# What purpose are the transistors serving in this CRT oscilloscope circuit?

So this is part of a CRT oscilloscope diagram, and I don't understand the function of the transistors on the bottom right, Q18 and Q19. Their outputs (41 and 42, crossing through the dotted line) go directly to the horizontal deflection plates. Furthermore, this diagram says the base voltage is 1.2V, but all three points on each are measuring 20-30VDC for me. Keep in mind this scope is broken, something's wrong with the horizontal sweep circuit. I just need to understand the purpose of these transistors and what kind of voltage they're supposed to be carrying to even attempt to diagnose this thing. Thanks a heap in advance.

Q18 and Q19 are the X-output amplifier.

A CRT may need hundreds of volts peak-peak to deflect the beam over the width of the screen. The signals generated by the sweep generator (From Q14 and Q15 via S103) will only generate only a few volts peak to peak.

Q18 and Q19 amplify the voltage to the levels required with a few other requirements:

1. The voltage to the X1 plate needs to sweep with the opposite polarity to the voltage on plate X2 (ie when one goes up the other goes down).

2. The mean voltage needs to be constant at 105V to avoid defocusing th eCRZT or causing astigmatism.

3. The bandwidth and pulse response of the amplifier needs to be appropriate to amplify the sawtooth waveform from the sweep generator with acceptable accuracy.

Most oscilloscopes use a similar arrangement with a differential output stage.

• I don't see with what voltage they're amplifying, with the base voltage a mere 1.2V and an emitter voltage of -10V... ? Where are they getting this high voltage from? Commented Feb 12, 2020 at 23:15
• @JaredCravens - Q14,Q15 are part of the sweep generator and generate a sawtooth waveform. It is probably a few volts peak to peak at the base of Q18 after being buffered by Q16. The 1.2V is the average DC voltage. The output stage is powered by the 180v power supply. The emitter of Q18,Q19 should be at about 0.6V. Commented Feb 12, 2020 at 23:50
• Okay, so does the output stage of Q18 and Q19 influence a change in this 180v to send the varying voltages to the deflection plates? Commented Feb 13, 2020 at 0:06
• Yes, the current in Q18,Q19 affect the voltage drop in R56, R57 and so the voltage across the deflection plates. Commented Feb 13, 2020 at 0:34
• "th eCRZT" => "the CRT"
– Mast
Commented Feb 13, 2020 at 7:52

You have a differential amplifier that adds two inputs:

• one input is the horizontal position
• the other input is the saw-tooth sweep

Gain is perhaps near 50. Q18 has about 5mA flowing from +180, through R57, through Q18, through R54 to -10V DC. Q19 is similarly biased. VR7 sets horizontal gain.

simulate this circuit – Schematic created using CircuitLab

• Wow Glen, thanks a million. This makes it so much easier for me to understand. These scope schematics are brutal. Can you tell me why it's utilizing two transistors in a row? And how we get from 180v to 105? Commented Feb 13, 2020 at 2:34
• I guess the compound transistors (Q16+Q18 and Q17+Q19) are used because the driving signals are somewhat high impedance. The input impedance of darlington pair is nice and high. With 5mA flowing through Q18 & Q19, you get a 75V drop across R57, and a 75V drop across R56. Those two are big power resistors. And Q18, Q19 are likely big as well, to dissipate heat (somewhat less than a watt) Commented Feb 13, 2020 at 2:50

What you have shown us is a classic oscilloscope schematic. I designed 'scopes in the 1960 and 70s. Judging by the load resistors (15k) of Q18 & 19 this must be a very low bandwidth, low performance scope with low sweep speeds. These transistors form the output stage of the X amplifier which drive the plates of the CRT. CRTs require around 5 -10 volts of signal per division. So to deflect the spot across the screen (assuming it has 10cm wide screen) requires somewhere between 50 and 100 volts pk-pk. This signal appears at the collectors of 'push-pull' stage Q18 & 19, so between 25 and 50 volts per side in antiphase. The gain of the output stage is roughly (15k +15k)/500 ohms or 60 X when the gain potentiometer is mid-position.

Q16 & 17 are emitter followers. This is necessary to drive the reflected capacitance from the output stage. Q18 & 19 will have a collector to base capacitance (Cob) of somewhere between 5 & 10 pF. This is multiplied by the stage gain. This is called the Miller capacitance. Hence Q16 & 17 have to drive around 300 to 600 pF

You will see the Y amplifier uses a different technique. It uses a cascode stage consisting of Q11 & 13 and Q12 & 14. Again it is a push-pull output stage but because the output transistors are used in a grounded base circuit there is no Miller capacitance, hence this kind of technique lends itself to higher bandwidths.

Any more questions, just ask!

• Wow, so much helpful information. Thanks so much Peter! I can't believe you designed these things. These old CRT scopes fascinate me, as well as the guys, like you, who were behind the golden age of electronics. Is there any way I could get ahold of you personally? I'm in desperate need to learn about these, as there isn't much in-depth information online about them. Also I'd love to ask you about your career if you had the time. Commented Feb 14, 2020 at 12:34

It's not really a Darlington pair as the two transistors are common emitter and the output is inverted. It is a simple amplifier section which provides gain and probably uses a higher voltage transistor for the deflection voltage. If you have 20 volts everywhere I'd check vr6 r54 r55.

I think that if one of the transistor i.e. either Q18 or Q19 is short you will get around 20-30 volts on all the three terminals of Q18 & Q19. Just check the transistors.