I am trying to make a simple ASK modulator circuit using an npn transistor. It works acceptably at lower frequencies. However at higher frequencies, it doesn't work properly. The amplitude of output wave when supposed to be 0, is nowhere near to 0. PFB the images showing the circuit and the simulation diagram-

enter image description here The software used is multisim on a windows machine.

enter image description here

Blue is the carrier wave of 27MHz The red digital wave is 150 khz that is to be ask modulated. Another red wave is the output wave whose amplitude is clearly non zero for the negative cycles of the digital wave.

Both the input signal's are 3V at peak levels.

Additionally, output wave's +ve amplitude is as much as carrier but not the -ve one.

I don't have much of domain knowledge in this field. I tried searching, but most of the text is out of my understandable levels. Kindly help me achieve the desired output.


Modified the digital signal to range from 0 - 3V. The waveform produced is like -

enter image description here

EDIT 2 -

As per Peter's suggestion, biased the carrier wave to range from 0-6 V. PFB the waveform. It's much better, but still not the ideal as shown in the texts.

enter image description here

  • \$\begingroup\$ What are the 27MHz signal and control (modulation) drive levels? I suspect I know what is going on, but we will need that information. \$\endgroup\$ Mar 30, 2019 at 14:46
  • \$\begingroup\$ the voltage levels are 3V for both carrier and pulse \$\endgroup\$ Mar 30, 2019 at 14:49
  • \$\begingroup\$ @PeterSmith i have also edited question to include the info \$\endgroup\$ Mar 30, 2019 at 15:01
  • 1
    \$\begingroup\$ Applying a bipolar signal to this transistor "switch" is very odd. Perhaps you expect complete switching ON/OFF of the 27 MHz carrier? Is your 150 kHz modulating signal bipolar, or something like a logic signal ranging from 0 to +3V? \$\endgroup\$
    – glen_geek
    Mar 30, 2019 at 15:26
  • \$\begingroup\$ Wouldn't the ouput capacitance of the bjt and that big emitter resistor create a voltage divider \$\endgroup\$
    – Linkyyy
    Mar 30, 2019 at 15:37

2 Answers 2


Your drive levels for both 27MHz and control appear to be +/- 3V, which is not the way to drive a grounded NPN.

Both signals should be positive with respect to the emitter at all times.

You could either bias the transistor at -3v (rather than ground from the emitter resistor) or keep both signals between 0 and 6V (the base drive doesn't really need that much drive, though).

Note that a transistor inserted with collector and emitter reversed will still produce transistor action (but at a very low current gain).

The reason that the negative part of the output carrier is more positive than the original drive when the switch is ON (base is high, collector is approaching -3V) is because you have forward biased the collector base junction which will drop about 0.6V although there will not be much current in the collector (which is what the plot shows).

When the switch is off (base at -3V) the emitter will be at 0V under normal conditions, but the capacitance of the transistor is very probably transferring sufficient energy to the 2k emitter resistor to get you the output seen in the plots.

There will always be a bit of reverse conduction in this setup due to transistor output capacitance (a TO-92 package has quite a bit). This will give (as noted) a voltage divider between the effective reactance of the collector - emitter path and the emitter resistor.

I modelled it up in LTSpice and the reverse current path (from the 27MHz signal) is quite clear.

I ran this with a couple of different carriers (27MHz and 2.7MHz) to show the effects of junction capacitance.

Current through emitter resistor carrier = 27MHz

Current in emitter resistor with carrier at 27MHz Current through emitter resistor carrier = 2.7MHz

Current through emitter resistor with carrier = 2.7MHz

To get close to theoretical response, apart from reducing the size of the emitter resistor (which you have done) look for RF transistors (therefore optimised for things such as junction capacitance). They are generally quite affordable.

Here is the result using a 470 ohm emitter resistor and a BFR183 RF transistor:

Circuit with BFR183

The BFR183 is 21p at qty 10 from Mouser (convert to your currency).

The HT12 encoder doesn't have a high current capability, so I would use a buffer such as a 74LVC06A between the encoder and the modulator.

  • \$\begingroup\$ Hi. As per your suggestion, i have modified the carrier signal to range from 0-6 and digital form 0-3. Thanks to these, the results are much better. However still there is a wave when base is at 0. PF the waveforms in part of the edit. \$\endgroup\$ Mar 30, 2019 at 16:17
  • \$\begingroup\$ See my update with reference to carrier frequency and capacitive feedthrough. \$\endgroup\$ Mar 30, 2019 at 16:19
  • \$\begingroup\$ THat's a great wave. I was able to get almost that by reducing the emitter R to 200. Regarding the noise thing, do you know of any other transistor of comparable price that can help achieve almost the ideal theoritical response. \$\endgroup\$ Mar 30, 2019 at 16:25
  • \$\begingroup\$ Thanks for answering my queries. Please allow me to ask one more. The digital waveform is actually supposed to come from ht12 encoder. Would the same circuit work fine in that case as well. There is no option to add encoder in multisim so couldn't check. \$\endgroup\$ Mar 30, 2019 at 17:10

Since product recommendations are restricted here, a generic modulator is described. These are available off-the-shelf.
The innards of a double-balanced-mixer (DBM) can perform this function (circuit described below):DBM schematic
However, the amplitude of the input is restricted to a small signal. If this is part of a transmitter chain, some amplification of its output will likely be required. Note that many commercial DBMs are intended for impedance levels of around 50 ohms, and won't work well at much higher resistance levels.
The three ports of a DBM might be called RF port, IF port, and LO port. Your input signal goes to RF port, Output signal is taken from LO port, and modulating signal is applied to IF port. A DBM must be have its IF port extend to DC (some don't).
In this case, the modulating signal is a 3V logic square wave of 150 kHz, having 250 ohm source resistance. Blue is the input signal (27 MHz), Red is the output signal, Green is the modulating signal:
waveforms, DBM input:(blue) output:(red) modulating logic:(grn) Note that the rise and fall times are fast, causing spurious emissions at frequencies other than 27 MHz. This is unavoidable with logic-style (ON/OFF) modulators, but some effort is often added to slow rise and fall times, so that modulation artifacts are limited to a frequency region close to 27 MHz.
A dead-simple method of modulation might simply turn off the DC supply to the 27 MHz oscillator, killing its oscillations - sometimes done in toys.

Another generic device type capable of ON/OFF modulation is an analog switch or another: video switch. Some of these may be unable to turn off completely, allowing some small-amplitude carrier to sneak through via internal capacitances. Some of these switches characterize their On/off attenuation in their data sheets, at various frequencies.

  • \$\begingroup\$ Thanks for your effort. However, this circuit looks intimidating to me and i was somehow able to achieve desired output by following Peter's answer. Also i would point that the dead simple method was in my mind, but actually the digital wave is coming from ht12 encoder so not applicable here. \$\endgroup\$ Mar 30, 2019 at 16:59

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