# Probe circuit for measuring higher voltage on oscilloscope?

The only oscilloscope I have is a tiny DSO138 (input section is on the top left of page 4 in the linked PDF). The scope has a limit of 100Vpp (actually displays only 80Vpp). At the highest attenuation (100x), there is 1V input to the first op-amp.

I want to see the output of a step down transformer. The transformer secondary shows 40V AC on my multimeter. Assuming this to be close to the RMS value, peak would be 56.5V and peak to peak at 113V. This is slightly higher than what I should put into the scope.

How can I go about attenuating the signal? Will a pair of 100K resistors as voltage divider work (with input set to 100x on the scope)? Will I need to attach any capacitors? I will probably never use it for more than 200Vpp ever, but there is an audio amplifier project coming up. If I can make a decent probe to view 100-200Vpp, it will be helpful.

All advice will be helpful. My electronics knowledge is rather limited (but I cannot go out and buy a proper probe or scope just yet). Also, I've never used a real scope.

Here is the input section of the scope:

edit: As far as I understood, standard scopes come with a 50Ohm and a 1M resistor inside them, and probes are 1x/10x/100x/1000x with reference to that, with 10x being most common. Is this XXXx multiplier with reference to the first op amp input voltage (somewhere around 0-1V)? That would mean my scope already has the 1x, 10x, and 100x probe circuits built in, right? Would that mean using an additional 10x passive divider give me grossly innacurate results? Are scopes mostly used for seeing waveforms in the 0-10V range?

Another question: The input opamp TL084 has a bias current of 30pA (max 200pA). With the input 100K on the lowest range, that would mean my lowest ranges are not protected upto 100Vpp. Only protected upto +=5V and +-50V, even though there is a 100K resistor on the input of the lowest range?

• You do not give us much to work with. Yes, oscilloscopes can be very expensive, and have different input loads. Most offer a 1 M ohm and a 50 ohm load option, so you chose a resistor of 100 or 1,000 times those values.
– user105652
Mar 31, 2018 at 0:39

OK, so to answer my own question.... Despite whatever is shown in the schematics, despite R4 actually being a 2M resistor that's soldered on the board, the input impedance of the DSO138 scope is 1 MegOhms (at all ranges, 1V, 0.1V and 10mV). Weird? Yes, and I don't have an answer to that.

I started out by getting ~18 Megohms worth of resistors and four 47p capacitors in series and made a probe (complete with pen holder and all) and realized it had become a close to 20x probe (compensated with a 150p capacitor at the BNC connector to give a square output with the 1KHz test freq). Attenuation was greater at DC (about 20x) than at 50Hz (about 18x) or 1KHz (about 16x).

Next, I got 9 Meg worth of resistors, accurately measured with the multimeter. Connected seven 100p caps in series, and connected those parallel to the resistors. The reading is almost exactly 10x at DC and the 1KHz test pulse. Hardly any compensation required.

The caps are required at anything other than DC. Even at 50Hz, output became distorted without caps, almost a triangle wave. So, not possible to measure voltage or do anything useful with only a resistor divider, even at 50Hz.

Tips: Ceramic SMD capacitors crack easily. Do not overheat or apply pressure due to board flex (I used a thin 5mmx40mm PCB). Better to connect the string of caps with flexible wires on both ends. A bunch of caps and a few series resistors are required to increase the voltage rating. 1/4 W through hole resistors are rated at 250V, caps are rated at 50V. I used 4 resistors and 4 caps for the 20x, so max 200Vpp. 4 resistors and 7 caps in the 10x, gives me max 350Vpp. Better to use a safety margin and not go to the limit, especially at lower frequencies.

Spice simulation does not match actual results exactly due to capacitance changes in the capacitors as well as due to construction (and the coaxial cable). I estimate DSO138 impedance to be 1Meg and input capacitance <100pF.

You should go and buy a 10:1 scope probe, preferably for the scope you have. As you can see, the input impedance of your scope is rather complex, so to have a matching voltage divider might be a challenge. For some ideas, look as this answer

But if your source has rather small impedance, and the expected frequency range is below few kHz, a simple 90 kOhm to 10 kOhm voltage divider will do the job.

You only need to make sure that the secondary coil of your transformer is not connected to any AC, and is floating.

• Thanks for the reply and links! By adding this 10x to the existing 100x, I am making it 1000x, right? Will this additional 90K/10K divider also work at audio frequencies? (They don't sell probes for this one, so I have to make.) Mar 31, 2018 at 1:35
• @Indraneel, you are asking too many questions. Just do a divider as shown above, and set the first "sensitivity selector" to "1V". And don't change it. Then mentally correct the scope scale by 10: instead of reading 5V/div, you read it as 50V/div. (or 20V/div, or 10V/div, depending on second "sensitivity selector"). Mar 31, 2018 at 2:03
• Ok, will do! (and not ask too many questions :( ) Mar 31, 2018 at 2:25

This is a common 10x probe (https://www.allaboutcircuits.com/uploads/articles/Typical_schematic_for_X10_Passive_Probe.jpg). It's an RC voltage divider using internal oscope resistor/capacitor as part of the circuit.

Here's some theory behind probe design for handling ranges of frequencies:

From the hvprobe.htm site:

A significant part of the design effort (and cost) of high voltage probes deals with the problem of how to go smoothly from a resistive divider at low frequencies, to a capacitive divider at high frequencies, while keeping a constant attenuation value at mid-frequencies. This isn't easy. Consider for example, that an overall shield is clearly needed and must properly prevent the high-Z end of the probe from simply acting as an antenna (as some HV probes do! i.e. ground the tip of the probe and still see large signals at the output). This shield acts as a capacitance to ground for the HV resistor, routing some of the high-frequency current which is supposed to go to the output, to ground. Hence at some middle frequency there's a dip! This is solved in various ways - with shields connected to the probe tip (but inside the ground), capacitors bypassing the resistor, special resistor construction, etc. Most solutions can just as easily cause a region with a response hump, as well as a dip, or even both. BTW, these problems are much harder if one seeks to make a probe with very low capacitive loading and high frequency response.