Level shift a 0-20 V pulse to +,-10 V without deteriorating the rise time

Question

I need a fast +,-10 V pulse. I thought I could trigger the base of a switching transistor with my fast rise time digital I/O and get a 0-20 V at the collector and then shift the DC to 0 volt and have a symmetrical +,-10 V pulse. The first part of creating the 20 V pulse is easy. But is there an easy way to shift its DC to 0 volt?

I can't use op-amps, because they would deteriorate the rise time of the pulse. I want the rise time to be kept below 10 ns. Comparators with such fast rise time and high supplies are I guess not available as an IC. So probably I need a discrete solution.

I want to use the generated pulse for measuring slew rate of an op-amp. Based on a suggestion in a comment, I used a gate driver:

Answer 1

Seems a MOSFET gate-driver will meet your requirements. There are many examples of complete one-chip solutions that accept a logic-level input, and convert this to a level-shifted version with a slight time delay (<100ns) , with very fast rise/fall times (less than 20ns in many cases) and capable of delivering several amps of current into a capacitive load.

Many suitable examples will be found when you search the various electronics suppliers using search phrases such as "MOSFET Gate driver", & "logic clock driver".

However, many of the more commonly available options have a maximum output voltage limit of +15V; to deliver +/-10V (20V) reliably with some margin I would choose a unit with max Vcc > 23V; the available options meeting this rating are less common. The newer chips designed to drive SiC MOSFETs tend to have output drive capability up to +30V, however, these are more expensive than the older, lower voltage devices.

Here are some examples to consider, sorted by max output V (ascending), note that "C-load" is the load capacitance connected to test the rise/fall times (as mentioned in the datasheet); I suspect the rise/fall times you will see will be shorter (faster) since your application does not have a large capacitive load.

#	OEM	Part no	max output V	delay time	rise/fall time	C-load
1	TI	UCC2732x	16V	25ns	20ns	1.8nF
2	ON-semi	NCP51530	20V	60ns	8ns	1nF
3	ON-semi	MC34151	20V	35ns	15ns	1nF
4	Diodes Inc	DGD0215/0216	22V	35ns	15ns	1nF
5	ON-semi	NCP51705	28V	25ns	8ns	1nF

Above: Table 1

The previous table shows non-isolated chips where the input and output share a common 0V connection; to get +/-10V output using these chips requires AC coupling the output (or input) and perhaps some DC restoration. Alternatively, there are gate driver chips that include an isolation barrier between the input and output; this will allow the input pulse to be ground referenced, while the output is referenced to a negative supply (ie: the output switches between -10V & +10V). Here are some examples to consider:

#	OEM	Part no	max output V	delay time	rise/fall time	C-load
1	Skyworks	Si8285	30V	40ns	5.5ns	200pF
2	Analog Devices	ADuM4146	30V	75ns	15ns	2000pF
3	Infineon	1EDI20N12AF	40V	120ns	10ns	1000pF

Above: Table 2

The figure below is from the datasheet for Infineon 1EDI20N12AF, showing how to connect the positive & negative supplies to the output section via pins VCC2 & GND2. The input section is fed from a separate logic-level supply via pins VCC1 & GND1.

If you select the Infineon device, to adapt it for your application I suggest doing the following:

1. The diagram shows an IGBT connected to the output; remove this and replace it with your circuit under test.
2. Connect the node marked "0V" to node "SGND".
3. Connect VCC2 to +10V, and GND2 to -10V.
4. Use good capacitors to filter the power supply, and locate these very close to the chip; minimise the current path loop area formed by the bypass capacitors and the chip.

Answer 2

So the rise time is your only concern and neither the delay nor the pulse stretching is your problem.

To give a rough idea, I'm leaving a very simple circuit below consisting of MOSFETs with very low C_iss or Q_G such as 2N7002 and BSS84:

As you can see, the rise time is almost unchanged. 2N7002 can withstand continuous gate-source voltage of 20V and pulsed gate-source voltage of 40V, so it's safe to drive with 20V directly. The problem with this circuit is, obviously, that it's lossy:

• Because R1 interacts with M1, it has to be kept low (Increasing it changes the output rise time). Average dissipation, according to LTspice, goes slightly above 1 W.
• The peak gate currents for 7002 can reach 1 Amps (I put 10 Ohms of gate resistance randomly but you can increase until the rise time goes beyond your requirements). Considering the 120-Ohm internal gate resistance of 7002, the peak losses will be high. LTspice gave a total avg loss of less than 50 mW for that part, though.

Answer 3

It looks like job for "gate driver". What about to take some 20V+ gate driver like Si8261 (or many others) ? Its rise/fall time is under 10ns, amplitude can be set by supply voltage. Thanks to galvanicaly isolated input you can tie its VDD to +10V and GND to -10V and your input signal still can be refered to your GND (0V).

Answer 4

We had a project where we needed to replace a rather power hungry dedicated levelshifter with an alternative. In the end we came up with the following idea:

^{simulate this circuit – Schematic created using CircuitLab}

It is an extention to the common zener diode levelshifter approach. The additional cap ensures that the AC part of the signal is transferred fast enough and not low-pass filtered by the parasitic RC-filter. It can be seen as an AC voltage dividers formed by the caps and a DC voltage divider formed by the resistors. The exact values are strongly dependent on the parasitics and input impedance and power budget. This circuit "abuses" the fact that the inputs have clamping/ESD diodes in place. R2 effectively limits the current during the "high" state but at the expense of a reduced signal amplitude because of the voltage divider. We use a SN74 based buffer to sanatize the signals afterwards and keep the trace parasitics in check, but this would be too slow in your case.

Answer 5

You could use a capacitor with clamping diodes at the output side to maintain the correct DC levels, but that only works if you have repeating pulses with sufficiently high frequency, not of there are long periods without activity. So it depends on your pulsed signal whether this is possible, but it's the simplest solution.

Or you could just use an opamp with +10V and -10V supply rails. That would work for any sequence of pulses.

As an alternative you can build your own opamp, with probably at least 3 or 4 transistors, but that's already a big project where many things can go wrong.