I'm working on an ECG/EMG circuit and instead of duplicating the input, I've used transistors to route the signal to a specific in-amp depending on the user's selection. This is what I came up with:

enter image description here

It works in simulation, but I'm not sure if it's going to work in real or not. The idea is pretty simple. The user pulls one of the SLCT pins high and that pin activates a certain set of transistors and these transistors route the signal to in in-smp of a certain device (ECG or EMG). SIG1, SIG2 are connected to the electrodes.

The base resistor (according to my weak knowledge) doesn't matter in this application since I don't care about the amplification or the collector current. I just want the whole signal to go through. The 1M ohm resistor is to prevent the signal from going to ground.

The transistor I'm using is 2N3904.

Am I doing it right, or there is something I messed up? Put in mind that the vital signals are too weak and noisy.

  • \$\begingroup\$ "prevent the signal from going to ground"? What does that mean? \$\endgroup\$
    – Hearth
    Oct 23 at 0:13
  • 2
    \$\begingroup\$ That probably won't work for this situation, the bias current for the transistors will affect the signal too much. Why don't you want the signal to go to both amplifiers in parallel? If it really is necessary to do that then an analog switch will be much better, cheaper and smaller. \$\endgroup\$ Oct 23 at 0:28
  • \$\begingroup\$ You want your inputs going directly into an inamp. \$\endgroup\$ Oct 23 at 0:34
  • 1
    \$\begingroup\$ @Moe i understand. Don't do that. \$\endgroup\$ Oct 23 at 1:20
  • 1
    \$\begingroup\$ Bipolar transistors can be used to do analog switching but the big problem is where the bias current goes. The bias current through the 240ohm resistors has to go into the signal source, that is a lot of current and will almost certainly compromise the input signal. \$\endgroup\$ Oct 23 at 1:24

Unfortunately, bipolar junction transistors (BJT) won't work as you intended here.

I believe you are assuming that a BJT, when switched on, is like a very low resistance between collector and emitter, but this is not the case. BJTs are ill suited to the task of forming switchable high/low impedance paths from input to output, because their base-emitter junction is effectively a diode, which ties the base and emitter potentials to within 0.7V of each other.

Therefore, in the configuration you have drawn, the emitter always "follows" the base signal, not the collector. in fact, it's called an emitter follower" for that reason, which you can read about in the Wikipedia article.

What you require is either some kind of MOSFET design, or better, an analogue switch, which does the job for you.

Even MOSFETs will require carefully crafted potentials. Even though they are more like the switchable impedance drain-to-source path you require, MOSFETS have a parastic body diode, which makes that simplified model of their behaviour only true for certain gate potentials. Also, for a MOSFET to be "on", it relies on the gate potential being very different from the potential of the signal being switched, which is not trivial to achieve.

For these reasons analogue switch ICs typically do not use BJTs as signal path switches. They make use of two or more MOSFETs to get the job done, and they have all the additional circuitry necessary to ensure that gate potentials are appropriate for any given signal potential.

Analogue switches, when "on" provide signals paths of just a few Ohms, or milliohms in some cases, and that path is fairly independent of the switching circuitry. However, due to the gate potential constraints I mentioned, they all require that the analogue signal potential stay between the power supply potentials. Analogue switch ICs usually have three power supply pins, ground, positive and negative. The signal being switched cannot ever exceed either the positive or negative supply voltages.

Another cause for concern is amplitude of the analogue signals themselves. If they require amplification, then that should be done as early as possible. Every stage they traverse in some system (even an analogue switch) is adding noise to the signal. By amplifying at a late stage, each prior stage is adding small amounts of noise to a small signal. By amplifying first, each subsequent stage is adding a small amount of noise to a large signal. Obviously, in the latter case signal-to-noise ratio is much better.

Lastly, don't rule out relays. They provide complete isolation of analogue signal from the switching system, and don't suffer the amplitude restrictions imposed by analogue switches.

So, I have two recommendations:

  1. Amplify first, switch later.

  2. Use an analogue switch IC or relay.

Some analogue switches you might consider are:

  • \$\begingroup\$ "Amplify first, switch later" made me think about something. Amplifying the signal then switching it using MOSFETS won't get the job done? since the gate is isolated, the bias voltage shouldn't affect the input vital signal, right? \$\endgroup\$
    – Moe
    Oct 24 at 2:00
  • \$\begingroup\$ The switched signal affects the Miller capacity (obviously, large signals more than small signals). I guess it work kind of okay if the voltage supply to the gate has sufficiently low impedance (look up the Rg of the MOSFET; it's not just just the gate voltage source which is the limit). Whenever the Miller capacity changes, the gate voltage changes as well until the gate voltage source has "ironed out" the Miller effect again. Now, IIRC, ECG signals are of pretty low frequency, but EMG signals go up into the kHz range. -- TL;DR version: it's...complicated. \$\endgroup\$
    – Klaws
    Oct 24 at 12:15

Can I use transistors to route/switch ECG signals?

Short answer: NO.

ECG signals are in the +/-0.5-1mV range, so too low for any transistor type circuit. The schematic you show will not work under any circumstances since you are unable to provide base current to actually switch on the transistors.

I'd suggest that the best possible solution is to use Mercury wetted reed relays. These provide excellent isolation and very low contact resistance. Here is one potential relay the COTO 7000 HG Notice here that this is the lowest possible contact resistance you can get, and more than able to handle the the extremely low signal levels your require:

enter image description here

  • \$\begingroup\$ As far as I know, the problem is not with the base current. The problem is that the signal is too weak that the collector current will follow the base current. So what if I amplified the signal first then replaced the BJTs with MOSFETs? \$\endgroup\$
    – Moe
    Oct 24 at 2:04
  • 1
    \$\begingroup\$ @Moe, assuming that the input impedance of the ECG/EMG is above 100k Ohm, you definitely have a problem with base current. The schema you show will not work under any conditions. Amplifying the signal of course gives you a greater amplitude to work with, but I'd assume you then would need amplitude controls on your ECG/EMG. Look at the datasheet for the 2N3904 to see how hopeless your situation is ...VCE(sat is about 200mV and Hfe is probably < 30 at the sort of currents you are discussing. \$\endgroup\$ Oct 24 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.