this should be close to midway between the rails.

Another test to try is, as I also mentioned, to use an LED and an in-series current limit resistor between the PTT output and the positive rail voltage. (Polarity correct, of course.) Without a microphone input source, you should be able to adjust the potentiometer in such a way that you can turn the LED on and also turn it off. If it stays on no matter what, then you've already determined there's a problem. Either with your wiring, parts, or else because the gain is too high and noise itself is causing continual triggering.

That's all I can recommend for now.

  • \$\begingroup\$ I am not using a microphone, I am using line-level input, which is a fair bit stronger than a microphone. This is because I am operating digital modes, which need precision (outside noise can cause issues, especially for weak signal modes). The PC sends line-level baseband to the VOX circuit, which keys up the radio, and passes the audio through SSB modulation and then is sent to the antenna. In receive mode, the radio receives some signal, inserts a carrier, and does AM demodulation. This audio is then sent back to the PC. \$\endgroup\$
    – Expectator
    Commented Feb 2, 2021 at 1:01
  • \$\begingroup\$ @Expectator Line level is \$−10\:\text{dBV}\$. So yeah, that's pretty high. Adding \$40\:\text{dB}\$ means \$+30\:\text{dBV}\$. Probably way over the top. Justme's suggestion about reducing the gain (which i included above, too) is probably in the cards for you. But you FIRST should perform the tests without the audio input applied and make sure that the basics of expected functioning occurs. By the way, did you try using line level into the transceiver?? Given the design details I saw, I don't think it should work well. Not without additional attenuation, anyway. \$\endgroup\$
    – jonk
    Commented Feb 2, 2021 at 2:02
  • \$\begingroup\$ @Expectator Also, a line input is usually on the order of \$10\:\text{k}\Omega\$. Not the \$1\:\text{k}\Omega\$ I saw for the transceiver's microphone input circuit. So jacking a \$600\:\Omega\$ line output into your transceiver is not really "bridging" and I'd expect a halving of the signal right away due to that alone, if applied. Still, that's too much, so I'd still expect difficulties unless your output volume control is set pretty low. \$\endgroup\$
    – jonk
    Commented Feb 2, 2021 at 2:11

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