I am trying to design a simple RF amplifier for a 1MHz AM signal (for low-power transmission). My modulator (JFET-based) outputs ~800mV peak-to-peak. My power supply is 9V, and I want near the largest possible amplification without distortion (class A operation) - my modulator outputs a reasonably-clean sine with next to 0 harmonic distortion, so I might avoid filtering.

This is what I came up with:


simulate this circuit – Schematic created using CircuitLab


As I understand, voltage gain is Rc/Re which is 6.8 here. That should mean an output of Vout = 800mV*6.8 = 5.44V, reasonably big and there is still a big room for non-precise Q point setting. I picked those two resistor values based on what I have at hand.

I want to set the Q-point so the output is at 4-5V. That should be at 0.74-0.59 mA. To enable a more precise setting, the lower resistor (1.7k) is a 10k pot.

As I understand, the BC238 should be capable to this amplification at this frequency, the datasheet has a 85MHz transition frequency at 0.5mA, so it should still have a bandwidth of f' = f / 6.8 = 12.5 MHz.

I built up this circuit on a piece of copper board I cut lands into with miniature powertool with a diamond cutting bit. I used an old analog signal generator to provide the test signal, set it to ~800mV 1MHz (rather imprecise due to it's analog nature, sadly), and monitored the output with a scope. With a 100MHz 10x probe on the 1x setting I got ~700mV, and on the 10x I got ~300mV. I am not sure what is the real output voltage (700mV or 3V). (this is solved, see edit 3)

In any case, the resulting signal is smaller than the expected output. I understand that internal capacitances in the transistor start to attenuate the signal at higher frequencies, but this drastic effect was unexpected for me. I feel like I might have overlooked something simple.

PS: I have already tried adding a small (6nF-1u, the later was an electrolyt cap) cap between ground and emitter. That kind-of fixed the gain problem (both on 1x and 10x probe settings) but distorted the signal highly, the output more resembled a sawtooth than a sine.

So questions:

  • what is / how to find my real output voltage?
  • how can I increase the maximum working frequency of this amplifier?
  • is there a simple but fatal design flaw in my circuit?

Edit 1

I measured output voltage at lower frequencies. This generator can only go down to .1MHz, so from there I collected 10 data points up to 1MHz. Measurements were rather inprecise (the voltage display on my scope changed rather quickly, possibly the generator also is not perfect), but should be accurate to +- 10mV. All measurements were done on the 10x setting.

| f  | mV  |
| .1 | 500 |
| .2 | 500 |
| .3 | 490 |
| .4 | 460 |
| .5 | 450 |
| .6 | 410 |
| .7 | 380 |
| .8 | 370 |
| .9 | 350 |
| 1  | 310 |

(all those number were the measured voltages with a x10 probe, so the real ones should be those x10)

The downward going trend is still apparent from this dataset. At 100kHz, I measured a 4.2V signal with the 1x setting, which I still do not understand.


Measured two more data points - f=10kHz I got 5.04V from 800mV input, and at f=1kHz I got 3V from 400mV input. Those numbers are all around what I have expected.

The measurement problem (difference with x1 and x10) might be because my probe loads the circuit differently, and I have a big (6.8k) output impedance, but I am not sure of that.

Edit3 - probing

I am pretty sure now that the measuring problem is due to the different loading on my circuit in the x1 vs x10 mode. I experimented with a 10k-10k voltage divider and a 10V signal (1MHz). With x10, I measure 500mV that equals 5V, which is around correct. With x1, I only get around 1.5V, so I should really only trust the x10 readings.

So the question remaining is that why is gain dropping so much in this circuit?

  • 1
    \$\begingroup\$ You have a signal generator so try getting it to work at audio frequencies. If it doesn't produce the goods at audio then it won't at RF. \$\endgroup\$
    – Andy aka
    Aug 1, 2020 at 11:26
  • \$\begingroup\$ Good idea! I have done a few tests already, and it seemed to work well with lower frequencies, but I'll do some more and update the post. \$\endgroup\$
    – Uncle Dino
    Aug 1, 2020 at 11:27
  • \$\begingroup\$ You need to simultaneously plot the input amplitude too else there is no way of checking gain. The signal generator needs to be connected to the circuit when you do this. Also check this at 1 kHz, 10 kHz and 50 kHz \$\endgroup\$
    – Andy aka
    Aug 1, 2020 at 12:23
  • \$\begingroup\$ @Andyaka imput amplitude was a constant 800mV (plus the same measuring error). I'll check with smaller frequencies. \$\endgroup\$
    – Uncle Dino
    Aug 1, 2020 at 12:36
  • \$\begingroup\$ OK, that's good. \$\endgroup\$
    – Andy aka
    Aug 1, 2020 at 12:37

2 Answers 2


Nothing fatal with the circuit. It's DC bias is reasonably set, and it will amplify. One problem is no effort to include even a simple representation of a load - any load will reduce gain.
Resistors chosen are too high. Many very wideband amplifiers work with 50 ohm sources and 50 ohm loads. They might pull tens of milliamps or more from their DC supply. 800mV input might be a bit high, risking amplifier non-linear behaviour. At 1MHz, 50 ohm impedance at input/output is perhaps overkill, but this amplifier's high impedance will result in providing very little output power.

Probing the transistor collector with a 'scope probe adds capacitance, which combines with the 6.8k collector resistor to form a low-pass filter - which attenuates higher frequencies. You only need a loading capacitance of about 20pf to cause noticeable attenuation at 1 MHz. A 1X 'scope probe might have more than 100pf capacitance. A 10X 'scope probe might have 11pf.

  • Estimate a resistance (and perhaps capacitance) of your intended load. Add this to your circuit's output, coupled through an appropriate-size capacitor. You will find that gain is adversely affected.
  • Try scaling amplifier resistors downward, biasing that transistor with more collector current.

If these measures can't produce enough gain @ 1MHz., consider a tuned amplifier, with the 6.8k resistor replaced by an inductor resonated by a parallel capacitor. Doing so may risk making an oscillator, but makes achieving a reasonable power gain easier. Tuning to 1MHz is not particularly easy for a beginner, and coupling power from the collector to a load impedance adds complication too.

  • \$\begingroup\$ I think this answered all of my questions. I might add an emitter follower stage as well, so that's why I did not really care about output impedance - but I forgot to take the probe's loading into account. I think the gain is now somewhat enough, if not, I can simply increase Rc more. I have already made a few 1MHz tuned circuit (had some troubble with it, possibly low Q), but my oscillator stage is working now. \$\endgroup\$
    – Uncle Dino
    Aug 1, 2020 at 15:30

Let us assume 30pF on the collector node. (scope probe and Cob of the transistor, plus resistor parasitic).

WIth 7,000 ohms and 30pF, the time constant is 210 nanoseconds, or about 800 KHz F3dB.

I would REDUCE the collector resistor by 3:1, to 2.4K or 2.7K.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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